Hc Verma II for Class 12 Science Physics Chapter 46 - The Nucleus

Answer:

Nuclear forces are short range strong attractive forces that act between two proton-proton, neutron-proton and neutron-neutron pairs. Two protons have strong nuclear forces between them and also exert electrostatic repulsion on each other. However, electrostatic forces are long ranged and have very less effect as compared to the strong nuclear forces.
So, in a nucleus (that is very small in dimension), there's no such significance of repulsive force as compared to the strong attractive nuclear force. On the other hand, an atom contains electrons revolving around its nucleus. These electrons are kept in their orbit by the strong electrostatic force that is exerted on them by the protons present inside the nucleus. Hence, a nucleus contains protons as well as neutrons.

Answer:

Neutrons are chargeless particles and they exert only short range nuclear forces on each other. If we have two pairs of neutrons and the separation between them is same in both the pairs. The force between the neutrons will be of same magnitude for the two pairs until there is some other influence on any of them.

Answer:

Inside the nucleus, two protons exert nuclear force on each other. These forces are short-ranged (a few fm), strong and attractive forces. They also exert electrostatic repulsive force (long-ranged). While discussing the behaviour of a hydrogen molecule, the nuclear force between the two protons is always neglected. This is because the separation between the two protons in the molecule is $~$70 pm which is much greater than the range of the nuclear force.

Answer:

Binding energy per nucleon of a nucleus is defined as the energy required to break-off a nucleon from it.
(a) As the binding energy per nucleon of iron is more than that of carbon, it is easier to take out a nucleon from carbon than iron.
(b) As the binding energy per nucleon of iron is more than that of lead. Therefore, it is easier to take out a nucleon from lead as compared to iron.

Answer:

If we assemble 6 protons and 6 neutrons to form ¹²C nucleus, 92.15 MeV (product of mass number and binding energy per nucleon of carbon-12) of energy is released. Therefore, the energy released in the formation of two carbon nuclei is 184.3 MeV. On the other hand, when 12 protons and 12 neutrons are combined to form a ²⁴Mg atom, 198.25 MeV of energy (binding energy) is released. Hence, in case of ​²⁴Mg nucleus, more energy is liberated.

Answer:

Cathode rays consist of electrons that are accelerated using electrodes. They do not carry high energy and do not harm human body. On the other hand, beta rays consist of highly energetic electrons that can even penetrate and damage human cells. Beta rays are produced by the decay of radioactive nuclei. If the two are travelling in space, they can be distinguished by the phenomenon named production of Bremsstrahlung radiation, which is produced by the deceleration of a high energy particle when deflected by another charged particle, leading to the emission of blue light. Only beta rays are capable of producing it. 

Answer:

When the nucleons of a nucleus are separated, a certain amount of energy is to be given to the nucleus, which is known as the binding energy.
Binding energy = [(Number of nucleons) $\times$ (Mass of a nucleon) - (Mass of the nucleus)]

 When the nucleons of a nucleus are separated, the increase in the total mass comes from the binding energy, which is given to the nucleus to break-off its constituent nucleons as energy is related to mass by the relation given below.
E = Δmc²

Answer:

In beta decay, a neutron from the nucleus is converted to a proton releasing an electron and an antineutrino or a proton is converted to a neutron releasing a positron and a neutrino.

$$\begin{gathered} \mathrm{i.e.} {\beta }^{-} \mathrm{decay}: n\to p+e+\overline{\nu } \\ {\beta }^{+} \mathrm{decay}: p\to n+e^{+}+\nu \end{gathered}$$

Since the number of valence electrons present in the parent atom do not change, the remaining atom does not get oppositely charged. Instead, due to a change in the atomic number, there's a formation of a new element.

Answer:

It is given that when a boron nucleus $\left({}_5{}^{10}\mathrm{B}\right)$ is bombarded by a neutron, an α-particle is emitted.

Let X nucleus be formed as a result of the bombardment.
According to the charge and mass conservation,

$$\begin{gathered} {}_5{}^{10}B+{}_0{}^{1}n\to X+{}_2{}^{4}He \\ \mathrm{Charge}: 5 0 \underline{3} 2 \\ \mathrm{Mass}: 10 1 \underline{7} 4 \end{gathered}$$

The mass number of X should be 7 and its atomic number should be 3.

$\therefore X={}_3{}^{7}Li$

Answer:

Gamma rays consist of photons that are produced when a nucleus from its excited state comes to its ground state releasing energy. Since gamma rays are chargeless and massless particles, the nucleus does not suffer any loss in mass during the gamma decay.

Answer:

Two photons having equal liner momentum have equal wavelengths as here for both the photons the direction and magnitude of linear momentum will be same. For the rest of the options, magnitude will be same but nothing can be said about the direction of the photons.
Hence the correct option is D.

Answer:

When three helium nuclei combine to form a carbon nucleus, energy is liberated. This energy is greater than the that liberated when these nuclei combine on their own. Hence, formation of carbon nucleus leads to much more stability as compared to the combination of three helium nuclei.

Answer:

(a) exact 12 u

In nuclear physics, a unit used for measurement of mass is unified atomic mass unit, which is denoted by u.

It is defined such that

1 u = $\frac{1}{12}\times$ (Mass of neutral carbon atom in its ground state)

Mass of neutral carbon atom in its ground state = 12 × 1 u = 12 u

Thus, the mass of neutral carbon atom in its ground state is exactly 12 u.

Answer:

(c) the number of nucleons in the nucleus

Mass number of a nucleus is defined as the sum of the number of neutron and protons present in the nucleus, i.e. the number of nucleons in the nucleus.

Answer:

(c) two extra neutrons and no extra electron

¹²C and ¹⁴C are the two isotopes of carbon atom that have same atomic number, but different mass numbers. This means that they have same number of protons and electrons, but different number of neutrons. Therefore, ​^12​C has 6 protons, 6 electrons and 6 neutrons, whereas ​¹⁴C has 6 electrons, 6 protons and 8 neutrons. 

Answer:

(d) sometimes more than and sometimes equal to its atomic number

Mass number of a nucleus is defined as the sum of the number of neutron and protons present in the nucleus, i.e. the number of nucleons in the nucleus, whereas atomic number is equal to the number of protons present. Therefore, the atomic number is smaller than the mass number. But in the nucleus (like that of hydrogen ¹H_​1), only protons are present. Due to this, the mass number is equal to the atomic number.

Answer:

(a) a straight line
The average nuclear radius (R) and the mass number of the element (A) has the following relation:
$$\begin{gathered} R=R_o{A}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ \raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$R_o$}\right.=A^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ \ln \left(\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$R_o$}\right.\right)=\frac{1}{3}\ln A \end{gathered}$$
Therefore, the graph of ln(R/R0) versus ln A is a straight line passing through the origin with slope 1/3.

Answer:

(d) Fpp < Fpn = Fnn

Protons and neutrons are present inside the nucleus and they exert strong attractive nuclear forces on each other, which are equal in magnitude. Due to their positive charge, protons repel each other. Hence the net attractive force between two protons gets reduced, but the nuclear force is stronger than the electrostatic force at a separation of 1 fm.

∴ Fpp < Fpn = Fnn

Here, Fpp, Fpn and Fnn denote the magnitudes of the net force by a proton on a proton, by a proton on a neutron and by a neutron on a neutron, respectively.

Answer:

(b) F_pp = F_pn = F_nn

Protons and neutrons are present inside the nucleus and they exert strong attractive nuclear force on each other. These forces are equal in magnitude, irrespective of the charge present on the nucleons.

