Hc Verma II for Class 11 Science Physics Chapter 42 - Photoelectric Effect And Wave–particle Duality

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Page No 363:

Answer:

No, we cannot find the mass of a photon by the definition p = mv. The equation p = mv is valid only for objects that move with a velocity that is much slower than the speed of light. The momentum of a relativistic particle like photon is given by $p c = \sqrt{E^{2} - m^{2} c^{4}}$ .
A photon has zero rest mass. Therefore, on putting m = 0 in the equation, we get
$p = \frac{E}{c}$ , which is the valid equation for a photon.

Page No 363:

Answer:

Let the source's area be A, and intensity of the source be I. The energy of each emitted photonis E. Then, the number of photons emitted in a given time will be $n = \frac{I}{A E}$ .
If the areas of the sources and the wavelengths of light emitted by the two sources are different, then the number of photons emitted will be different.

Page No 363:

Answer:

(a) In relativity, the relative speed of two objects $(v_{r e l})$ moving in the same direction with speeds u and v is given by
   $v_{rel} = \frac{u - v}{1 - \frac{u v}{c^{2}}} . . . (1)$
As the photons are moving with the speed of light, u = c and v = c.
On substituting the values of u and v in equation (1), we get:
$v_{r e l}$ = 0
Thus, relative velocity of a photon with respect to another photon will be 0, when they are going in the same direction.
(b) In relativity, relative speed of two objects moving in opposite directions with speeds u and v is given by
    $v_{rel} = \frac{u + v}{1 + \frac{u v}{c^{2}}} . . . (2)$
   We know that a photon travels with the speed of light. Therefore, u = c and v = c
   On substituting the values of u and v in equation (2), we get:
   $v_{r e l} = c$
   Thus, the relative velocity of a photon with respect to another photon will be equal to the speed of light when they are going in opposite directions.

Page No 363:

Answer:

Photons are electrically neutral. Hence, they are not deflected by electric and magnetic fields.

Page No 363:

Answer:

As the hot body is placed in a closed room maintained at a lower temperature, there will be transfer of heat in the room through convection and radiation. Heat radiation also consists of photons; therefore, photons will be emitted by the hot body. Hence, the number of photons in the room will increase.

Page No 363:

Answer:

Relativistic equation of energy:
$E^{2} = p^{2} c^{2} + m^{2} c^{4} . . . (1)$
Here, p²c² = kinetic energy of photon
m₀²c⁴ = internal energy of photon
We know photons have zero rest mass. Therefore, m₀ = 0.
Substituting the value of m₀ = 0 in equation (1), we get:
$E = p c$
Thus, the energy of a photon should be called its kinetic energy.

Page No 363:

Answer:

No, it does not violate the principle of conservation of momentum. In the photon-electron collision, the energy and momentum are conserved.

Page No 363:

Answer:

Photoelectrons are emitted from a metal's surface if the frequency of incident radiation is more than the threshold frequency of the given metal surface. As yellow light does not eject photoelectrons from a metal it means that the threshold frequency of the metal is more than the frequency of yellow light. Since the frequency of orange light is less than the frequency of yellow light, therefore it will not be able to eject photoelectrons from the metal's surface. The frequency of green light is more than the frequency of yellow light. Hence, when it is incident on the metal surface, it will eject electrons from the metal.

Page No 363:

Answer:

Photosynthesis starts when a plant is exposed to visible light. The visible light's photons possess just enough energy to excite the electrons of molecules of the plant without causing damage to its cells. Infrared rays have less frequency than visible light. Due to this, the energy of the photons of infrared rays are not sufficient to initiate photosynthesis. Therefore, photosynthesis does not start if plants are exposed only to infrared light.

Page No 363:

Answer:

When light of wavelength less than λ₀ is incident on the metal surface, the free electrons of the metal will gain energy and come out of the metal surface. As the metal plate is insulated (it is not connected with the battery), the free electrons of the metal will not be replaced by the other electrons. Hence, photoelectron emission will stop after some time.

Page No 363:

Answer:

From Einstein's mass- energy equation,
E = mc²
$\Rightarrow E = \frac{m_{0} c^{2}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$
Relativistic momentum,
$p = m v \Rightarrow p = \frac{m_{0} c^{2}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$
Combining the above equations, we get:
$E^{2} = {m_{0}}^{2} c^{4} + p^{2} c^{2}$
From the above equation, it is clear that for p = E/c to be valid, the rest mass of the body should be zero. As electrons do not have zero rest mass, this equation is not valid for electrons.

Page No 363:

Answer:

de-Broglie wavelength,
      $λ = \frac{h}{m v}$ ,

      where h = Planck's constant
            $m$ = mass of the particle
            v = velocity of the particle
(a) It is given that the speed of an electron and proton are equal.
     It is clear from the above equation that
     $λ \propto \frac{1}{m}$
As mass of a proton, m_p > mass of an electron, m_e, the proton will have a smaller wavelength compared to the   electron.

(b) $λ = \frac{h}{p}$ $(∵ p = m v)$
     So, when the proton and the electron have same momentum, they will have the same wavelength.
(c) de-Broglie wavelength $(λ)$ is also given by
      $λ = \frac{h}{\sqrt{2 m E}}$ ,
      where E = energy of the particle.
Let the energy of the proton and electron be E.
Wavelength of the proton,
$λ_{p} = \frac{h}{\sqrt{2 m_{p} E}} . . . (1)$
Wavelength of the electron,
$λ_{e} = \frac{h}{\sqrt{2 m_{e} E}} . . . (2)$
Dividing (2) by (1), we get:
$\frac{λ_{e}}{λ_{p}} = \frac{\sqrt{m_{e}}}{\sqrt{m_{p}}} \Rightarrow \frac{λ_{e}}{λ_{p}} < 1 \Rightarrow λ_{e} < λ_{p}$
It is clear that the proton will have smaller wavelength compared to the electron.

Page No 363:

Answer:

Colour is a characteristic of electromagnetic waves. Electrons behave as a de-Broglie wave because of their velocity. A de-Broglie wave is not an electromagnetic wave and is one dimensional. Hence, no colour is shown by an electron.

Page No 363:

Answer:

(d) force × distance time

Planck's constant,
h = $\frac{E}{v} = \frac{Force \times distance}{frequency}$
$\Rightarrow$ h = force $\times$ distance time

Page No 363:

Answer:

Two photons having equal linear momenta have equal wavelengths is correct. As in the rest of the options magnitude of momentum or energy can be same because energy and momentum are inversely proportional to wavelength. But the direction of propagation of the photons can be different.
Hence the correct option is D.

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Answer:

(a) both p and E increase

From the de-Broglie relation, wavelength,
$λ = \frac{h}{p} . . . (1)$
$\Rightarrow$ $p = \frac{h}{λ}$
Here, h = Planck's constant
p = momentum of electron
It is clear from the above equation that $p \propto \frac{1}{λ}$ .
Thus, if the wavelength $(λ)$ is decreased, then momentum $(p)$ will be increase.
Relation between momentum and energy:
$p = \sqrt{2 m E}$
Here, E = energy of electron
m = mass of electron
Substituting the value of p in equation (1), we get:
$λ = \frac{h}{\sqrt{2 m E}} \Rightarrow \sqrt{E} = \frac{h}{λ \sqrt{2 m}} \Rightarrow E = \frac{h^{2}}{2 m λ^{2}}$
Thus, on decreasing $λ$ , the energy will increase.

