SolveItNCERT · CBSE Boards

NCERT Solutions · Class 12 Physics Dual Nature of Radiation and Matter

9 exercises · 2 still being checked

Exercises 11.1–11.9

  1. Exercise 11.1

    Find the
    (a)
    maximum frequency, and
    (b)
    minimum wavelength of X-rays produced by $\displaystyle 30$ kV electrons.
    NCERT’s answer
    (a)
    7.$\displaystyle 24$ × \(\displaystyle 10^{18}\) Hz (b) $\displaystyle 0.041$ nm
    Each electron accelerated through potential \(\displaystyle V\) gains kinetic energy \(\displaystyle eV\). In X-ray production (inverse photoelectric effect), the maximum photon energy — corresponding to minimum wavelength — occurs when all this kinetic energy converts to a single photon: \(\displaystyle eV = h\nu_{max} = \dfrac{hc}{\lambda_{min}}\).
    Given \(\displaystyle V = 30\ \mathrm{kV} = 3\times10^{4}\ \mathrm{V}\):
    \[eV = (1.6\times10^{-19}\ \mathrm{C})(3\times10^{4}\ \mathrm{V}) = 4.8\times10^{-15}\ \mathrm{J}\]
    (a)
    Maximum frequency:
    \[\nu_{max} = \frac{eV}{h} = \frac{4.8\times10^{-15}}{6.63\times10^{-34}} \approx 7.24\times10^{18}\ \mathrm{Hz}\]
    (b)
    Minimum wavelength:
    \[\lambda_{min} = \frac{hc}{eV} = \frac{(6.63\times10^{-34})(3\times10^{8})}{4.8\times10^{-15}} \approx 4.14\times10^{-11}\ \mathrm{m}\]
    Maximum frequency \(\displaystyle \approx 7.24\times10^{18}\ \mathrm{Hz}\); minimum wavelength \(\displaystyle \approx 4.14\times10^{-11}\ \mathrm{m} = 41.4\ \mathrm{pm}\).

  2. Exercise 11.2

    The work function of caesium metal is $\displaystyle 2.14$ eV. When light of frequency $\displaystyle 6$ ×\(\displaystyle 10^{14}\)Hz is incident on the metal surface, photoemission of electrons occurs. What is the
    (a)
    maximum kinetic energy of the emitted electrons,
    (b)
    Stopping potential, and
    (c)
    maximum speed of the emitted photoelectrons?
    NCERT’s answer
    (a)
    0.$\displaystyle 34$ eV = $\displaystyle 0.54$ × \(\displaystyle 10^{-19}\)J (b) $\displaystyle 0.34$ V (c) $\displaystyle 344$ km/s
    Einstein's photoelectric equation: \(\displaystyle K_{max} = h\nu - \phi_0\), stopping potential \(\displaystyle K_{max} = eV_0\), and \(\displaystyle K_{max} = \tfrac{1}{2}m_e v_{max}^2\).
    Photon energy:
    \[h\nu = (6.63\times10^{-34})(6\times10^{14}) = 3.978\times10^{-19}\ \mathrm{J} = 2.486\ \mathrm{eV}\]
    (a)
    Maximum kinetic energy:
    \[K_{max} = 2.486\ \mathrm{eV} - 2.14\ \mathrm{eV} \approx 0.346\ \mathrm{eV} = 5.54\times10^{-20}\ \mathrm{J}\]
    (b)
    Stopping potential:
    \[V_0 = \frac{K_{max}}{e} \approx 0.35\ \mathrm{V}\]
    (c)
    Maximum speed:
    \[v_{max} = \sqrt{\frac{2K_{max}}{m_e}} = \sqrt{\frac{2(5.54\times10^{-20})}{9.11\times10^{-31}}} \approx 3.49\times10^{5}\ \mathrm{m/s}\]
    \(\displaystyle K_{max}\approx 0.35\ \mathrm{eV}\); \(\displaystyle V_0\approx 0.35\ \mathrm{V}\); \(\displaystyle v_{max}\approx 3.49\times10^{5}\ \mathrm{m/s}\).

  3. Exercise 11.3

    The photoelectric cut-off voltage in a certain experiment is $\displaystyle 1.5$ V. What is the maximum kinetic energy of photoelectrons emitted?
    NCERT’s answer
    1.$\displaystyle 5$ eV = $\displaystyle 2.4$ × \(\displaystyle 10^{-19}\) J
    The stopping (cut-off) voltage \(\displaystyle V_0\) relates to the maximum kinetic energy of photoelectrons by \(\displaystyle K_{max} = eV_0\), since the stopping potential is defined as the retarding potential that just brings the fastest photoelectrons to rest.\[K_{max} = eV_0 = (1.6\times10^{-19}\ \mathrm{C})(1.5\ \mathrm{V}) = 2.4\times10^{-19}\ \mathrm{J}\]\(\displaystyle K_{max} = 1.5\ \mathrm{eV} = 2.4\times10^{-19}\ \mathrm{J}\).