∴ F_pp = F_pn = F_nn
Here, F_pp, F_pn and F_nn denote the magnitudes of the nuclear force by a proton on a proton, by a proton on a neutron and by a neutron on a neutron, respectively.

Answer:

(b) F_e >> F_n

Two protons exert strong attractive nuclear force and repulsive electrostatic force on each other. Nuclear forces are short range forces existing in the range of a few fms. Therefore, at a separation of 10 nm, the electromagnetic force is greater than the nuclear force, i.e. F_e >> F_n. 

Answer:

(d) varies in a way that depends on the actual value of A

Binding energy per nucleon in a nucleus first increases with increasing mass number (A) and reaches a maximum of 8.7 MeV for A (50−80). Then, again it slowly starts decreasing with the increase in A and drops to the value of 7.5 MeV.

Answer:

(d) It is the sum of the kinetic energies of all the nucleons present in the nucleus. 

Binding energy of a nucleus is defined as the energy required to break the nucleus into its constituents. It is also measured as the Q-value of the breaking of nucleus, i.e. the difference between the rest energies of reactants (nucleus) and the products (nucleons) or the difference between the kinetic energies of the products and the reactants. 

Answer:

(c) more than half the active nuclei decay

The average life is the mean life time for a nuclei to decay.
It is given as 
$\tau =\frac{1}{\lambda }=\frac{{{\rm T}}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{0.693}$
Here, $\tau , \lambda \mathrm{and} {{\rm T}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ are the average life, decay constant and half life-time of the active nuclei, respectively. The value of the average lifetime comes to be more than the average lifetime. Therefore, in one average life, more than half the active nuclei decay.

Answer:

(d) Photon 

The atomic number and mass number of a nucleus is defined as the number of protons and the sum of the number of protons and neutrons present in the nucleus, respectively. Since in the decay, neither the atomic number nor the mass number change, it cannot be a beta-decay (release of electron, proton or neutron). Hence, the particle emitted can only be a photon.

Answer:

(c) a neutron in the nucleus decays emitting an electron

Negative beta decay is given as
$n\to p+e^{-}+\overline{\nu }$  
Neutron decays to produce proton, electron and anti-neutrino.

Answer:

(b) 12 h

A freshly prepared radioactive source emits radiation of intensity that is 64 times the permissible level. This means that it is possible to work safely till 6 half-lives (as 2⁶ = 64) of the radioactive source. Since the half-life of the source is 2h, the minimum time after which it would be possible to work safely with this source is 12 h.

Answer:

(b) (ln 2/λ) and 1/λ

The half-life of a radioactive sample ($t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$) is defined as the time elapsed before half the active nuclei decays.

Let the initial number of the active nuclei present in the sample be N₀.
$$\begin{gathered} \frac{N_o}{2}=N_o{e}^{-\lambda t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}} \\ \Rightarrow t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{\ln 2}{\lambda } \end{gathered}$$

Average life of the nuclei, $t_{av}=\frac{S}{N_o}=\frac{1}{\lambda }$
Here, S is the sum of all the lives of all the N nuclei that were active at t = 0 and $\lambda$ is the decay constant of the sample.

Answer:

(b) proton

If an alpha particle is bombarded on a nitrogen (N-14) nucleus, an oxygen (O-17) nucleus and a proton are released.
According to the conservation of mass and charge,
${}_2{}^{4}He+{}_7{}^{14}N\to {}_6{}^{17}O+{}_1{}^{1}p$
So, the emitted particle is a proton.

Answer:

(a) 10 g

⁵⁷Co is undergoing beta decay, i.e. electron is being produced. But an electron has very less mass (9.11$\times$10^-31 kg) as compared to the Co atom. Therefore, after 570 days, even though the atoms undergo large beta decay, the weight of the material in the container will be nearly 10 g.

Answer:

(c) x

When the train is stationary, the separation between the alpha particle and recoiling uranium nucleus is x in time t after the decay. Even if the decay is taking place in a moving train and the separation is measured by the passenger sitting in it, the separation between the alpha particle and nucleus will be x. This is because the observer is also moving with the same speed with which the alpha particle and recoiling nucleus are moving, i.e. they all are in the same frame that is moving at a uniform speed.

Answer:

(c) a heavy nucleus bombarded by thermal neutrons breaks up

In a nuclear reactor, a large fissile atomic nucleus like uranium-235 absorbs a thermal neutron and undergoes a nuclear fission reaction. The heavy nucleus splits into two or more lighter nuclei releasing gamma radiation and free neutrons.

Answer:

(c) Density

Radius of a nucleus with mass number A is given as
 $R=R_o{A}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$ 
Here, R_o = 1.2 fm
∴ Volume of the nucleus = $\frac{4\pi R^3}{3}=\frac{4\pi {R_o}^{3}A}{3}$ 

This depends on A. With an increase in A, V increases proportionally.
Mass of the nucleus $\simeq$ Am_N
Here, m_N is the mass of a nucleon.
Therefore, mass of the nucleus also increases with the increasing mass number.
Binding energy also depends on mass number (number of nucleons) as it is the difference between the total mass of the constituent nucleons and the nucleus. Therefore, it also varies with the changing mass number.

On the other hand,
Density = $\frac{\mathrm{Mass}}{\mathrm{Volume}}=\frac{A{m}_N}{{\displaystyle \raisebox{1ex}{$4\pi R^3$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=\frac{A{m}_N}{{\displaystyle \raisebox{1ex}{$4\pi {R_o}^{3}A$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=\frac{m_N}{{\displaystyle \raisebox{1ex}{$4\pi {R_o}^3$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=\frac{3m_N}{{\displaystyle 4\pi {R_o}^3}}$
This is independent of A and hence does not change as mass number increases.

Answer:

(c) a neutron does not exert electric repulsion
(d) Coulomb forces have longer range compared to the nuclear forces

This is because in heavy nuclei, the N/Z ratio becomes larger in order to maintain their stability and reduce instability caused due to the repulsion among the protons.The neutrons exert only attractive short-range nuclear forces on each other as well as on the neighbouring protons, whereas the protons exert attractive short-range nuclear forces on each other as well as the electrostatic repulsive force. Thus, the nuclei with high mass number, in order to be stable, have large neutron to proton ratio (N/Z).

Answer:

(c) neutron has large rest mass than the proton.

A nucleus is made up of two fundamental particles- neutrons and protons. If a nucleus has more number of neutrons than what is needed to have stability, then neutrons decay into protons and if there's an excess of protons, then they decay to form neutrons. Since a neutron has larger rest mass than a proton, the Q-value of its decay reaction is positive and a free neutron decays to a proton, while an isolated proton cannot decay to a neutron as the Q-value of its decay reaction is negative. Hence, it is physically not possible.

Answer:

(d) The active nucleus changes to one of its isobars after the beta decay.

In a beta decay, either a neutron is converted to a proton or a proton is converted to a neutron such that the mass number does not change. Also,the number of the nucleons present in the nucleus remains the same. Thus, the active nucleus gets converted to one of its isobars after beta decay.

Answer:

(d) γ-decay

​In alpha particle decay, the unstable nucleus emits an alpha particle reducing its proton number Z by 4 and neutron number N by 2 such that the element gets changed.
 ${}_Z{}^{A}X\to {}_{Z-2}{}^{A-4}Y+{}_2{}^{4}He$
During  β⁻-decay​,  a neutron is converted to a proton​, an electron and an antineutrino, i.e. an active nucleus gets converted to one of its isobars and hence the element gets changed. 
${}_Z{}^{A}X\to {}_{Z+1}{}^{A}Y+e+\overline{\nu }$
 During  β⁺-decay​,  a proton in the nucleus is converted to a neutron​, a positron and a neutrino in order to maintain the stability of the nucleus, i.e. an active nucleus gets converted to one of its isobars and hence the element gets changed.
${}_Z{}^{A}X\to {}_{Z-1}{}^{A}Y+{\beta }^{+}+\nu$
When a nucleus is in higher excited state or has excess of energy, it comes to the ground state in order to become stable and release energy in the form of electromagnetic radiation called gamma ray. Hence, the element in gamma decay doesn't change.