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Answer:

(c) n_r > n_b

The two bulbs are of equal power. It means that they consume equal amount of energy per unit time.
Now, as the frequency of blue light $(f_{b})$ is higher than the frequency of red light $(f_{r})$ , $h f_{b} > h f_{r}$ .
Hence, the energy of a photon of blue light is more than the energy of a photon of red light.
Thus, a photon of blue light requires more energy than a photon of red light to be emitted.
For the same energy given to the bulbs in a certain time, the number of photons of blue light will be less than that of red light.
$∴$ n_r > n_b (As the amount of energy emitted from the two bulb is same)

Page No 363:

Answer:

(c) for a photon but not for an electron

The equation E = pc is valid for a particle with zero rest mass. The rest mass of a photon is zero, but the rest mass of an electron is not zero. So, the equation will be valid for photon, and not electron.

Page No 363:

Answer:

(a) v ≥ v₀

As the work function of the metal is hv₀, the threshold frequency of the metal is v₀.

For photoelectric effect to occur, the frequency of the incident light should be greater than or equal to the threshlod frequency of the metal on which light is incident.

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Answer:

(c) λ ≤ λ₀

As the work-function of the metal is hc/λ₀, its threshold wavelength is λ₀.
For photoelectric effect, the wavelength of the incident light should be less than or equal to the threshold wavelength of the metal on which light is incident.

Page No 364:

Answer:

(b) the photoelectrons are emitted but are re-absorbed by the emitter metal

In an experiment on photoelectric effect, the photons incident at the metal plate cause photoelectrons to be emitted. The metal plate is termed as "emitter". The electrons ejected are collected at the other metal plate called "collector". When the potential of the collector is made negative with respect to the emitter (or the stopping potential is applied), the electrons emitted from the emitter are repelled by the collector. As a result, some electrons go back to the cathode and the current decreases.

Page No 364:

Answer:

(c) become more than double

According to Einstein's equation of photoelectric effect,
$e V_{0} = h v - φ \Rightarrow V_{0} = \frac{h v - φ}{e} . . . (1)$
Here, V₀ = stopping potential
v = frequency of light
$φ$ = work function

Let the new frequency of light be 2ν and the corresponding stopping potential be V₀'.
Therefore,
$e V_{0}' = 2 h v - φ V_{0}' = \frac{2 h v - φ}{e} . . . (2)$

Multiplying both sides of equation (1) by 2, we get:
$2 V_{0} = \frac{2 h v - 2 φ}{e} . . . (3)$
Now if we compare (2) and (3), it can be observed that:
$\frac{2 h v - φ}{e} > \frac{2 h v - 2 φ}{e} \Rightarrow V_{0}' > 2 V_{0}$
It is clear from the above equation that if the frequency of light in a photoelectric experiment is doubled, the stopping potential will be more than doubled.

Page No 364:

Answer:

(b) A is true but B is false.

Saturated current varies directly with the intensity of light. As the intensity of light is increased, a large number of photons fall on the metal surface. As a result, a large number of electrons interact with the photons. As a result, the number of emitted electrons increases and, hence, the current also increases.
At the same time, the frequency of the light source also increases.Also, with the increase in frequency of light, the stopping potential increases as well. This will reduce the current. The combined effect of these two is that the current will remain the same
Hence, A is true.
From the Einstein's photoelectric equation.
$K_{m a x} = h v - φ$
Where K_max = kinetic energy of electron
v = frequency of light
$φ$ = work function of metal
It is clear from the above equation. As the frequency of light source is doubled, kinetic energy of electron increases. But, it becomes more than the double.
Hence, B is false.

Page No 364:

Answer:

(c) will remain constant

As the source is removed farther from the emitting metal, the intensity of light will decrease. As the stopping potential does not depend on the intensity of light, it will remain constant.

Page No 364:

Answer:

From the given curves ,curve (d) is correct.

As the relation between intensity (I) of light and distance (r) is
$I \propto \frac{1}{r^{2}}$
As the distance between the source and the metal is increased, it will result in decrease in the intensity of light. As the saturation current is directly proportional to the intensity of light ( $i \propto I$ ), it can be concluded that current varies as $i \propto \frac{1}{r^{2}}$ . Thus, curve d is correct.

Page No 364:

Answer:

(c) is related to the shortest wavelength

For photoelectric effect to be observed, wavelength of the incident light $(λ)$ should be less than the threshold wavelength $(λ_{0})$ of the metal.
Einstein's photoelectric equation:
$e V_{0} = \frac{h c}{λ_{0}} - φ$
Here, V₀ = stopping potential
             $λ_{0}$ = threshold wavelength
             h = Planck's constant
             $φ$ = work-function of metal
It is clear from the above equation that stopping potential is related to the shortest wavelength (threshold wavelength).

Page No 364:

Answer:

(c) λ_e > λ_p

Let m_e and m_p be the masses of electron and proton, respectively.
Let the applied potential difference be V.

Thus, the de-Broglie wavelength of the electron,
$λ_{e} = \frac{h}{\sqrt{2 m_{e} e V}} . . . (1)$
And de-Broglie wavelength of the proton,
$λ_{p} = \frac{h}{\sqrt{2 m_{p} e V}} . . . (2)$
Dividing equation (2) by equation (1), we get:
$\frac{λ_{p}}{λ_{e}} = \frac{\sqrt{m_{e}}}{\sqrt{m_{p}}}$
m_e < m_p
$∴$ $\frac{λ_{p}}{λ_{e}} < 1$
$\Rightarrow λ_{p} < λ_{e}$

Page No 364:

Answer:

(a) the number of photons emitted by the source in unit time increases
(b) the total energy of the photons emitted per unit time increases

When the intensity of a light source in increased, a large number of photons are emitted from the light source. Hence, option (a) is correct.
Due to increase in the number of photons, total energy of the photons emitted per unit time also increases. Hence, option (b) is also correct.
Increase in the intensity of light increases only the number of photons, not the energy of photons, Hence, option (c) is incorrect.
The speed of photons is not affected by the intensity of light, Hence, option (d) is incorrect.

Page No 364:

Answer:

(a) there is a minimum frequency below which no photoelectrons are emitted
(b) the maximum kinetic energy of photoelectrons depends only on the frequency of light and not on its intensity
(c) even when the metal surface is faintly illuminated the photoelectrons leave the surface immediately

Photoelectric effect can be explained on the basis of quantum nature of light. According to the quantum nature of light, energy in light is not uniformly spread. It is contained in packets or quanta known as photons.
Energy of a photon, E = hv, where h is Planck's constant and v is the frequency of light.
Above a particular frequency, called threshold frequency, energy of a photon is sufficient to emit an electron from the metal surface and below which, no photoelectron is emitted, as the energy of the photon is low. Hence, option (a) supports the quantum nature of light.
Now, kinetic energy of an electron,
$K = h v_{0} - φ$
Thus, kinetic energy of a photoelectron depends only on the frequency of light (or energy). This shows that if the intensity of light is increased, it only increases the number of photons and not the energy of photons. Kinetic energy of photons can be increased by increasing the frequency of light or by increasing the energy of photon, which supports E = hv and, hence, the quantum nature of light. Hence, option (b) also supports the quantum nature of light.

