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NEET-XII-Physics

42: Photoelectric Effect and Wave Particle Duality

with Solutions - page 3

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  • Qstn #10
    The frequency and intensity of a light source are doubled. Consider the following statements.
    (A) The saturation photocurrent remains almost the same.
    (B) The maximum kinetic energy of the photoelectrons is doubled.
    (a) A and B are true.
    (b) A is true but B is false.
    (c) A is false but B is true.
    (d) A and B are false.
    digAnsr:   b
    Ans : (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}_{max}=hv-\phi ``
    Where Kmax = kinetic energy of electron
    v = frequency of light
    `` \phi `` = 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.
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  • Qstn #11
    A point source of light is used in a photoelectric effect. If the source is removed farther from the emitting metal, the stopping potential
    (a) will increase
    (b) will decrease
    (c) will remain constant
    (d) will either increase or decrease
    digAnsr:   c
    Ans : (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.
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  • Qstn #12
    A point source causes photoelectric effect from a small metal plate. Which of the following curves may represent the saturation photocurrent as a function of the distance between the source and the metal?
    Figure
    Ans : 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.
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  • Qstn #13
    A non-monochromatic light is used in an experiment on photoelectric effect. The stopping potential
    (a) is related to the mean wavelength
    (b) is related to the longest wavelength
    (c) is related to the shortest wavelength
    (d) is not related to the wavelength
    digAnsr:   c
    Ans : (c) is related to the shortest wavelength
    For photoelectric effect to be observed, wavelength of the incident light `` \left(\lambda \right)`` should be less than the threshold wavelength `` \left({\lambda }_{0}\right)`` of the metal.
    Einstein's photoelectric equation:
    `` e{V}_{0}=\frac{hc}{{\lambda }_{0}}-\phi ``
    Here, V0 = stopping potential
    `` {\lambda }_{0}`` = threshold wavelength
    h = Planck's constant
    `` \phi `` = work-function of metal
    It is clear from the above equation that stopping potential is related to the shortest wavelength (threshold wavelength).
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  • Qstn #14
    A proton and an electron are accelerated by the same potential difference. Let λe and λp denote the de Broglie wavelengths of the electron and the proton, respectively.
    (a) λe = λp
    (b) λe < λp
    (c) λe > λp
    (d) The relation between λe and λp depends on the accelerating potential difference.
    digAnsr:   c
    Ans : (c) λe > λp
    Let me and mp be the masses of electron and proton, respectively.
    Let the applied potential difference be V.
    Thus, the de-Broglie wavelength of the electron,
    `` {\lambda }_{e}=\frac{h}{\sqrt{2{m}_{\,\mathrm{\,e\,}}eV}}...\left(1\right)``
    And de-Broglie wavelength of the proton,
    `` {\lambda }_{p}=\frac{h}{\sqrt{2{m}_{\,\mathrm{\,p\,}}eV}}...\left(2\right)``
    Dividing equation (2) by equation (1), we get:
    `` \frac{{\lambda }_{\,\mathrm{\,p\,}}}{{\lambda }_{\,\mathrm{\,e\,}}}=\frac{\sqrt{{m}_{\,\mathrm{\,e\,}}}}{\sqrt{{m}_{\,\mathrm{\,p\,}}}}``
    me < mp
    `` \therefore `` `` \frac{{\lambda }_{p}}{{\lambda }_{e}}<1``
    `` \Rightarrow {\lambda }_{p}<{\lambda }_{e}``
    Page No 364:
  • #
    Section : iii
  • Qstn #1
    When the intensity of a light source in increased,
    (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
    (c) more energetic photons are emitted
    (d) faster photons are emitted
    digAnsr:   a,b,c,d
    Ans : (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.
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  • Qstn #2
    Photoelectric effect supports quantum nature of light because
    (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
    (d) electric charge of the photoelectrons is quantised
    digAnsr:   a,b,c
    Ans : (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}-\phi ``
    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.
    Page No 364:
  • Qstn #3
    A photon of energy hv is absorbed by a free electron of a metal with work-function hv - φ.