Answer:

(a) α-decay
(b) β⁺-decay

​In alpha particle decay, the unstable nucleus emits an alpha particle reducing its proton number (atomic number) Z as well as neutron number N by 2.
 ${}_Z{}^{A}X\to {}_{Z-2}{}^{A-4}Y+{}_2{}^{4}He$
 During  β⁻-decay​,  a neutron is converted to a proton​, an electron and an antineutrino. Thus, there is an increase in the  atomic number.
${}_Z{}^{A}X\to {}_{Z+1}{}^{A}Y+e^{-}+\overline{\nu }$
During  β⁺-decay​,  a proton in the nucleus is converted to a neutron​, a positron and a neutrino in order to maintain the stability of the nucleus. Thus, there is a decrease in the atomic number. ​
${}_Z{}^{A}X\to {}_{Z-1}{}^{A}Y+{\beta }^{+}+\nu$
When a nucleus is in higher excited state or has excess of energy, it comes to the lower state in order to become stable and release energy in the form of electromagnetic radiation called gamma ray. The element in the gamma decay doesn't change.
Therefore, alpha and beta plus decay suffer decrease in atomic number.

Answer:

(d) gamma rays

Magnetic force acts on a charged particle, due to which it deflects from its path. The magnitude of this force is measured as $\left|\overrightarrow{F}\right|=\left|q(\overrightarrow{v}\times \overrightarrow{B})\right|$.
Here, q is the charge on the particle that is moving with speed v in a uniform magnetic field B.
Since alpha, beta-plus and beta-minus are charged particles, they suffer deflection due to the field applied. On the other hand, gamma rays are photons and due to zero charge, they do not suffer any deflection.

Answer:

(d) Gamma rays

Alpha rays, beta-plus and beta-minus rays carry charged particles that show particle behaviour. On the other hand, gamma rays carry photons that show particle as well as wave behaviour. Hence, only gamma rays are electromagnetic waves.

Answer:

(d) Coulomb repulsion does not allow the nuclei to come very close.

Lithium atom contains 3 protons and 3 neutrons in the nucleus and 3 valence electrons. When two lithium nuclei are brought together, they repel each other. The attractive nuclear forces being short-range are insignificant as compared to the electrostatic repulsion. Thus, the nuclei do not combine to form carbon atom because of coulomb repulsion.

Answer:

(b) the binding energy per nucleon decreases on an average as A increases
(c) if the nucleus breaks into two roughly equal parts, energy is released

Binding energy per nucleon varies in a way that it depends on the actual value of mass number (A). As the mass number (A) increases, the binding energy also increases and reaches its maximum value of 8.7 MeV for A (50−80) and for A > 100. The binding energy per nucleon decreases as A increases and the nucleus breaks into two or more atoms of roughly equal parts so as to attain stability and binding energy of mass number between 50−80.

Answer:

Given:
Mass of the nucleus, M = Am_p
Volume of the nucleus, V = $\frac{4}{3}\pi {R_0}^{3}A$

Density of the matter, $d=\frac{M}{V}=\frac{A{m}_p}{{\displaystyle \frac{4}{3}}\pi {R_0}^{3}A}$
                                               $$\begin{gathered} =\frac{3m_p}{4\times \pi {R_0}^3} \\ =\frac{3\times 1.007276}{4\times 3.14(1.1{)}^3} \\ = 3\times {10}^{17} \mathrm{kg}/{\mathrm{m}}^3 \end{gathered}$$

Specific gravity of the nuclear matter = $\frac{\mathrm{Density} \mathrm{of} \mathrm{matter} }{\mathrm{Density} \mathrm{of} \mathrm{water}}$
 $\therefore$ Specific gravity = $\frac{3\times {10}^{17}}{{10}^3}$ = 3 $\times$ 10¹⁴ kg/m³

Answer:

Given:
Mass of the neutron star, M = 4.0 × 10³⁰ kg
Density of nucleus, d = 2.4$\times$10¹⁷${}^{}$

Density of nucleus, $d=\frac{M}{V}$

Here, V is the volume of the nucleus.

$$\begin{gathered} \therefore V=\frac{M}{d}=\frac{4\times {10}^{30}}{2.4\times {10}^{17}} \\ =\frac{1}{0.6}\times {10}^{13} \\ =\frac{1}{6}\times {10}^{14} \end{gathered}$$

If R is the radius, then the volume of the neutron star is given by
$$\begin{gathered} V=\frac{4}{3}\pi R^3 \\ \therefore \frac{1}{6}\times {10}^{14}=\frac{4}{3}\times \mathrm{\pi }\times R^3 \\ \Rightarrow R^3=\frac{1}{6}\times \frac{3}{4}\times \frac{1}{\mathrm{\pi }}\times {10}^{14} \\ \Rightarrow R^3=\frac{1}{8}\times \frac{100}{\mathrm{\pi }}\times {10}^{12} \\ \therefore R=\frac{1}{2}\times {10}^4\times 3.17 \\ =1.585\times {10}^4=15 \mathrm{km} \end{gathered}$$

Answer:

Given:
Binding energy of α particle = 28.2 MeV
Let x be the mass of α particle.
We know an α particle consists of 2 protons and 2 neutrons.

Binding energy, $B=\left(Z{m}_p+N{m}_n-M\right)c^2$

Here, m_p = Mass of proton
m_n = Mass of neutron
Z= Number of protons
N = Number of neutrons
c = Speed of light

On substituting the respective values, we have
$$\begin{gathered} 28.2 =(2\times 1.007276+2\times 1.008665 -x)c^2 \\ \Rightarrow x=4.0016 \mathrm{u} \end{gathered}$$

Answer:

Given:
Mass of ⁷Li = 7.0160 u
Mass of ⁴He = 4.0026 u.
Reaction:
$\mathrm{L}{i}^7+p\mathit{\to }\alpha \mathit{+}\alpha \mathit{+}E\mathit{,}$
Energy release $\left(E\right)$ is given by
$$\begin{gathered} E=\left[m\left({}^7\mathrm{Li}\right)+\left(m_p\right)-2\times m\left({}^4\mathrm{He}\right)\right]c^2 \\ =\left[\left(7.0160 \mathrm{u}+1.007276 \mathrm{u}\right)-2\left(4.0026 \mathrm{u}\right)\right]c^2 \\ =(8.023273 \mathrm{u}-8.0052 \mathrm{u}) {\mathrm{c}}^2 \\ =0.018076 \times 931 \mathrm{MeV} \\ =16.83 \mathrm{MeV} \end{gathered}$$

Answer:

Given:
Atomic mass of Au, A = 196.96
Atomic number of Au, Z = 79
Number of neutrons, N = 118
Binding energy, $B=(\mathrm{Z}{m}_p+N{m}_n-M)c^2$                                                                
Here, m_p = Mass of proton
M = Mass of nucleus
m_n = Mass of neutron
c = Speed of light
On substituting the respective values, we get
$$\begin{gathered} B=\left[\right(79\times 1.007276+118\times 1.008665) \mathrm{u}-196.96 \mathrm{u}]c^2 \\ =\left(198.597274-196.96\right)\times 931 \mathrm{MeV} \\ =1524.302094 \mathrm{MeV} \end{gathered}$$
Binding energy per nucleon $=\frac{1524.3}{197}=7.737 \mathrm{MeV}$