  4. Exercise 11.4

    Monochromatic light of wavelength $\displaystyle 632.8$ nm is produced by a helium-neon laser. The power emitted is $\displaystyle 9.42$ mW.
    (a)
    Find the energy and momentum of each photon in the light beam,
    (b)
    How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and
    (c)
    How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
    NCERT’s answer
    (a)
    3.$\displaystyle 14$ × \(\displaystyle 10^{-19}\)J, $\displaystyle 1.05$ × \(\displaystyle 10^{-27}\) kg m/s (b) $\displaystyle 3$ × \(\displaystyle 10^{16}\) photons/s (c) $\displaystyle 0.63$ m/s
    Each photon carries energy \(\displaystyle E = h\nu = \dfrac{hc}{\lambda}\) and momentum \(\displaystyle p = \dfrac{h}{\lambda} = \dfrac{E}{c}\). The photon flux follows from total power divided by energy per photon, \(\displaystyle n = P/E\). For the hydrogen atom, matching momentum gives \(\displaystyle v = p/m_p\).
    (a)
    With \(\displaystyle \lambda = 632.8\ \mathrm{nm} = 6.328\times10^{-7}\ \mathrm{m}\):
    \[E = \frac{hc}{\lambda} = \frac{(6.63\times10^{-34})(3\times10^{8})}{6.328\times10^{-7}} \approx 3.14\times10^{-19}\ \mathrm{J}\]
    \[p = \frac{h}{\lambda} = \frac{6.63\times10^{-34}}{6.328\times10^{-7}} \approx 1.05\times10^{-27}\ \mathrm{kg\,m/s}\]
    (b)
    Photons per second, with \(\displaystyle P = 9.42\ \mathrm{mW} = 9.42\times10^{-3}\ \mathrm{W}\):
    \[n = \frac{P}{E} = \frac{9.42\times10^{-3}}{3.14\times10^{-19}} \approx 3.0\times10^{16}\ \mathrm{photons/s}\]
    (c)
    Treating the hydrogen atom's mass as the proton mass \(\displaystyle m_p\), equal momentum requires:
    \[v = \frac{p}{m_p} = \frac{1.05\times10^{-27}}{1.67\times10^{-27}} \approx 0.63\ \mathrm{m/s}\]
    (a) \(\displaystyle E\approx3.14\times10^{-19}\ \mathrm{J}\), \(\displaystyle p\approx1.05\times10^{-27}\ \mathrm{kg\,m/s}\); (b) \(\displaystyle \approx3.0\times10^{16}\) photons/s; (c) \(\displaystyle v\approx0.63\ \mathrm{m/s}\).

  5. Exercise 11.5

    In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be $\displaystyle 4.12$ × \(\displaystyle 10^{-15}\) V s. Calculate the value of Planck’s constant.
    NCERT’s answer
    6.$\displaystyle 59$ × \(\displaystyle 10^{-34}\) J s
    NCERT_Solution_Class12_Physics_Ch11_Q11-5Einstein's equation \(\displaystyle eV_0 = h\nu - \phi_0\) rearranges to \(\displaystyle V_0 = \dfrac{h}{e}\nu - \dfrac{\phi_0}{e}\), a straight line in \(\displaystyle V_0\) vs \(\displaystyle \nu\) with slope \(\displaystyle \dfrac{h}{e}\).Given slope \(\displaystyle = 4.12\times10^{-15}\ \mathrm{V\,s}\): \[h = (\text{slope})\times e = (4.12\times10^{-15})(1.6\times10^{-19}) \approx 6.59\times10^{-34}\ \mathrm{J\,s}\]\(\displaystyle h \approx 6.59\times10^{-34}\ \mathrm{J\,s}\), close to the standard value \(\displaystyle 6.63\times10^{-34}\ \mathrm{J\,s}\).

  6. Exercise 11.6

    The threshold frequency for a certain metal is $\displaystyle 3.3$ × \(\displaystyle 10^{14}\) Hz. If light of frequency $\displaystyle 8.2$ × \(\displaystyle 10^{14}\) Hz is incident on the metal, predict the cut- off voltage for the photoelectric emission.
    NCERT’s answer
    2.$\displaystyle 0$ V
    The work function corresponds to the threshold frequency, \(\displaystyle \phi_0 = h\nu_0\), and Einstein's equation gives \(\displaystyle eV_0 = h(\nu - \nu_0)\).With \(\displaystyle \nu_0 = 3.3\times10^{14}\ \mathrm{Hz}\) and \(\displaystyle \nu = 8.2\times10^{14}\ \mathrm{Hz}\): \[eV_0 = h(\nu-\nu_0) = (6.63\times10^{-34})(8.2\times10^{14} - 3.3\times10^{14}) = (6.63\times10^{-34})(4.9\times10^{14})\] \[eV_0 \approx 3.249\times10^{-19}\ \mathrm{J}\] \[V_0 = \frac{3.249\times10^{-19}}{1.6\times10^{-19}} \approx 2.03\ \mathrm{V}\]Cut-off voltage \(\displaystyle V_0 \approx 2.03\ \mathrm{V}\).

  7. Exercise 11.7

    The work function for a certain metal is $\displaystyle 4.2$ eV. Will this metal give hotoelectric emission for incident radiation of wavelength $\displaystyle 330$ nm?