Photoelectrons are emitted from a metal surface even if the metal surface is faintly illuminated; it means that less photons will interact with the electrons. However, few electrons absorb energy from the incident photons and come out from the metal. This shows the quantum nature of light. Hence, (c) also supports the quantum nature of light.

Electric charge of the photoelectrons is quantised; but this statement does not support the quantum nature of light.

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Answer:

(d) It may come out with kinetic energy less than hv − φ.

When light is incident on the metal surface, the photons of light collide with the free electrons. In some cases, a photon can give all the energy to the free electron. If this energy is more then the work-function of the metal,then there are two possibilities. The electron can come out of the metal with kinetic energy hv − φ or it may lose energy on collision with the atoms of the metal and come out with kinetic energy less than hv − φ. Thus, it may come out with kinetic energy less than hv − φ.

Page No 364:

Answer:

(b) photoelectric emission may or may not take place
(d) the stopping potential will decrease

For photoelectric effect to be observed, wavelength of incident light should not be more than the largest wavelength called threshold wavelength $(λ_{0})$ . If the wavelength of light in an experiment on photoelectric effect is doubled and if it is equal to or less than the threshold wavelength, then photoelectric emission will take place. If it is greater than the threshold wavelength, photoelectric emission will not take place. The photoelectric emission may or may not take place.Photoelectric emission depends on the wavelength of incident light.

Hence, option (b) is correct and (a) is incorrect.

From Einstein's photoelectric equation,
$e V_{0} = \frac{h c}{λ_{0}} - φ$ ,
where V₀ = stopping potential
             $λ_{0}$ = threshold wavelength
             h = Planck's constant
             $φ$ = work-function of metal
It is clear that
$V_{0} \propto \frac{1}{λ_{0}}$
Thus, if the wavelength of light in an experiment on photoelectric effect is doubled, its stopping potential will become half.

Page No 364:

Answer:

(a) the intensity of the source is increased

When the intensity of the source is increased, the number of photons emitted from the source increases. As a result, a large number of electrons of the metal interact with these photons and hence, the number of electrons emitted from the metal increases. Thus, the photocurrent in an experiment of photoelectric effect increases. The photocurrent does not depend on the exposure time. Hence, option (a) is correct.

Page No 364:

Answer:

(b) The kinetic energy of the electrons will increase.

As there is no effect of electric field on the number of photons emitted, the photoelectric current will remain same. Hence, option (a) is incorrect.

When an electric field is applied, then electric force will act on the electron moving opposite the direction of electric field, which will increase the kinetic energy of the electron. Hence, option (b) is correct.

As the kinetic energy of the electron is increasing, its stopping potential will increase. Hence, option (c) is incorrect.

Threshold wavelength is the characteristic property of the metal and will not change. Hence, (d) is incorrect.

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Answer:

(a) move with the same speed
(c) move with the same kinetic energy
(d) have fallen through the same height

Let m₁ be the mass of the heavier particle and m₂ be the mass of the lighter particle.
If both the particles are moving with the same speed v, de Broglie wavelength of the heavier particle,
$λ_{1} = \frac{h}{m_{1} v}$ ...(1)
de Broglie wavelength of the lighter particle,
$λ_{2} = \frac{h}{m_{2} v}$ ...(2)
Thus, from equations (1) and (2), we find that if the particles are moving with the same speed v, then $λ_{1} < λ_{2}$ .
Hence, option (a) is correct.

If they are moving with the same linear momentum, then using the de Broglie relation
$λ = \frac{h}{p}$
We find that both the bodies will have the same wavelength. Hence, option (b) is incorrect.

If K is the kinetic energy of both the particles, then de Broglie wavelength of the heavier particle,
$λ_{1} = \frac{h}{\sqrt{2 m_{1} K}}$
de Broglie wavelength of the lighter particle,
$λ_{2} = \frac{h}{\sqrt{2 m_{2} K}}$
It is clear from the above equation that if $m_{1} > m_{2}$ , then $λ_{1} < λ_{2}$ .
Hence, option (c) is correct.

When they have fallen through the same height h, then velocity of both the bodies,
v = $\sqrt{2 g h}$
Now,
$λ_{1} = \frac{h}{m_{1} \sqrt{2 g h}}$
$λ_{2} = \frac{h}{m_{2} \sqrt{2 g h}}$
m₁>m₂
$∴$ $λ_{1} < λ_{2}$
Hence, option (d) is correct.

Page No 365:

Answer:

Given:
Range of wavelengths, $λ_{1}$ = 400 nm to $λ_{2}$ = 780 nm
Planck's constant, h = 6.63 $\times$ 10^{$-$ 34} Js
Speed of light, c = 3 $\times$ 10⁸ m/s
Energy of photon,
$E = h v$
$ν = \frac{c}{λ}$
$∴ E = h ν = \frac{h c}{λ}$
Energy $(E_{1})$ of a photon of wavelength $(λ_{1})$ :
$E_{1} = \frac{h c}{λ_{1}} = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{400 \times 10^{- 9}} = \frac{6.63 \times 3}{4} \times 10^{- 9} = 4.97725 \times 10^{- 19} = 5 \times 10^{- 19} J$

Energy (E₂) of a photon of wavelength ( $λ_{2}$ ):
$E_{2} = \frac{6.63 \times 3}{7.8} \times 10^{- 19} = 2.55 \times 10^{- 9} J$
So, the range of energy is 2.55 × 10⁻¹⁹ J to 5 × 10⁻¹⁹ J.

Page No 365:

Answer:

Given:
Wavelength of light, $λ$ = 500 nm
Planck's constant, h = 6.63 $\times$ 10^{$- 34$} J-s
Momentum of a photon of light,
$p = \frac{h}{λ} = \frac{6.63 \times 10^{- 34}}{500 \times 10^{- 9}} = \frac{6.63}{5} \times 10^{- 27} = 1.326 \times 10^{- 27} = 1.33 \times 10^{- 27} kg - m / s$

Page No 365:

Answer:

Given:
Wavelength of absorbed photon, $λ_{1}$ = 500 nm
Wavelength of emitted photon, $λ_{2}$ = 700 nm
Speed of light, c = 3 $\times 10^{8}$ m/s
Planck's constant, h = 6.63 $\times$ 10^{$- 34$} Js
Energy of absorbed photon,
$E_{1} = \frac{h c}{λ_{1}} = \frac{h \times 3 \times 10^{8}}{500 \times 10^{- 9}}$
Energy of emitted photon,
$E_{2} = \frac{h c}{λ_{2}} = \frac{h \times 3 \times 10^{8}}{700 \times 10^{- 9}}$
Energy absorbed by the atom in the process:
$E_{1} - E_{2} = h c [\frac{1}{λ_{1}} - \frac{1}{λ_{2}}] = 6.63 \times 3 [\frac{1}{5} - \frac{1}{7}] \times 10^{- 19} = 6.63 \times 3 \times \frac{2}{35} \times 10^{- 19} = 1.136 \times 10^{- 19} J$