Answer:

(a)
Given:
Atomic mass of ²³⁸U, m(²³⁸U) = 238.0508 u
Atomic mass of ²³⁴Th, m(²³⁴Th)  = 234.04363 u
Atomic mass of ⁴He, m(⁴He) = 4.00260 u
When ²³⁸U emits an α-particle, the reaction is given by
${\mathrm{U}}^{238}\to {\mathrm{Th}}^{234}+{\mathrm{He}}^4$
Mass defect, Δm = [m(²³⁸U)$-$(m(²³⁴Th)+m(⁴He))]
Δm = [238.0508 $-$ (234.04363 + 4.00260) = 0.00457 u
Energy released (E) when ²³⁸U emits an α-particle is given by
$$\begin{gathered} E=∆m c^2 \\ E = [0.00457 \mathrm{u}]\times 931.5 \mathrm{MeV} \\ \Rightarrow E = 4.25467 \mathrm{MeV} = 4.255 \mathrm{MeV} \end{gathered}$$

(b)

When two protons and two neutrons are emitted one by one, the reaction will be
${\mathrm{U}}^{233}\to {\mathrm{Th}}^{234}+2n+2p$

$$\begin{gathered} \mathrm{Mass} \mathrm{defect}, ∆m=m\left({\mathrm{U}}^{238}\right)-[m\left({\mathrm{Th}}^{234}\right)+2\left(m_{\mathrm{n}}\right)+2\left(m_p\right)] \\ ∆m=238.0508 \mathrm{u}-[234.04363 \mathrm{u}+2(1.008665) \mathrm{u}+2(1.007276) \mathrm{u}] \\ ∆m= 0.024712 \mathrm{u} \end{gathered}$$

Energy released (E) when ²³⁸U emits two protons and two neutrons is given by

$$\begin{gathered} E=∆m{c}^2 \\ E=0.024712 \times 931.5 \mathrm{MeV} \\ E=23.019=23.02 \mathrm{MeV} \end{gathered}$$

Answer:

Given:
Atomic mass of ²²³Ra, m(²²³Ra) = 223.018 u
Atomic mass of ²⁰⁹Pb, m(²⁰⁹Pb) = 208.981 u 
Atomic mass of ¹⁴C, m(¹⁴C) = 14.003 u

Reaction: 

${}^{223}\mathrm{Ra}{\to }^{209}\mathrm{Pb}+{}^{14}\mathrm{C}$

$$\begin{gathered} \mathrm{Energy}, E=\left[m\left({}^{223}Ra\right)-\left(m\left({}^{209}Pb\right)+m\left({}^{14}\mathrm{C}\right)\right)\right]c^2 \\ = \left[223.018 \mathrm{u}-\left(208.981+14.003\right) \mathrm{u}\right] c^2 \\ =0.034\times 931 \mathrm{MeV} \\ =31.65 \mathrm{MeV} \end{gathered}$$

Answer:

Given:
Mass of an atom with Z protons and N neutrons = M_Z,_N
Mass of hydrogen atom = M_H
As hydrogen contains only protons, the reaction will be given by
$$\begin{gathered} E_{Z\mathit{,}N}\to E_{\mathrm{Z}-1,N}+p_1 \\ \Rightarrow E_{Z\mathit{,}N}\to E_{z-1,N}{+}^1{\mathrm{H}}_1 \end{gathered}$$
∴ Minimum energy needed to separate a proton, $∆E=(M_{\mathrm{Z}-1,\mathrm{N}}+M_{\mathrm{H}}-M_{\mathrm{Z},\mathrm{N}})c^2$

Answer:

Before the separation of neutron, let the mass of the nucleus be M_Z,N.
Then, after the separation of neutron, the mass of the nucleus will be M_Z,_N-1.
The reaction is given by
$E_{Z\mathit{,}N}=E_{Z,N-1}+{}_0{}^1\mathrm{n}$
If $M_{\mathrm{N}}$ is the mass of the neutron, then the energy needed to separate the neutron b$\left(∆E\right)$ will be
$∆E$ = (Final mass of nucleus + Mass of neutron − Initial mass of the nucleus)c²
$∆E=(M_{Z,\mathrm{N}-1}+M_{\mathrm{N}}-M_{Z,\mathrm{N}})c^2$

Answer:

Given:
Atomic mass of ³²P, m(³²P) = 31.974 u
Atomic mass of ³²S, m(³²S) = 31.972 u
Reaction:
${\mathrm{P}}^{32}\to {\mathrm{S}}^{32}{+}_1{\mathrm{v}}^0{+}_{-1}{\mathrm{\beta }}^0$

Energy of antineutrino and β-particle, E = [m(³²P)$-$m(³²S)]c²
                                                         = (31.974 u− 31.972 u)c²
                                                         = 0.002 × 931 = 1.862 MeV

Answer:

Given:
Half-life period of free neutron beta-decays to a proton, $T_{1/2}$ = 14 minutes
 
Half-life period, T_1/2  = $\frac{0.6931}{\lambda }$

Here, $\lambda$
= Decay constant

$$\begin{gathered} \therefore \lambda =\frac{0.693}{14\times 60} \\ =8.25\times {10}^{-4} {\mathrm{s}}^{-1} \end{gathered}$$

If m_p is the mass of proton, let m_n and m_e be the mass of neutron and mass of electron, respectively.

 $$\begin{gathered} \therefore \; Energ\mathrm{y} \mathrm{liberated}, E=[m_n-\left(m_p+m_e\right)]c^2 \\ =[1.008665 \mathrm{u}-\left(1.007276+0.0005486\right) \mathrm{u}]c^2 \\ =0.0008404\times 931 \mathrm{MeV} \\ =782 \mathrm{keV} \end{gathered}$$

Answer:

$$\begin{gathered} \left(\mathrm{a}\right) {}_{88}{}^{226}\mathrm{Ra}\to {}_2{}^4\mathrm{\alpha }+{}_{86}{}^{222}\mathrm{Rn} \\ \left(\mathrm{b}\right) {}_8{}^{19}\mathrm{O}\to {}_9{}^{19}\mathrm{F}+\overline{e}+\overline{v} \\ \left(\mathrm{c}\right) {}_{13}{}^{25}\mathrm{Ar}\to {}_{12}{}^{25}\mathrm{Mg}+e^{+}+v \end{gathered}$$

Answer:

Given:
Maximum kinetic energy of the positron, K = 0.650 MeV

(a) Neutrino and positron are emitted simultaneously.
∴ Energy of neutrino = 0.650 − Kinetic energy of the given positron
                             = 0.650 − 0.150
                             = 0.5 MeV = 500 keV

(b) Momentum of the neutrino, $P=\frac{E}{c}$
 Here, E = Energy of neutrino
  c = Speed of light
 
  $$\begin{gathered} \Rightarrow P=\frac{500\times 1.6\times {10}^{-19}}{3\times {10}^8}\times {10}^3 \\ =2.67\times {10}^{-22} {\mathrm{Kgms}}^{-1} \end{gathered}$$

Answer:

(a) Decay of potassium-40 by β⁻emission is given by
     ${}_{19}\mathrm{K}^{40}{\to }_{20}{\mathrm{Ca}}^{40}+{\mathrm{\beta }}^{-}+\overline{\mathrm{v}}$
    Decay of potassium-40 by β⁺ emission is given by
    ${}_{19}\mathrm{K}^{40}{\to }_{18}{\mathrm{Ar}}^{40}+{\mathrm{\beta }}^{+}+\mathrm{v}$
    Decay of potassium-40 by electron capture is given by
    ${}_{19}\mathrm{K}^{40}+{\mathrm{e}}^{-}{\to }_{18}{\mathrm{Ar}}^{40}+\mathrm{v}$

(b) 
  Q_value in the β⁻ decay is given by
  Q_value = [m(₁₉K⁴⁰) − m(₂₀Ca⁴⁰)]c²
            = [39.9640 u − 39.9626 u]c²
           = 0.0014 $\times$ 931 MeV
           = 1.3034 MeV
Q_value in the β⁺ decay is given by
 Q_value = [m(₁₉K⁴⁰) − m(₂₀Ar⁴⁰) − 2m_e]c²
           = [39.9640 u − 39.9624 u −  0.0021944 u]c²
           = (39.9640 − 39.9624) 931 MeV − 1022 keV
           = 1489.96 keV − 1022 keV
           = 0.4679 MeV  
Q_value in the electron capture is given by
Q_value = [ m(₁₉K⁴⁰) − m(₂₀Ar⁴⁰)]c²
          = (39.9640 − 39.9624)uc²
          = 1.4890 = 1.49 MeV

Answer:

The nuclear process taking place is shown below.
$$\begin{gathered} {}_8{}^6\mathrm{Li}+n \to {}_3{}^7\mathrm{Li} \\ {}_3{}^7\mathrm{Li}+n\to {}_3{}^8\mathrm{Li}\to {}_4{}^8\mathrm{Be}+\overline{v}+e^{-} \\ {}_4{}^8\mathrm{Be}\to {}_2{}^4\mathrm{He}+{}_2{}^4\mathrm{He} \end{gathered}$$

Answer:

Given:
Mass of ¹¹C, m(¹¹C) = 11.0114 u
Mass of ¹¹B, m(¹¹B) = 11.0093 u
Energy liberated in the β⁺ decay (Q) is given by
$Q=\left[m\left({}^{11}C\right)-m\left({}^{11}B\right)-2m_e\right]c^2$
   = (11.0114 u − 11.0093 u $-$ 2$\times$0.0005486 u)c²
   = 0.0010028 $\times$ 931 MeV
   = 0.9336 MeV = 933.6 keV
For maximum KE of the positron, energy of neutrino can be taken as zero.
∴ Maximum KE of the positron = 933.6 keV

Answer:

Given:
Atomic mass of ²²⁸Th, m(²²⁸Th) = 228.028726 u
Atomic mass of ²²⁴Ra, m(²²⁴Ra) = 224.020196 u
Atomic mass of ${}_2{}^4\mathrm{H}$, m(${}_2{}^4\mathrm{H}$) = 4.00260 u
Mass of ²²⁴Ra = 224.020196 × 931 + 0.217 MeV = 208563.0195 MeV

Kinetic energy of alpha particle, K = $\left[m\left({}^{228}\mathrm{Th}\right)-\left[m\left({}^{224}\mathrm{Ra}\right)+m\left({}_2{}^4\mathrm{H}\right)\right]\right]$c²
                                                        = (228.028726 × 931) − [(208563.0195 + 4.00260 × 931]
                                                        = 5.30383 MeV = 5.304 MeV

Answer:

 Given:
 Atomic mass of ¹²N, m(¹²N) = 12.018613 u
 ¹²N → ¹²C* + e⁺ + v
 ¹²C* → ¹²C + γ (4.43 MeV)

 Net reaction is given by
 ¹²N → ¹²C + e⁺ + v + γ (4.43 MeV)
 
Q_value  of the ${\beta }^{+}$ decay will be
 Q_value= [ m(¹²N) $-$ (m(¹²C*) + 2m_e)]c²
           = [12.018613 $\times$931 MeV $-$ (12$\times$931 + 4.43) MeV $-$ (2$\times$511) keV]
           = [11189.3287 $-$ 11176.43 $-$ 1.022] MeV
           = 11.8767 MeV = 11.88 MeV

The maximum kinetic energy of beta particle will be 11.88 MeV, assuming that neutrinos have zero energy.

Answer:

Given:
Decay constant of ${}_{80}{}^{197}\mathrm{Hg}$, $\lambda$ = 1.8 × 10⁻⁴ s^$-1$

(a)
Half-life, $T_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0.693}{\lambda }$

 $$\begin{gathered} \Rightarrow T_{1/2}=\frac{0.693}{1.8\times {10}^{-4}} \\ = 3850 \mathrm{s}=64 \mathrm{minutes} \end{gathered}$$      

(b)     
$$\begin{gathered} \mathrm{Average} \mathrm{life}, T_{av}=\frac{T_{1/2}}{0.693} \\ =\frac{64}{0.693} \\ =92 \mathrm{minutes} \end{gathered}$$

(c)
Number of active nuclei of mercury at t = 0
                                                                     = N₀
                                                                    = 100
Active nuclei of mercury left after conversion of 25% isotope of mercury into gold = N =  75

Now, $\frac{N}{N_0}=e^{-\lambda t}$
Here, N = Number of inactive nuclei
            N₀ = Number of nuclei at t = 0
            $\lambda$ = Disintegration constant

On substituting the values, we get
$\frac{75}{100}=e^{-\lambda t}$
$$\begin{gathered} \Rightarrow 0.75={\mathrm{e}}^{-\lambda x} \\ \Rightarrow \mathrm{In} 0.75=-\lambda t \\ \Rightarrow t=\frac{\mathrm{In} 0.75}{-0.00018} \\ =1600 \mathrm{s} \end{gathered}$$

Answer:

Given:
Half-life of ¹⁹⁹Au, T_1/2= 2.7 days

Disintegration constant, $\lambda =\frac{0.693}{T_{{\displaystyle 1/2}}}$ =  $\frac{0.693}{2.7\times 24\times 60\times 60}=2.97\times {10}^{-6 }{\mathrm{s}}^{-1}$

Number of atoms left undecayed, N = $\frac{1\times {10}^{-6}\times 6.023\times {10}^{23}}{198}$

Now, activity, $A_0=\lambda N$
  
= $\frac{1\times {10}^{-6}\times 6.023\times {10}^{23}}{198}\times 2.97\times {10}^{-6}$
   = 0.244 Ci

(b) After 7 days,
Activity, $A=A_0{e}^{-\lambda t}$
Here, A₀ = 0.244 Ci
$$\begin{gathered} \therefore A =0.244\times e^{-2.97\times {10}^{-6}\times 7\times 3600\times 24} \\ =0.244\times e^{17962.56\times {10}^{-4}} \\ =0.040 \mathrm{Ci} \end{gathered}$$

Answer:

Given:
Half-life of radioactive ¹³¹I, T_1/2 = 8 days
Activity of the sample at t = 0, A₀ = 20 µCi
Time, t = 4 days

(a) Disintegration constant, $\lambda =\frac{0.693}{T_{1/2}}=0.0866$
Activity (A) at t = 4 days is given by
$$\begin{gathered} A=A_0{e}^{-\lambda t} \\ \Rightarrow A=20\times {10}^{-6}\times e^{-0.0866\times 4} \\ =1.41\times {10}^{-6} \mathrm{Ci}=14 \mathrm{\mu Ci} \end{gathered}$$

(b) Decay constant is a constant for a radioactive sample and depends only on its half-life.
$$\begin{gathered} \lambda =\frac{0.693}{[8\times 24\times 3600]} \\ =1.0026\times {10}^{-6} {\mathrm{s}}^{-1} \end{gathered}$$