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    NCERT’s answer
    No, because ν < \(\displaystyle ν_{o}\)
    Photoelectric emission occurs only if the incident photon energy \(\displaystyle E = \dfrac{hc}{\lambda}\) meets or exceeds the work function \(\displaystyle \phi_0\).For \(\displaystyle \lambda = 330\ \mathrm{nm} = 3.30\times10^{-7}\ \mathrm{m}\): \[E = \frac{hc}{\lambda} = \frac{(6.63\times10^{-34})(3\times10^{8})}{3.30\times10^{-7}} \approx 6.03\times10^{-19}\ \mathrm{J}\] \[E \approx \frac{6.03\times10^{-19}}{1.6\times10^{-19}} \approx 3.77\ \mathrm{eV}\]Since \(\displaystyle E (3.77\ \mathrm{eV}) < \phi_0 (4.2\ \mathrm{eV})\), the photon energy is insufficient.No — light of wavelength $\displaystyle 330$ nm cannot cause photoelectric emission from this metal, since its photon energy (\(\displaystyle \approx3.77\ \mathrm{eV}\)) is less than the work function ($\displaystyle 4.2$ eV).

  8. Exercise 11.8

    Light of frequency $\displaystyle 7.21$ × \(\displaystyle 10^{14}\) Hz is incident on a metal surface. Electrons with a maximum speed of $\displaystyle 6.0$ × \(\displaystyle 10^{5}\) m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
    NCERT’s answer
    4.$\displaystyle 73$ × \(\displaystyle 10^{14}\) Hz
    Einstein's photoelectric equation: \(\displaystyle h\nu = \phi_0 + K_{max} = h\nu_0 + \tfrac{1}{2}m_e v_{max}^2\), so \(\displaystyle \nu_0 = \nu - \dfrac{\tfrac{1}{2}m_e v_{max}^2}{h}\).Maximum kinetic energy: \[K_{max} = \tfrac{1}{2}m_e v_{max}^2 = \tfrac{1}{2}(9.11\times10^{-31})(6.0\times10^{5})^2 \approx 1.64\times10^{-19}\ \mathrm{J}\]Photon energy: \[h\nu = (6.63\times10^{-34})(7.21\times10^{14}) \approx 4.78\times10^{-19}\ \mathrm{J}\]Threshold energy: \[h\nu_0 = h\nu - K_{max} = 4.78\times10^{-19} - 1.64\times10^{-19} \approx 3.14\times10^{-19}\ \mathrm{J}\] \[\nu_0 = \frac{3.14\times10^{-19}}{6.63\times10^{-34}} \approx 4.74\times10^{14}\ \mathrm{Hz}\]Threshold frequency \(\displaystyle \nu_0 \approx 4.74\times10^{14}\ \mathrm{Hz}\).

  9. Exercise 11.9

    Light of wavelength $\displaystyle 488$ nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is $\displaystyle 0.38$ V. Find the work function of the material from which the emitter is made. $\displaystyle 11.10$ What is the de Broglie wavelength of
    (a)
    a bullet of mass $\displaystyle 0.040$ kg travelling at the speed of $\displaystyle 1.0$ km/s,
    (b)
    a ball of mass $\displaystyle 0.060$ kg moving at a speed of $\displaystyle 1.0$ m/s, and
    (c)
    a dust particle of mass $\displaystyle 1.0$ × \(\displaystyle 10^{-9}\) kg drifting with a speed of $\displaystyle 2.2$ m/s ? $\displaystyle 11.11$ Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

    Disagrees with the book

    This working does not reach the answer printed in NCERT. One of the two is wrong and it has not yet been settled which — check it before you rely on it.

    NCERT’s answer
    2.$\displaystyle 16$ eV = $\displaystyle 3.46$ × \(\displaystyle 10^{-19}\)J $\displaystyle 11.10$ (a) $\displaystyle 1.7$ × \(\displaystyle 10^{-35}\) m (b) $\displaystyle 1.1$ × \(\displaystyle 10^{-32}\) m (c) $\displaystyle 3.0$ × \(\displaystyle 10^{-23}\)m $\displaystyle 11.11$ λ = h/p = h/(hν/c) = c/ν
    Einstein's equation: \(\displaystyle \phi_0 = h\nu - eV_0 = \dfrac{hc}{\lambda} - eV_0\).Photon energy for \(\displaystyle \lambda = 488\ \mathrm{nm} = 4.88\times10^{-7}\ \mathrm{m}\): \[E = \frac{hc}{\lambda} = \frac{(6.63\times10^{-34})(3\times10^{8})}{4.88\times10^{-7}} \approx 4.08\times10^{-19}\ \mathrm{J} \approx 2.55\ \mathrm{eV}\]Work function: \[\phi_0 = E - eV_0 = 2.55\ \mathrm{eV} - 0.38\ \mathrm{eV} \approx 2.17\ \mathrm{eV}\]Work function \(\displaystyle \phi_0 \approx 2.17\ \mathrm{eV}\ (\approx 3.47\times10^{-19}\ \mathrm{J})\).