Page No 365:

Answer:

Given:
Power of the sodium vapour lamp, P = 10 W
Wavelength of sodium light, $λ$ = 590 nm
Electric energy consumed by the bulb in one second = 10 J
Amount of energy converted into light = 60 %
∴ Energy converted into light = $\frac{60}{100} \times 10 = 6 J$
Energy needed to emit a photon from the sodium atom,
$E' = \frac{h c}{λ} E' = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{590 \times 10^{- 9}} E' = \frac{6.63 \times 3}{590} \times 10^{- 17} J$

Number of photons emitted,
$n = \frac{6}{\frac{6.63 \times 3}{590} \times 10^{- 17}} n = \frac{6 \times 590}{6.63 \times 3} \times 10^{17}$
n = 1.77 × 10¹⁹

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Answer:

Here,
Intensity of light, I = 1.4 × 10³ W/m²,
Wavelength of light, $λ$ = 500 nm = 500 $\times$ 10^{$- 9$} m
Distance between the Sun and Earth, l = 1.5 $\times$ 10¹¹ m
Intensity,
$I = \frac{Power}{Area} = 1.4 \times 10^{3} W / m^{2}$
Let n be the number of photons emitted per second.
∴ Power, P = Energy emitted/second
$P = \frac{n h c}{λ}$ ,
where $λ$ = wavelength of light
          h = Planck's constant
            c = speed of light
Number of photons/m² = $\frac{n h c}{λ \times A} = \frac{n h c}{λ \times 1}$ = I
$∴ n = \frac{I \times λ}{h c} = \frac{1.4 \times 10^{3} \times 500 \times 10^{- 9}}{6.63 \times 10^{- 34} \times 3 \times 10^{8}} = 3.5 \times 10^{21}$

(b) Consider number of two parts at a distance r and r + dr from the source.
Let dt' be the time interval in which the photon travels from one part to another.
Total number of photons emitted in this time interval,
$N = n d t = (\frac{P λ}{h c \times A}) \frac{d r}{c}$
These points will be between two spherical shells of radius 'r' and r + dr. It will be the distance of the 1st point from the sources.
In this case,
$l = 1.5 \times 10^{11} m Wavelength, λ = 500 nm = 500 \times 10^{- 9} m \frac{P}{4 π r^{2}} = 1.4 \times 10^{3} ∴ No . of photons / m^{3} = \frac{P}{4 π r^{2}} \frac{λ}{h c^{2}} = 1.4 \times 10^{3} \times \frac{500 \times 10^{- 9}}{6.63 \times 10^{- 34} \times 9 \times 10^{16}} = 1.2 \times 10^{13}$

(c) Number of photons emitted = (Number of photons / s-m²) × Area
                                       $= (3.5 \times 10^{21}) \times 4 π l^{2} = 3.5 \times 10^{21} \times 4 \times (3.14) \times (1.5 \times 10^{11})^{2} = 9.9 \times 10^{44}$

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Answer:

Here,
$Wavelength of monochromatic light, λ = 663 \times 10^{- 9} m Angle of incidence, θ = 60 ° Number of photons per second, n = 1 \times 10^{19} Momentum of photon, p = \frac{h}{λ}, w here h is P lanck' s constant . p = \frac{6.63 \times 10^{- 34}}{663 \times 10^{- 9}} = 10^{- 27}$

Force exerted on the wall,
$F = n \times [p cosθ - (- p cosθ)] = 2 n p cosθ = 2 \times 1 \times 10^{19} \times 10^{- 27} \times \frac{1}{2} = 1.0 \times 10^{- 8} N$

Page No 365:

Answer:

Power of the incident beam, P = 10 watt
Relation between wavelength $(λ)$ and momentum (p):
$λ = \frac{h}{p}, w here h is P lanck' s constant \Rightarrow p = \frac{h}{λ} On dividing both side s by t, we get : \frac{p}{t} = \frac{h}{λ t} . . . (1)$
Energy,
$E = \frac{h c}{λ} \Rightarrow \frac{E}{t} = \frac{h c}{λ t}$
Let P be the power. Then,
$P = \frac{E}{t} = \frac{h c}{λ t} P = \frac{p c}{t} [U sing equation (1)] \Rightarrow \frac{P}{c} = \frac{p}{t} F orce, F = \frac{p}{t} = \frac{P}{c} (Since F = \frac{M omentum}{T ime}) Force, F = \frac{7}{10} (absorbed) + 2 \times \frac{3}{10} (reflected) F = \frac{7}{10} \times \frac{P}{c} + 2 \times \frac{3}{10} \times \frac{P}{c} F = \frac{7}{10} \times \frac{10}{3 \times 10^{8}} + 2 \times \frac{3}{10} \times \frac{10}{3 \times 10^{8}} F = \frac{13}{3} \times 10^{- 8} = 4.33 \times 10^{- 8} N$

Page No 365:

Answer:

Given:
Mass of the mirror, m = 20 g = 20 × 10⁻³ kg
The weight of the mirror will be balanced if the force exerted by the photons will be equal to the weight of the mirror.
Now,
Relation between wavelength $(λ)$ and momentum (p):
$p = \frac{h}{λ} On dividing both side s by t, we get : \frac{P}{t} = \frac{h}{λ t} . . . (1)$
Energy,
$E = \frac{h c}{λ} \Rightarrow \frac{E}{t} = \frac{h c}{λ t}$
Let P be the power. Then,
$P = \frac{E}{t} = \frac{h c}{λ t} P = \frac{p c}{t} [U sing equation (1)] \Rightarrow \frac{P}{c} = \frac{p}{t} F orce, F = \frac{P}{t} = \frac{P}{c} (Since F = \frac{M omentum}{T ime})$

Thus, rate of change of momentum = Power/c
As the light gets reflected normally,
Force exerted = 2 (Rate of change of momentum) = 2 × Power/c
$30 % of (\frac{2 \times Power}{c}) = m g \Rightarrow Power = \frac{20 \times 10^{- 3} \times 10 \times 3 \times 10^{8} \times 10}{2 \times 3} = 100 MW$

Page No 365:

Answer:

Given:
Power of the light bulb, P = 100 W
Radius of the spherical chamber, R = 20 cm = 0.2 m
It is given that 60% of the energy supplied to the bulb is converted to light.
Therefore, power of light emitted by the bulb, P' = 60 W
Force,
$F = \frac{P}{c}$ ,
where c is the speed of light
$F = \frac{60}{3 \times 10^{8}} = 2 \times 10^{- 7} N$
Pressure = $\frac{Force}{Area}$
$= \frac{2 \times 10^{- 7}}{4 \times 3.14 \times (0.2)^{2}} (A = 4 π r^{2}) = \frac{1}{8 \times 3.14} \times 10^{- 5} = 0.039 \times 10^{- 5} = 3.9 \times 10^{- 7} = 4 \times 10^{- 7} N / m^{2}$