Answer:

Given:
Decay constant, $\lambda$ = 4.9 × 10⁻¹⁸ s⁻¹

(a) Average life of uranium $\left(\tau \right)$ is given by
$$\begin{gathered} \tau =\frac{1}{\lambda } \\ =\frac{1}{4.9\times {10}^{-18}} \\ =\frac{1}{4.9}\times {10}^{18} \mathrm{s} \\ =\frac{{10}^{16}}{4.9\times 365\times 24\times 36} \mathrm{years} \\ =\frac{{10}^{16}}{4.9\times 365\times 24\times 36} \mathrm{years} \\ = 6.47\times {10}^{-7}\times {10}^{16} \mathrm{years} \\ =6.47\times {10}^9 \mathrm{years} \end{gathered}$$

(b) Half-life of uranium $\left(T_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)$ is given by
$$\begin{gathered} T_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0.693}{\lambda }=\frac{0.693}{4.9\times {10}^{-18}} \\ =\frac{0.693}{4.9}\times {10}^{18} \mathrm{s} \\ =0.1414\times {10}^{18} \mathrm{s} \\ =\frac{0.1414\times {10}^{18}}{365\times 24\times 3600} \\ =\frac{1414\times {10}^{12}}{365\times 24\times 36} \\ = 4.48\times {10}^{-3}\times {10}^{12} \\ =4.5\times {10}^9 \mathrm{years} \end{gathered}$$

(c) Time, t = 9 × 10⁹ years
 Activity (A) of the sample, at any time t, is given by
  $A=\frac{A_0}{{\displaystyle 2^{\frac{t}{T_{1/2}}}}}$
  Here, A₀ = Activity of the sample at t = 0
$\therefore \frac{A_0}{A}=2^{\frac{9\times {10}^9}{4.5\times {10}^9}}=2^2=4$

Answer:

Given:
Initial rate of decay, A₀ = 500,
Rate of decay after 50 minutes, A = 200
Time, t = 50 min
            = 50 $\times$ 60
            = 3000 s

(a)
Activity, A = A₀e^−λt
Here, $\lambda$ = Disintegration constant
$\therefore$ 200 = 500 × e^{−50×60×λ}
$$\begin{gathered} \Rightarrow \frac{2}{5}=e^{-3000\lambda } \\ \Rightarrow \mathrm{In}\frac{2}{5}=-3000\lambda \\ \Rightarrow \lambda =3.05\times {10}^{-4} {\mathrm{s}}^{-1} \end{gathered}$$

(b)
$$\begin{gathered} \mathrm{Half}-\mathrm{life}, T_{1/2}=\frac{0.693}{\lambda } \\ =2272.13 \mathrm{s} \\ =38 \min \end{gathered}$$

Answer:

Given:
Initial count rate of radioactive sample, A₀ = 4 × 10⁶ disintegration/sec
Count rate of radioactive sample after 20 hours, A = 1 × 10⁶ disintegration/sec
Time, t = 20 hours
Activity of radioactive sample $\left(A\right)$ is given by
$$\begin{gathered} A=\frac{A_0}{2^{{\displaystyle \frac{t}{{\displaystyle \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}}} \\ \mathrm{Here}, T_{1/2}=\mathrm{Half}-\mathrm{life} \mathrm{period} \\ \mathrm{On} \mathrm{substituting} \mathrm{the} \mathrm{values} \mathrm{of} A_0 \mathrm{and} A, \mathrm{we} \mathrm{have} \\ 2^{\frac{t}{{\displaystyle \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}=2^2 \\ \Rightarrow \frac{t}{{\displaystyle \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=2 \\ \Rightarrow \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=t/2=20 \mathrm{h}/2=10 \mathrm{h} \end{gathered}$$

100 hours after the beginning,
$$\begin{gathered} \mathrm{Count} \mathrm{rate}, A"=\frac{A_0}{2^{{\displaystyle \frac{t}{{\displaystyle \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}}} \\ \Rightarrow A"=\frac{4\times {10}^6}{2^{100/10}} \\ =0.390625\times {10}^4 \\ =3.9\times {10}^3 \mathrm{disintegrations}/\sec \end{gathered}$$
     

Answer:

Given:
Half-life of radium, T_1/2 = 1602 years
Atomic weight of radium = 226 g/mole
Atomic weight of chlorine = 35.5 g/mole
Now,
1 mole of RaCl₂ = 226 + 71 = 297 g
297 g = 1 mole of RaCl₂
0.1 g = $\frac{1}{297}\times 0.1 \mathrm{mole} \mathrm{of} {\mathrm{RaCl}}_2$
Total number of atoms in 0.1 g of RaCl₂, N $=\frac{0.1\times 6.023\times {10}^{23}}{297} = 0.02027\times {10}^{22}$
∴ No of atoms, N = 0.02027 $\times$ 10²²

  $$\begin{gathered} \mathrm{Disintegration} \mathrm{constant}, \lambda =\frac{0.693}{T_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ =\frac{0.693}{1602\times 365\times 24\times 3600} \\ =1.371\times {10}^{-11} \end{gathered}$$

Activity of radioactive sample, A = $\lambda N$
                                                      = $1.371\times {10}^{-11}\times 2.027\times {10}^{20}$
                                                      = $2.8\times {10}^{9 }\mathrm{disintegrations}/\mathrm{second}$

Answer:

Given:
Half-life of radioisotope, $T_{1/2}$ = 10 hrs
Initial activity, A₀ = 1 Ci
Disintegration constant, $\lambda =\frac{0.693}{10\times 3600} {\mathrm{s}}^{-1}$
Activity of radioactive sample, $A=A_0{e}^{-\lambda t}$
Here, A₀ = Initial activity
$\lambda$ = Disintegration constant
t = Time taken

After 9 hours,
Activity, $A=A_0{e}^{-\lambda t}=1\times e^{-\frac{0.693}{10\times 3600}\times 9}=0.536 \mathrm{Ci}$

∴ Number of atoms left, N = $\frac{A}{\lambda }$ = $\frac{0.536\times 10\times 3.7\times {10}^{10}\times 3600}{0.693}=103.023\times {10}^{13}$
After 10 hrs, 

$$\begin{gathered} Activity, A\mathit{"}=A_0{e}^{-\lambda t} \\ =1\times e^{\frac{-0.693}{10}\times 10}=0.5 \mathrm{Ci} \end{gathered}$$

Number of atoms left after the 10^th hour $\left(N"\right)$ will be
$$\begin{gathered} A"=\lambda N" \\ N"=\frac{A"}{\lambda } \\ =\frac{0.5\times 3.7\times {10}^{10}\times 3.600}{0.693/10} \\ =26.37\times {10}^{10}\times 3600=96.103\times {10}^{13} \end{gathered}$$

Number of disintegrations  = (103.023 − 96.103) × 10¹³
                                       = 6.92 × 10¹³

Answer:

Given:
Half-life of ³²P source, $T_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$= 14.3 days
Time, t = 30 days = 1 month
Here, the selling rate of a radioactive isotope is decided by its activity.

∴ Selling rate = Activity of the radioactive isotope after 1 month
Initial activity, A₀ = 800 disintegration/sec
Disintegration constant $\left(\lambda \right)$ is given by
$\lambda =\frac{0.693}{T_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$ = $\frac{0.693}{14.3} {\mathrm{days}}^{-1}$
Activity $\left(A\right)$ is given by
A = A₀e^−λt
Here, $\lambda$ = Disintegration constant   
$\therefore$ Activity of the radioactive isotope after one month (selling rate of the radioactive isotope) $\left(A\right)$ is given below.     
$$\begin{gathered} A=800\times e^{\frac{-0.693}{14.3}\times 30} \\ =800\times 0.233669 \\ =186.935=\mathrm{Rs} 187 \end{gathered}$$

Answer:

According to the question, when the β⁺ decays to half of its original amount, the emission rate of γ-rays will drop to half. For this, the sample will take 270 days.
Therefore, the required time is 270 days.