    (a) The electron is sure to come out.
    (b) The electron is sure to come out with kinetic energy hv - φ.
    (c) Either the electron does not come out or it comes our with kinetic energy hv - φ.
    (d) It may come out with kinetic energy less than hv - φ.
    digAnsr:   d
    Ans : (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:
  • Qstn #4
    If the wavelength of light in an experiment on photoelectric effect is doubled,
    (a) photoelectric emission will not take place
    (b) photoelectric emission may or may not take place
    (c) the stopping potential will increase
    (d) the stopping potential will decrease
    digAnsr:   b,d,a
    Ans : (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 `` \left({\lambda }_{0}\right)``. 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{hc}{{\lambda }_{0}}-\phi ``,
    where V0 = stopping potential
    `` {\lambda }_{0}`` = threshold wavelength
    h = Planck's constant
    `` \phi `` = work-function of metal
    It is clear that
    `` {V}_{0}\propto \frac{1}{{\lambda }_{0}}``
    Thus, if the wavelength of light in an experiment on photoelectric effect is doubled, its stopping potential will become half.
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  • Qstn #5
    The photo current in an experiment on photoelectric effect increases if
    (a) the intensity of the source is increased
    (b) the exposure time is increased
    (c) the intensity of the source is decreased
    (d) the exposure time is decreased
    digAnsr:   a
    Ans : (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:
  • Qstn #6
    The collector plate in an experiment on photoelectric effect is kept vertically above the emitter plate. A light source is put on and a saturation photocurrent is recorded. An electric field is switched on that has a vertically downward direction.
    (a) The photocurrent will increase.
    (b) The kinetic energy of the electrons will increase.
    (c) The stopping potential will decrease.
    (d) The threshold wavelength will increase.
    digAnsr:   b,a,c,d
    Ans : (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.
    Page No 364:
  • Qstn #7
    In which of the following situations, the heavier of the two particles has smaller de Broglie wavelength? The two particles
    (a) move with the same speed
    (b) move with the same linear momentum
    (c) move with the same kinetic energy
    (d) have fallen through the same height
    digAnsr:   a,c,d,b
    Ans : (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,
    `` {\lambda }_{1}=\frac{h}{{m}_{1}v}`` ...(1)
    de Broglie wavelength of the lighter particle,
    `` {\lambda }_{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 `` {\lambda }_{1}<{\lambda }_{2}``.
    Hence, option
    (a) is correct.
    If they are moving with the same linear momentum, then using the de Broglie relation
    `` \lambda =\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,
    `` {\lambda }_{1}=\frac{h}{\sqrt{2{m}_{1}K}}``
    de Broglie wavelength of the lighter particle,
    `` {\lambda }_{2}=\frac{h}{\sqrt{2{m}_{2}K}}``
    It is clear from the above equation that if `` {m}_{1}>{m}_{2}``, then `` {\lambda }_{1}<{\lambda }_{2}``.
    Hence, option
    (c) is correct.
    When they have fallen through the same height h, then velocity of both the bodies,
    v = `` \sqrt{2gh}``
    Now,
    `` {\lambda }_{1}=\frac{h}{{m}_{1}\sqrt{2gh}}``
    `` {\lambda }_{2}=\frac{h}{{m}_{2}\sqrt{2gh}}``
    m1>m2
    `` \therefore `` `` {\lambda }_{1}<{\lambda }_{2}``
    Hence, option
    (d) is correct.
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  • #
    Section : iv
  • Qstn #1
    Visible light has wavelengths in the range of 400 nm to 780 nm. Calculate the range of energy of the photons of visible light.
    Ans : Given:
    Range of wavelengths, `` {\lambda }_{1}`` = 400 nm to `` {\lambda }_{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=hv``
    `` \nu =\frac{c}{\lambda }``
    `` \therefore E=h\nu =\frac{hc}{\lambda }``
    Energy `` \left({E}_{1}\right)`` of a photon of wavelength `` \left({\lambda }_{1}\right)``:
    `` ``
    `` {E}_{1}=\frac{hc}{{\lambda }_{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}\,\mathrm{\,J\,}``
    `` ``
    Energy (E2) of a photon of wavelength (`` {\lambda }_{2}``):
    `` ``
    `` {E}_{2}=\frac{6.63\times 3}{7.8}\times {10}^{-19}``
    `` =2.55\times {10}^{-9}\,\mathrm{\,J\,}``
    So, the range of energy is 2.55 × 10-19 J to 5 × 10-19 J.
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