Page No 365:

Answer:

Given:
Radius of the sphere, r = 1 cm
Intensity of light, I = 0.5 Wcm⁻²
Let A be the effective area of the sphere perpendicular to the light beam.
So, force exerted by the light beam on the sphere is given by,
$F = \frac{P}{c} = \frac{A I}{c}$
$F = \frac{π \times (1)^{2} \times 0.5}{3 \times 10^{8}} = \frac{3.14 \times 0.5}{3 \times 10^{8}} = 0.523 \times 10^{- 8} = 5.2 \times 10^{- 9} N$

Page No 365:

Answer:

Consider a sphere of centre O and radius OP. As shown in the figure, the radius OP of the sphere is making an angle θ with OZ. Let us rotate the radius about OZ to get another circle on the sphere. The part of the sphere between the circle is a ring of area $2 π r^{2} \sin θ d θ$ .

Consider a small part of area $∆ A$ of the ring at point P.
Energy of the light falling on this part in time $∆ t$ ,
$∆ U = I ∆ t (∆ A \cos θ)$

As the light is reflected by the sphere along PR, the change in momentum,
$∆ p = 2 \frac{∆ U}{c} \cos θ = \frac{2}{c} I ∆ t (∆ A \cos^{2} θ)$
Therefore, the force will be
$\frac{∆ p}{∆ t} = \frac{2}{c} I ∆ A \cos^{2} θ$

The component of force on $∆ A$ , along ZO, is
$\frac{∆ p}{∆ t} \cos θ = \frac{2}{c} I ∆ A \cos^{3} θ$

Now, force action on the ring,
$d F = \frac{2}{c} I (2 π r^{2} \sin θ d θ) \cos^{3} θ$

The force on the entire sphere,
$F = \int_{0}^{\frac{π}{2}} \frac{4 π r^{2} I}{c} \cos^{3} θ \sin θ d θ = - \int_{0}^{\frac{π}{2}} \frac{4 π r^{2} I}{c} \cos^{3} θ d (\cos θ) = \frac{π r^{2} I}{c}$
This is same as given in the previous problem.

Page No 365:

Answer:

When an electron undergoes an inelastic collision with a photon, we can apply the principle of conservation of energy to this collision. So,
$p c + m_{e} c^{2} = \sqrt{p^{2} c^{2} + {m_{e}}^{2} c^{4}}$ ...(i)
Here, h = Planck's constant
c = the speed of light
m_e= rest mass of electron
pc = energy of the photon

Squaring on both side of equation (i),
$(p c + m_{e} c^{2})^{2} = p^{2} c^{2} + {m_{e}}^{2} c^{4} \Rightarrow p^{2} c^{2} + {m_{e}}^{2} c^{4} + 2 (p c) (m_{e} c^{2}) = p^{2} c^{2} + {m_{e}}^{2} c^{4} \Rightarrow 2 (p c) (m_{e} c^{2}) = 0 or p c = 0 (as' m' and' c' are non zero)$
This gives vanishing energy of photon which is not possible.

Page No 365:

Answer:

Given:
Distance between the two neutral particles, r = 1 m
Electric potential energy,
E₁= $\frac{k q^{2}}{r}$ = $k q^{2}$ ,
where k = $\frac{1}{4 {πε}_{0}}$
Energy of photon,
E₂ = $\frac{h c}{λ}$ ,
where $λ$ = wavelength of light
h = Planck's constant
c = speed of light
Here, E₁ = E₂
$∴ k q^{2} = \frac{h c}{λ} \Rightarrow λ = \frac{h c}{k q^{2}}$
For wavelength, λ, to be maximum, charge q should be minimum.
$q = e = 1.6 \times 10^{- 19} C Maximum wavelength, λ = \frac{h c}{k q^{2}} = \frac{6.63 \times 3 \times 10^{- 34} \times 10^{8}}{9 \times 10^{2} \times (1.6)^{2} \times 10^{- 38}} = 0.863 \times 10^{3} = 863 m$

Next smaller wave length,
$λ = \frac{6.63 \times 3 \times 10^{- 34} \times 10^{8}}{9 \times 10^{4} \times 4 \times (1.6)^{2} \times 10^{- 38}} = \frac{863}{4} = 215.74 m$

Page No 365:

Answer:

Given:
Wavelength of light, λ = 350 nm = 350 × 10⁻⁹ m
Work-function of cesium, ϕ = 1.9 eV
From Einstein's photoelectric equation,
$E = ϕ + Kinetic energy of electron \Rightarrow K E = E - ϕ \Rightarrow K E = \frac{h c}{λ} - ϕ, w here λ = wavelength of light h = P lan ck' s constant$
Maximum kinetic energy of electrons,
$E_{\max} = \frac{h c}{λ} - ϕ E_{\max} = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{350 \times 10^{- 9} \times 1.6 \times 10^{- 19}} - 1.9 E_{\max} = \frac{6.63 \times 3 \times 10^{2}}{350 \times 1.6} - 1.9 E_{\max} = 1.65 eV = 1.6 eV$

Page No 365:

Answer:

Given:
Work function of a metal, W₀ = 2.5 × 10⁻¹⁹ J
Frequency of light beam, v = 6.0 × 10¹⁴ Hz

(a) Work function of a metal,
   W₀ = hv₀,
   where h = Planck's constant
             v₀₌threshold frequency
$∴$ v₀= $\frac{W_{0}}{h}$
$\Rightarrow v_{0} = \frac{2.5 \times 10^{- 19}}{6.63 \times 10^{- 34}} = 3.77 \times 10^{14} Hz = 3.8 \times 10^{14} Hz$

(b) Einstein's photoelectric equation:
     $e V_{0} = h v - W_{0}$ ,
where v = frequency of light
          V₀₌Stopping potential
          e = charge on electron
$∴ V_{0} = \frac{h v - W_{0}}{e} = \frac{6.63 \times 10^{- 34} \times 6 \times 10^{14} - 2.5 \times 10^{- 19}}{1.6 \times 10^{- 19}} = \frac{3.97 \times 10^{- 19} - 2.5 \times 10^{- 19}}{1.6 \times 10^{- 19}} = 0.91 V$

Page No 365:

Answer:

Work function of a photoelectric material, ϕ = 4 eV = 4 × 1.6 × 10⁻¹⁹ J
Stopping potential, V₀ = 2.5 V
Planck's constant, h = 6.63 $\times 10^{- 34} Js$

(a) Work function of a photoelectric material,
   $ϕ = \frac{h c}{λ_{0}}$
    Here, λ₀ = threshold wavelength of light
                c = speed of light
$∴ λ_{0} = \frac{h c}{ϕ} λ_{0} = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{4 \times 1.6 \times 10^{- 19}} λ_{0} = \frac{6.63 \times 3}{64} \times \frac{10^{- 27}}{10^{- 9}} λ_{0} = 3.1 \times 10^{- 7} m λ_{0} = 310 nm$