Answer:

(a) The reaction is given by
    ${\mathrm{C}}_6\to {\mathrm{B}}_5+{\mathrm{\beta }}^{+}+\mathrm{v}$
 It is a β⁺ decay since atomic number is reduced by 1.

(b) Half-life of the decay scheme, T _1/2   = 20.3 minutes

Disintegration constant,  $\lambda =\frac{0.693}{T_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$ = $\frac{0.693}{20.3} {\min }^{-1}$

If t is the time taken by the mixture in converting, let the total no. of atoms be 100N_0.

Answer:

Given:
Number of tritium atoms, N₀ = 4 × 10²³
Half-life of tritium nuclei, $T_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$= 12.3 years

Disintegration constant,  $\lambda =\frac{0.693}{T_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}=\frac{0.693}{12.3}$ years^$-1$

Activity of the sample $\left(A\right)$ is given by 
A₀ = $\frac{dN}{dt}$ = $\lambda N_0$
$$\begin{gathered} \Rightarrow A_0=\frac{0.693}{t_{1/2}}{N}_0 \\ =\frac{0.693}{12.3}\times 4\times {10}^{23} \mathrm{disintegration}/\mathrm{year} \\ =\frac{0.693\times 4\times {10}^{23}}{12.3\times 3600\times 24\times 365} \mathrm{disintegration}/\sec \\ =7.146\times {10}^{14} \mathrm{disintegration}/\sec \end{gathered}$$

(b) Activity of the sample, A = 7.146$\times$10¹⁴ disintegration/sec

   $$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{decays} \mathrm{in} \mathrm{the} \mathrm{next} 10 \mathrm{h}\mathrm{ours}=7.146\times {10}^{14}\times 10\times 3600 \\ =257.256\times {10}^{17} \\ =2.57\times {10}^{19} \end{gathered}$$

(c) Number of atoms left undecayed, $N=N_0{e}^{-\lambda t}$
   Here, N₀ = Initial number of atoms
   $\therefore N=4\times {10}^{23}\times e^{\frac{-0.693}{12.3}\times 6.15}=2.83\times {10}^{23}$
 
Number of atoms disintegrated = (N_{0 $-$} N) = (4$-$2.83) × 10²³ = 1.17 × 10²³

Answer:

Given:
Counts received per second = 50000 Counts/second
Number of active nuclei, N = 6 × 10¹⁶
Total counts radiated from the source, $\frac{dN}{\mathrm{dt}}$ = Total surface area × 50000 counts/cm²
                                                                     = 4 × 3.14 × 1 × 10⁴ × 5 × 10⁴
                                                                     = 6.28 × 10⁹ Counts
We know
$\frac{dN}{dt}=\lambda N$

Here, λ = Disintegration constant
$$\begin{gathered} \therefore \lambda =\frac{6.28\times {10}^9}{6\times {10}^{16}} \\ =1.0467\times {10}^{-7} \\ =1.05\times {10}^{-7}{\mathrm{s}}^{-1} \end{gathered}$$

Answer:

Given:
Half-life of ²³⁸U, t_1/2 = 4.47 × 10⁹ years
$$\begin{gathered} \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{atoms} \mathrm{present} \mathrm{in} \mathrm{the} \mathrm{rock} initially, N_0=\left(\frac{6.023\times {10}^{23}\times 2}{238}+\frac{6.023\times {10}^{23}\times 0.6}{206}\right) \\ =\left(\frac{12.046}{238}+\frac{3.62}{206}\right)\times {10}^{20} \\ =\left(0.0505+0.0175\right)\times {10}^{20} \\ =0.0680\times {10}^{20} \end{gathered}$$

Now, N = N₀e^{$-\lambda t$}
Here, $\lambda$ = Disintegration constant
t = Life of the rock
$$\begin{gathered} \Rightarrow N=N_0{e}^{\frac{-0.693}{t_{1/2}}\times t} \\ \Rightarrow 0.0505=0.0680e^{\frac{-0.693}{4.47\times {10}^9}\times t} \\ \Rightarrow \ln \left(\frac{0.0505}{0.0680}\right)=\frac{-0.6931}{4.47\times {10}^9}\times t \\ \Rightarrow t=1.92\times {10}^9 \mathrm{years} \end{gathered}$$

Answer:

Given:
Initial activity of charcoal, A₀ = 15.3 disintegrations per gram per minute
Half-life of charcoal, $T_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$= 5730 years
Final activity of charcoal after a few years, A = 12.3 disintegrations per gram per minute

Disintegration constant, $\lambda =\frac{0.693}{T_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$ = $\frac{0.693}{5370 } {\mathrm{y}}^{-1}$

Let the sample take a time of t years for the activity to reach 12.3 disintegrations per gram per minute.

Activity of the sample, $A=A_0{e}^{-\lambda t}$
$$\begin{gathered} A=A_0{e}^{-\frac{0.693}{5730}\times t} \\ \Rightarrow \mathrm{In}\frac{12.3}{15.3}=\frac{-0.693}{5730}t \\ \Rightarrow 0.218253=\frac{0.693}{5730}\times t \\ \Rightarrow t=1804.3 \mathrm{years} \end{gathered}$$

Answer:

Given:
Half-life time of tritium, $T_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ = 12.5 years
Disintegration constant, $\lambda =\frac{0.693}{12.5 } \mathrm{per} \mathrm{year}$
Let A₀ be the activity, when the bottle was manufactured.
Activity after 8 years $\left(A\right)$ is given by
$A=A_0{e}^{\frac{-0.693}{12.5}\times 8}$     ...(1)
Let us consider that the mountaineering had taken place t years ago.
Then, activity of the bottle $\left(A\text{'}\right)$ on the mountain is given by
$A\text{'}=A_0{e}^{-\lambda t}$
Here, A' = (Activity of the bottle manufactured 8 years ago) × 1.5%
$A\text{'}=A_0{e}^{\frac{-0.693}{12.5}\times 8}\times 0.015 ...\left(2\right)$
Comparing (1) and (2)
$$\begin{gathered} \frac{0.693}{12.5} t=\frac{-0.6931\times 8}{12.5}+\mathrm{In} [0.015] \\ \Rightarrow \frac{-0.693}{12.5} t=\frac{-0.693}{12.5}\times 8-4.1997 \\ \Rightarrow 0.693 t=58.040 \\ \Rightarrow t=83.75 \mathrm{years} \end{gathered}$$

Answer:

(a) For t = 0,
$\ln \left(\frac{R_{\mathit{0}}}{R}\right)=\mathrm{In}\left(\frac{30\times {10}^9}{30\times {10}^9}\right)=0$
For t = 25 s,
$\ln \left(\frac{R_{\mathit{0}}}{R_{\mathit{2}}}\right)=\mathrm{In}\left(\frac{30\times {10}^9}{16\times {10}^9}\right)=0.63$
For t = 50 s,
$\mathrm{In}\left(\frac{{\mathit{R}}_{\mathit{0}}}{{\mathit{R}}_{\mathit{3}}}\right)=\mathrm{In}\left(\frac{30\times {10}^9}{8\times {10}^9}\right)=1.35$
For t = 75 s,
$\ln \left(\frac{R_0}{R_4}\right)=\mathrm{In}\left(\frac{30\times {10}^9}{3.8\times {10}^9}\right)=2.06$
For t = 100 s,
$\mathrm{In}\left(\frac{R_0}{R_5}\right)=\mathrm{In}\left(\frac{30\times {10}^9}{2\times {10}^9}\right)=2.7$
The required graph is shown below.