(b) From Einstein's photoelectric equation,
   $E = ϕ + e V_{0} On substituting the respective values, we get: \frac{h c}{λ} = 4 \times 1.6 \times 10^{- 19} + 1.6 \times 10^{- 19} \times 2.5 \Rightarrow λ = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{6.5 \times 1.6 \times 10^{- 19}} \Rightarrow λ = \frac{6.63 \times 3 \times 10^{- 26}}{1.6 \times 10^{- 19} \times 6.5} \Rightarrow λ = 1.9125 \times 10^{- 7} = 191 nm$

Page No 365:

Answer:

Given:
Wavelength of light, $λ$ = 400 nm = 400 $\times$ 10^{$- 9$} m
Work function of metal, $ϕ$ = 2.5 eV
From Einstein's photoelectric equation,
$Kinetic energy = \frac{h c}{λ} - ϕ$
Here, c = speed of light
h = Planck's constant
$∴ K . E . = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{4 \times 10^{- 7} \times 1.6 \times 10^{- 19}} - 2.5 eV = 0.605 eV$
Also, K.E. = $\frac{p^{2}}{2 m}$ ,
where p is momentum and m is the mass of an electron.
$∴ p^{2} = 2 m \times K . E . \Rightarrow p^{2} = 2 \times 9.1 \times 10^{- 31} \times 0.605 \times 1.6 \times 10^{- 19} \Rightarrow p = 4.197 \times 10^{- 25} kg - m / s$

Page No 365:

Answer:

Given:
Wavelength of light, $λ$ = 400 nm = 400 $\times$ 10^{$- 9$} m
Stopping potential, V₀ = 1.1 V
From Einstein's photoelectric equation,
$\frac{h c}{λ} = \frac{h c}{λ_{0}} + e V_{0}$ ,
where h = Planck's constant
          c= speed of light
          $λ$ = wavelength of light
        $λ_{0}$ = threshold wavelength
        $V_{0}$ = stopping potential
On substituting the respective values in the above formula, we get:
$\frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{400 \times 10^{- 9}} = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{λ_{0}} + 1.6 \times 10^{- 19} \times 1.1 \Rightarrow 4.97 \times 10^{- 19} = \frac{19.89 \times 10^{- 26}}{λ_{0}} + 1.76 \times 10^{- 19} \Rightarrow 4.97 = \frac{19.89 \times 10^{- 7}}{λ_{0}} + 1.76 \Rightarrow \frac{19.89 \times 10^{- 7}}{λ_{0}} = 4.97 - 1.76 = 3.21 \Rightarrow λ_{0} = \frac{19.89 \times 10^{- 7}}{3.21} = 6.196 \times 10^{- 7} m = 620 nm$

Page No 365:

Answer:

(a)

When λ = 350, V_s = 1.45
and when λ = 400, V_s = 1
$∴ \frac{h c}{350} = w + 1.45 . . . (1) and \frac{h c}{400} = w + 1 . . . (2)$
Subtracting (2) from (1) and solving to get the value of h, we get:

h = 4.2 × 10⁻¹⁵ eV-s

(b) Now, work function,
$w = \frac{12240}{350} - 1.45 = 2.15 e v (c) w = \frac{n c}{λ} \Rightarrow λ_{threshold} = \frac{h c}{w} \Rightarrow λ_{threshold} = \frac{1240}{1.15} \Rightarrow λ_{threshold} = 576.8 nm$

Page No 365:

Answer:

Given:
Electric field of the monochromatic beam, E = 1.2 × 10¹⁵ times per second
Frequency, v = $\frac{1.2 \times 10^{15}}{2} = 0.6 \times 10^{15} Hz$
Work function of the metal surface, $ϕ$ = 2.0 eV
From Einstein's photoelectric equation, kinetic energy,
$K = h v - ϕ_{0} \Rightarrow K = \frac{6.63 \times 10^{- 34} \times 0.6 \times 10^{15}}{1.6 \times 10^{- 19}} - 2 = 0.486 eV$
Thus, the maximum kinetic energy of a photon is 0.486 eV.

Page No 365:

Answer:

Given:
$Electric field, E = E_{0} \sin [(1.57 \times 10^{7} m^{- 1}) (x - c t)]$
Work function, $ϕ$ = 1.9 eV
On comparing the given equation with the standard equation, $E = E_{0} \sin (k x - w t)$ , we get:
$ω = 1.57 \times 10^{7} \times c$
Now, frequency,
$v = \frac{1.57 \times 10^{7} \times 3 \times 10^{8}}{2 π} Hz$
From Einstein's photoelectric equation,
$e V_{0} = h v - ϕ$
Here, V₀ = stopping potential
           e = charge on electron
           h = Planck's constant
On substituting the respective values, we get:
$e V_{0} = 6.63 \times 10^{- 34} \times \frac{1.57 \times 3 \times 10^{15}}{2 π \times 1.6 \times 10^{- 19}} - 1.9 eV \Rightarrow e V_{0} = 3.105 - 1.9 = 1.205 eV \Rightarrow V_{0} = \frac{1.205 \times 1.6 \times 10^{- 19}}{1.6 \times 10^{- 19}} = 1.205 V$
Thus, the value of the stopping potential is 1.205 V.

Page No 366:

Answer:

Given:
$E = 100 \sin [(3 \times 10^{- 15} s^{- 1}) t] \sin [(6 \times 10^{- 15} s^{- 1}) t] = 100 \times \frac{1}{2} \cos [(9 \times 10^{15} s^{- 1}) t] - \cos [(3 \times 10^{15} s^{- 1}) t]$
The values of angular frequency $ω$ are 9 × 10¹⁵ and 3 × 10¹⁵ .
Work function of the metal surface, $ϕ$ = 2 eV
Maximum frequency,
$v = \frac{ω_{m a x}}{2 π}$ = $\frac{9 \times 10^{15}}{2 π}$ Hz
From Einstein's photoelectric equation, kinetic energy,
K = hv $-$ $ϕ$
$\Rightarrow$ K = $6.63 \times 10^{- 34} \times \frac{9 \times 10^{15}}{2 π} \times \frac{1}{1.6 \times 10^{- 19}}$ $-$ 2 eV
$\Rightarrow$ K = 3.938 eV

Page No 366:

Answer:

Given:
Intensity of light, I = 5 mW
Number of photons emitted per second, n = 8 × 10¹⁵
Stopping potential, V₀ = 2 V
Energy, E = $h v$ = $\frac{I}{n} = \frac{5 \times 10^{- 3}}{8 \times 10^{15}}$
From Einstein's photoelectric equation, work function,
$W_{0} = h v - e V_{0}$
Here, h = Planck's constant
e = 1.6 $\times$ 10^{$- 19$}C
On substituting the respective values, we get:
$W_{0} = \frac{5 \times 10^{- 3}}{8 \times 10^{15}} - 1.6 \times 10^{- 19} \times 2 = 6.25 \times 10^{- 19} - 3.2 \times 10^{- 19} = 3.05 \times 10^{- 19} = \frac{3.05 \times 10^{- 19}}{1.6 \times 10^{- 15}} = 1.906 eV$

Page No 366:

Answer:

We have to take two cases.
$C a s e (I) When stopping potential, V_{0} = 1.656 volts Frequency, v = 5 \times 10^{14} Hz C a s e (II) When stopping potential, V_{0} = 0 Frequency, v = 1 \times 10^{14} Hz (b) From Einstein's equation, e V_{0} = h v - W_{0} On substituting the values of case (1) and case (2), we get: 1.656 e = h \times 5 \times 10^{14} - W_{0} . . . (1) 0 = 5 \times h \times 1 \times 10^{14} - 5 \times W_{0} . . . (2) Subtracting equation (2) from (1), we get : W_{0} = \frac{1.656}{4} eV = 0.414 eV$
(a) Putting the value of W₀ in equation (2), we get:
$\Rightarrow 5 W_{0} = 5 h \times 10^{14} \Rightarrow 5 \times 0.414 = 5 \times h \times 10^{14} \Rightarrow h = 4.414 \times 10^{- 15} eVs Or \frac{h}{e} = 4.414 \times 10^{- 15} Vs$

Page No 366:

Answer:

Given:
Work function, W₀ = 0.6 eV
Now, work function,
$W_{0} = \frac{h c}{λ}$ ,
where, h = Planck's constant
$λ$ = wavelength of light
c = speed of light
$∴ λ = \frac{h c}{W_{0}} = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{0.6 \times 1.6 \times 10^{- 19}} = 20.71 \times 10^{- 7} m = 2071 nm$

Page No 366:

Answer:

Given:
Wavelength of light, λ = 400 nm
Power, P = 5 W
Energy of photon,
E = $\frac{h c}{λ} = (\frac{1242}{400}) eV$
Number of photons, n = $\frac{P}{E}$
$n = \frac{5 \times 400}{1.6 \times 10^{- 19} \times 1242}$

Number of electrons = 1 electron per 10⁶ photons
Number of photoelectrons emitted,
$n' = \frac{5 \times 400}{1.6 \times 1242 \times 10^{- 19} \times 10^{6}}$
Photo electric current,
I = Number of electron $\times$ Charge on electron
$I = \frac{5 \times 400}{1.6 \times 1242 \times 10^{- 19} \times 10^{6}} \times 1.6 \times 10^{- 19} = 1.6 \times 10^{- 6} A = 1.6 μA$

Page No 366:

Answer:

Given:
Radius of the silver ball, r = 4.8 cm
Wavelength of the ultra violet light, λ = 200 nm = 2 × 10⁻⁷ m
Total energy of light, E = 1.0 × 10⁻⁷ J
We are given that one photon out of every ten thousand is able to eject a photoelectron.
Energy of one photon,
$E' = \frac{h c}{λ}$ ,
where h = Planck's constant
    c = speed of light
$λ$ = wavelength of light
On substituting the respective values in the above formula, we get:
$E' = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{2 \times 10^{- 7}} = 9.945 \times 10^{- 19}$
Number of photons,
$n = \frac{E}{E'} = \frac{1 \times 10^{- 7}}{9.945 \times 10^{- 19}} = 1 \times 10^{11}$

Number of photoelectrons
$= \frac{1 \times 10^{11}}{10^{4}} = 1 \times 10^{7}$
The amount of positive charge developed due to the outgoing electrons,
$q = 1 \times 10^{7} \times 1.6 \times 10^{- 19} = 1.6 \times 10^{- 12} C$

Potential developed at the centre as well as on surface,
$V = \frac{K q}{r}$ ,
where K = $\frac{1}{4 {πε}_{0}}$
$∴ V = \frac{9 \times 10^{9} \times 1.6 \times 10^{- 12}}{4.8 \times 10^{- 2}} = 0.3 V$

Potential inside the silver ball remains constant. Therefore, potential at the centre of the sphere is 0.3 V.

Page No 366:

Answer:

Given:
Separation between the collector and emitter, d = 10 cm
Work function, ϕ = 2.39 eV
Wavelength range, λ₁ = 400 nm to λ₂ = 600 nm
Magnetic field B will be minimum if energy is maximum.
For maximum energy, wavelength λ should be minimum.
Einstein's photoelectric equation:
$E = \frac{h c}{λ} - ϕ H ere, h = Planck's constant λ = Wavelength of light c = speed of light ∴ E = \frac{1242}{400} - 2.39 = 3.105 - 2.39 = 0.715 eV$

The beam of ejected electrons will be bent by the magnetic field. If the electrons do not reach the other plates, there will be no current.
When a charged particle is sent perpendicular to a magnetic field, it moves along a circle of radius,
$r = \frac{m v}{q B}$ ,
where m = mass of charge particle
            B = magnetic field
            v = velocity of particle
            q = charge on the particle

Radius of the circle should be equal to r = d, so that no current flows in the circuit.
$\Rightarrow r = \sqrt{\frac{2 m E}{q B}} (∵ m v = \sqrt{2 m E}) \Rightarrow 0.1 = \sqrt{\frac{2 \times 9.1 \times 10^{- 31} \times 1.6 \times 10^{- 19} \times 0.715}{1.6 \times 10^{- 19} \times B}} \Rightarrow B = \sqrt{\frac{2 \times 9.1 \times 1.6 \times 0.715 \times 10^{- 50}}{1.6 \times 10^{- 20}}} \Rightarrow B = \sqrt{\frac{2 \times 9.1 \times 1.6 \times 0.715}{1.6}} \times 10^{- 5} \Rightarrow B = 2.85 \times 10^{- 5} T$

Page No 366:

Answer:

Given:
Fringe width, y = 1 mm $\times$ 2 = 2 mm
Work function, W₀ = 2.2 eV
D = 1.2 m
d = 0.24 mm
Fringe width,
$y = \frac{λ D}{d}$ ,
where $λ$ = wavelength of light
$∴ λ = \frac{2 \times 10^{- 3} \times 0.24 \times 10^{- 3}}{1.2} = 4 \times 10^{- 7} m$
$Energy, E = \frac{h c}{λ} = \frac{4.14 \times 10^{- 15} \times 3 \times 10^{8}}{4 \times 10^{- 7}} = 3.105 eV$
From Einstein's photoelectric equation,
$e V_{0} = E - W_{0}$ ,
where V₀ is the stopping potential and e is charge of electron.
$∴ e V_{0} = 3.105 - 2.2 = 0.905 eV V_{0} = \frac{0.905}{1.6 \times 10^{- 19}} \times 1.6 \times 10^{- 19} V = 0.905 V$

Page No 366:

Answer:

Given:
Work function of copper, ϕ = 4.5 eV,
Wavelength of monochromatic light, λ = 200 nm
From Einstein's photoelectric equation, kinetic energy,
$K = E - ϕ = \frac{h c}{λ} - ϕ, w here, h = Planck's constant c = speed of light ∴ K = \frac{1242}{200} - 4.5 = 6.21 - 4.5 = 1.71 eV$