(b) Slope of the graph = 0.028
∴ Decay constant, $\lambda$ = 0.028 min^$-1$
The half-life period$\left(T_{\frac{1}{2}}\right)$ is given by
    $$\begin{gathered} T_{\frac{1}{2}}=\frac{0.693}{\lambda } \\ =\frac{0.693}{0.028}=25 \min \end{gathered}$$

Answer:

Given:
Half-life period of ⁴⁰K, $T_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ = 1.30 × 10⁹ years
Count given by 1 g of pure KCI, A = 160 counts/s

Disintegration constant, $\lambda =\frac{0.693}{T_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$
Now, activity, A = λN

$$\begin{gathered} \Rightarrow 160=\frac{0.693}{t_{1/2}} \times N \\ \Rightarrow 160=\left(\frac{0.693}{1.30\times {10}^9\times 365\times 86400}\right)\times N \\ \Rightarrow N=\frac{160\times 1.30\times 365\times 86400\times {10}^9}{0.693} \\ \Rightarrow N=9.5\times {10}^{18} \end{gathered}$$
6.023 × 10²³ atoms are present in 40 gm.

$$\begin{gathered} \mathrm{Thus,} \mathrm{9.5\; \times \; 1018} \mathrm{atoms} \mathrm{will} \mathrm{be} \mathrm{present} in\frac{40\times 9.5\times {10}^{18}}{6.023\times {10}^{23}} \mathrm{gm}. \\ =\frac{4\times 9.5\times {10}^{-4}}{6.023} \mathrm{gm} \\ =6.309\times {10}^{-4} \\ =0.00063 \mathrm{gm} \end{gathered}$$
Relative abundance of ⁴⁰K  in natural potassium = (2 × 0.00063 × 100)% = 0.12%

Answer:

Given:
Decay constant of electron capture = 0.257 per day

(a) The reaction is given as
 ${}_{80}{}^{197}\mathrm{Hg}+\mathrm{e}\to {}_{79}{}^{197}\mathrm{Au}+\mathrm{v}$
The other particle emitted in this reaction is neutrino v.

(b) Moseley's law is given by
$\sqrt{v}$ = a(Z − b)

We  know
$v=\frac{c}{\lambda }$
Here, c = Speed of light
        $\lambda$ = Wavelength of the K_α X-ray

$$\begin{gathered} \sqrt{\frac{c}{\lambda }}=4.95\times {10}^7(79-1) \\ =4.95\times {10}^7\times 78 \\ \Rightarrow \frac{c}{\lambda }=(4.95\times 78{)}^2\times {10}^{14} \\ \Rightarrow \lambda =\frac{3\times {10}^8}{149073.21\times {10}^{14}} \\ =20 \mathrm{pm} \end{gathered}$$

Answer:

Given:
Half life period of isotope = t_1/2

Disintegration constant, $\lambda =\frac{0.693}{t_{{\displaystyle 1/2}}}$

Rate of Radio active decay $\left(R\right)$ is given by,
$R=\frac{d\mathrm{N}}{dt}$
We are to show that after time t >> t_1/2 the number of active nuclei is constant.
$$\begin{gathered} {\left(\frac{dN}{dt}\right)}_{Present}=R={\left(\frac{dN}{dt}\right)}_{decay} \\ \therefore R = {\left(\frac{dN}{dt}\right)}_{decay} \end{gathered}$$
Rate of radioactive decay, $R=\lambda N$
Here, λ = Radioactive decay constant
N = Constant number
$$\begin{gathered} R=\frac{0.693}{t_{1/2}}\times N \\ \Rightarrow R{t}_{1/2}=0.693N \\ \Rightarrow N=\frac{R{t}_{1/2}}{0.693} \end{gathered}$$
This value of N should be constant.

Answer:

Let the number of atoms present at t = 0 be N₀.
Let N be the number of radio-active isotopes present at time t.
Then,
N = N₀e^−λt
Here, $\lambda$ = Disintegration constant
∴ Number of radioactive isotopes decayed = N₀ − N = N₀ − N₀e^−λt
                                                                     = N₀ (1−e^−λt)   ...(1)
Rate of decay $\left(R\right)$ is given by
 R = λN₀   ...(2)
Substituting the value of N₀ from equation (2) to equation (1), we get
$$\begin{gathered} N=N_0(1-e^{-\mathrm{\lambda }t}) \\ =\frac{R}{\lambda }(1-e^{-\mathrm{\lambda }t}) \end{gathered}$$

Answer:

Given:
Initial no of atoms, N₀ = 1 mole = 6 × 10²³ atoms
Half-life of the radioactive material, T_1/2 = 14.3 days
Time taken by the plant to settle down, t = 70 h

Disintegration constant, $\lambda =\frac{0.693}{t_{1/2}}$ = $\frac{0.693}{14.3\times 24}$ h^$-$1
N = N₀e^−λt
   $$\begin{gathered} =6\times {10}^{23}\times e^{\frac{-0.693\times 70}{14.3\times 24}} \\ =6\times {10}^{23}\times 0.868 \\ =5.209\times {10}^{23} \end{gathered}$$

 $$\begin{gathered} \mathrm{Activity,} R=\frac{dN}{dt}=5.209\times {10}^{23}\times \frac{0.693}{14.3\times 24} \\ =\frac{0.0105\times {10}^{23}}{3600}\mathrm{dis}/\mathrm{hr} \\ =2.9\times {10}^{-6}\times {10}^{23} \mathrm{dis}/\sec \\ =2.9\times {10}^{17}\mathrm{dis}/\sec \end{gathered}$$

$$\begin{gathered} \mathrm{Fraction} \mathrm{of} \mathrm{activity} \mathrm{transmitted} = \left(\frac{1 \mathrm{\mu Ci}}{2.9\times {10}^{17}}\right)\times 100\% \\ =\left[\left(\frac{1\times 3.7\times {10}^4}{2.9\times {10}^{17}}\right)\times 100\right]\% \\ =1.275\times {10}^{-11}\% \end{gathered}$$

Answer:

Given:
Volume of the vessel, V = 125 cm³ = 0.125 L
Half-life time of tritium, t_1/2 = 12.3 y = 3.82 × 10⁸ s
Pressure, P = 500 kpa = 5 atm
Temperature, T = 300 K,

Disintegration constant,  $\lambda =\frac{0.693}{t_{1/2}}$
                                      =$$\begin{gathered} \frac{0.693}{3.82\times {10}^8}=0.1814\times {10}^{-8} \\ =1.81\times {10}^{-9} {\mathrm{s}}^{-1} \end{gathered}$$

No. of atoms left undecayed, N = n × 6.023 × 10²³
                                             $$\begin{gathered} =\frac{5\times 0.125}{8.2\times {10}^{-2}\times 3\times {10}^2}\times 6.023\times {10}^{23} \left(\because n=\frac{PV}{RT}\right) \\ =1.5\times {10}^{22} \mathrm{atoms} \end{gathered}$$

Activity, A = λN
       = 1.81 × 10⁻⁹ × 1.5 × 10²² = 2.7 × 10¹³ disintegration/sec
   
$\therefore A=\frac{2.7\times {10}^{13}}{3.7\times {10}^{10}} \mathrm{Ci}=729.72 \mathrm{Ci}$