Thus, at least 1.7 eV is required to stop the electron. Therefore, minimum kinetic energy will be 2 eV.
It is given that electric potential of 2 V is required to accelerate the electron. Therefore, maximum kinetic energy
$= (2 + 1.7) eV = 3.7 eV$

Page No 366:

Answer:

Given:
Charge density of the metal plate, $σ$ = 1.0 × 10⁻⁹ Cm⁻²
Work function of the cesium metal, φ = 1.9 eV
Wavelength of monochromatic light, $λ$ = 400 nm = 400 $\times$ 10^{$- 9$} m
Distance between the metal plates, d = 20 cm = 0.20 m
Electric potential due to a charged plate,
V = E × d,
where E, the electric field due to the charged plate, is $\frac{σ}{ε_{0}}$ and
d is the separation between the plates.
$∴ V = \frac{σ}{ε_{0}} \times d = \frac{1 \times 10^{- 9} \times 20}{8.85 \times 10^{- 12} \times 100} (∵ ε_{0} = 8.65 \times 10^{- 12} C^{2} N^{- 1} - m^{- 2}) = 22.598 V = 22.6 V From Einstein's photoelectric equation, e V_{o} = h v - W_{0} = \frac{h c}{λ} - W On substituting the respective values, we get: V_{0} = \frac{4.14 \times 10^{- 15} \times 3 \times 10^{8}}{4 \times 10^{- 7}} - 1.9 = 3.105 - 1.9 = 1.205 eV = 1.205 V$

As V₀ is much less than 'V', the minimum energy required to reach the charged plate must be equal to 22.7eV.
For maximum KE, 'V' must have an accelerating value.
Hence maximum kinetic energy,
$K . E . = V_{0} + V = 1.205 + 22.6 = 23.8005 eV$

Page No 366:

Answer:

Electric field of the metal plate,
$E = \frac{σ}{ε_{0}} = \frac{1 \times 10^{- 9}}{8.85 \times 10^{- 12}} = 113 V / m Acceleration, a = \frac{q E}{m}, w here q = charge on electron E = electric field m = mass of electron a = \frac{1.6 \times 10^{- 19} \times 113}{9.1 \times 10^{- 31}} = 19.87 \times 10^{12} t = \sqrt{\frac{2 y}{a}} = \sqrt{\frac{2 \times 20 \times 10^{- 2}}{19.87 \times 10^{12}}} = 1.41 \times 10^{- 7} s From Einstein' s photoelectric equation, K . E . = \frac{h c}{λ} - W = 1.2 eV = 1.2 \times 1.6 \times 10^{- 19} J [∵ in problem 31 : KE = 1.2 eV] ∴ Velocity, v = \sqrt{\frac{2 KE}{m}} = \sqrt{\frac{2 \times 1.2 \times 1.6 \times 10^{- 19}}{4.1 \times 10^{- 31}}} \sqrt{\frac{2 \times 1.2 \times 1.6 \times 10^{- 19}}{4.1 \times 10^{- 31}}} = 0.665 \times 10^{- 6} m / s$

∴ Horizontal displacement,
$S = v \times t S = 0.665 \times 10^{- 6} \times 1.4 \times 10^{- 7} S = 0.092 m = 9.2 cm$

Page No 366:

Answer:

Given:
Work function of the cesium plate, φ = 1.9 eV
Wavelength of radiation, λ = 250 nm
Energy of a photon,
$E = \frac{h c}{λ}, w here h = P lanck's constant c = speed of light ∴ E = \frac{1240}{250} = 4.96 eV$
From Einstein's photoelectric equation, kinetic energy of an electron,
$K = E - ϕ \Rightarrow K = \frac{h c}{λ} - ϕ [H ere, h is Planck's constant and c i s t h e speed of light] \Rightarrow K = 4.96 eV - 1.9 eV = 3.06 eV .$
For non-positive velocity of each photo electron, the velocity of a photoelectron should be equal to minimum velocity of the plate.
∴ Velocity of the photoelectron,
$v = \sqrt{\frac{2 K}{m}} (m = mass of electron) ∴ v = \sqrt{\frac{2 \times 3.06 \times 1.6 \times 10^{- 19}}{9.1 \times 10^{- 31}}} = 1.04 \times 10^{6} {ms}^{- 1}$

Page No 366:

Answer:

From Einstein's photoelectric equation,
$e V_{0} = \frac{h c}{λ} - ϕ \Rightarrow V_{0} = (\frac{h c}{λ} - ϕ) \frac{1}{e}$
Here, V₀ = stopping potential
            h = Planck's constant
            c = speed of light
            $ϕ$ = work function

The particle will move in a circle when the stopping potential is equal to the potential due to the singly charged ion at that point so that the particle gets the required centripetal force for its circular motion.
$\Rightarrow \frac{K e}{2 d} = (\frac{h c}{λ} - ϕ) \frac{1}{e} \Rightarrow \frac{K e^{2}}{2 d} = \frac{h c}{λ} - ϕ \Rightarrow \frac{h c}{λ} = \frac{{Ke}^{2}}{2 d} + ϕ = \frac{K e^{2} + 2 d ϕ}{2 d} \Rightarrow λ = \frac{(h c) (2 d)}{k e^{2} + 2 d ϕ} \Rightarrow λ = \frac{2 h d c}{\frac{1}{4 π \in_{0}} e^{2} + 2 d ϕ} \Rightarrow λ = \frac{8 π \in_{0} h c d}{e^{2} + 8 π \in_{0} d ϕ}$

Page No 366:

Answer:

Given:
Wavelength of light beam, $λ$ = 400 nm
Work function of metal plate, $ϕ$ = 2.2 eV
Energy of the photon,
$E = \frac{h c}{λ}, w here h = Planck's constant c = speed of light ∴ E = \frac{1240}{400} = 3.1 eV$

This energy will be supplied to the electrons.
Energy lost by the electron in the first collision
= 3.1 eV × 10%
= 0.31 eV

Now, the energy of the electron after the first collision = 3.1 $-$ 0.31 = 2.79 eV

Energy lost by electron in the second collision
= 2.79 eV× 10%
= 0.279 eV
Total energy lost by the electron in two collisions
= 0.31 + 0.279 = 0.589 eV
Using Einstein's photoelectric equation, kinetic energy of the photoelectron when it comes out from the metal,
$K = E - ϕ - energy lost due to collisions$
= (3.1 − 2.2 − 0.589) eV
= 0.31 eV

(b) Similarly for the third collision, the energy lost = (2.79 $-$ 0.279) eV × 10%
= 0.2511 eV
Energy of the electron after the third collision = 2.790 $-$ 0.2511 = 2.5389

Energy lost in the fourth collision = 2.5389 × 10%

Energy of the electron after the fourth collision = 2.5389 $-$ 0.25389 = 2.28501

This value is very close to the work function of the metal plate. After the fifth collision, the energy of the electron becomes less than the work function of the metal.

Therefore, the electron can suffer maximum four collisions before it becomes unable to come out of the metal.

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