11.1 Find the (a) maximum frequency, and (b) minimum wavelength of X-rays produced by 30 kV electrons.
Solution :
Here, $$V = 30 kV = 30 \times 10^{3}volts$$
(a) Maximum frequency of X-rays produced = maximum energy of an accelerated electron.
$$E = q.V$$
For electron
$$E = eV$$
$$eV = h f_{max}$$
$$f_{max} = \frac{eV} {h}$$
$$f_{max} = \frac{1.6 \times 10^{-19} \times 30 \times 10^{3}} {6.62 \times 10^{-34}}$$
$$f_{max} = 7.24 \times 10^{18}Hz$$
(b) Minimum wavelength of X-rays produced is
$$\lambda_{min}= \frac{c}{f_{max}}$$
$$\lambda_{min}= \frac{3 \times 10^{8}}{7.24 \times 10^{18}}$$
$$\lambda_{min}= 0.414 \times 10^{-10}m$$
$$\lambda_{min}= 0.414 A^{o}$$
11.2 The work function of caesium metal is 2.14 eV. When Light of frequency $6 \times 10^{14}Hz$ 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?
Solution:
$$f = 6 \times 10^{14}Hz$$ ,
$$W_{0} = 2.14 eV$$
(a) Maximum kinetic energy of the emitted electrons is
$$E = W_{0} + K.E_{max}$$
$$K.E_{max} = hf - W_{0}$$
$$K.E_{max}= 6.62 \times 10^{-34} \times 6 \times 10^{14} J - 2.14 eV$$
$$K.E_{max}= \frac{6.62 \times 10^{-34} \times 6 \times 10^{14}}{1.6 \times 10^{-19}}eV - 2.14 eV$$
$$K.E_{max}=2.489-2.14eV$$
$$K.E_{max}= 0.34eV$$
(b) If $V_{0}$ is the stopping potential, then, $$eV_{0}=K.E_{max}$$
Or
$$eV_{0} = 0.34 eV$$
$$V_{0} = 0.34V$$
(c) maximum speed of the emitted photoelectrons :
$$\frac{1}{2}mv^{2}_{max} = 0.34 eV$$
$$\frac{1}{2}mv^{2}_{max} = 0.34 \times 1.6 \times 10^{-19}J$$
$$v^{2}_{max} = \frac{2 \times 0.34 \times 1.6 \times 10^{-19}J}{m}$$
$$\therefore v_{max} = {\sqrt{\frac{2 \times 0.34 \times 1.6 \times 10^{-19} }{9.1 \times 10^{-31}}}}$$
$$v_{max}=347 \times 10^{3}m/s$$
$$v_{max}=347 km/s$$
11.3 The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
Solution :
Here, stopping potential, $V_{0} = 1.5 V$
Maximum kinetic energy of emitted photoelectrons is
$$K.E_{max} = eV_{0}$$
$$K.E_{max}= e \times 1.5 V$$
$$K.E_{max}= 1.5 eV$$
Or
$$K.E_{max}= 1.5 \times 1.6 \times 10^{-19}J$$
$$K.E_{max}= 2.4 \times 10^{-19}J$$
11.4 Monochromatic .Light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 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?
Solution :
Here,
$$\lambda = 632.8 nm = 632.8 \times 10^{-9}m$$
$$P = 9.42 mW = 9.42 \times 10^{-3}W$$
(a) Energy of each photon in the light beam is
$$E = \frac{hc}{\lambda}$$
$$E = \frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{632.8 \times 10^{-9}}$$
$$E = 3.14 \times 10^{-19}J$$
Momentum of each photon in the light beam is
$$p = \frac{h}{\lambda} = \frac{6.62 \times 10^{-34}}{632.8 \times 10^{-9}}$$
$$p= 1.05 \times 10^{-27}kgm/s$$
(b) Let Number of photons falling per second on the target is n
Energy of each photon = $3.14 \times 10^{-19}J$
Energy of n photon = $n \times 3.14 \times 10^{-19}J$
$$Power = \frac{Total Energy}{time}$$
$$Power = \frac{nE}{1}$$
$$P = nE$$
$$n = \frac{P}{E} = \frac{9.42 \times 10^{-3}}{3.14 \times 10^{-19}}$$
$$n = 3 \times 10^{16} photons/sec$$
(c) momentum of a hydrogen atom is equal to the momentum of a photon. Therefore, required speed (v) of the hydrogen atom is
$$Momentum= mass \times velocity $$
$$v = \frac{momentum}{m_{H}}$$
$$v= \frac{1.05 \times 10^{-27}}{1.66 \times 10^{-27}}$$
$$v = 0.63 m/s$$
11.5 In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be $4.12 \times 10^{-15} Vs$. Calculate the value of Planck’s constant.
Solution :
$$Slope \ of \ V_{0}-f \ graph = 4.12 \times 10^{-15}Vs$$
Now,
$$slope \ of \ V_{0}-f \ graph = \frac{h} {e}$$
$$ h= e \times \ slope \ of \ V_{0}-f graph$$
$$h = 1.6 \times 10^{-19} \times 4.12 \times 10^{-15}$$
$$h = 6.592 \times 10^{-34}Js$$
11.6 The threshold frequency for a certain metal is $3.3 \times 10^{14}Hz$. If .light of frequency $8.2 \times 10^{14}Hz$ is incident on the metal, predict the cut-off voltage for the photoelectric emission.
Solution :
Here
$$f_{0} = 3.3 \times 10^{14} Hz$$
$$f = 8.2 \times 10^{14} Hz$$
We know :
$$E = W_{0} + K.E_{max}$$
$$K.E_{max} = hf - W_{0}$$
$$eV_{0} = hf - W_{0}$$
$$eV_{0} = hf - hf_{0}$$
$$eV_{0} = h(f-f_{0})$$
$$V_{0} = \frac{h(f-f_{0}}{e}$$
$$V_{0} = \frac{6.22 \times 10^{-34} \times (8.2 \times 10^{14}-3.3 \times 10^{14} )}{1.6 \times 10^{-19}}$$
$$V_{0} = \frac{6.22 \times 10^{-34} \times (8.2-3.3) \times 10^{14}}{1.6 \times 10^{-19}}$$
$$V_{0} = \frac{6.22 \times 10^{-34} \times 4.9 \times 10^{14}}{1.6 \times 10^{-19}}$$
$$V_{0} = 2.03 V$$
11.7 The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?
Solution :
$$W_{0} = 4.2 eV$$
$$\lambda = 330nm = 330\times 10^{-9}m$$
Energy of photon of incident light is
$$E = \frac{hc}{\lambda}$$
$$E= \frac{6.62 \times 10^{-34} \times 3 \times10^{8} }{330 \times 10^{-9}}$$
$$E=6.018 \times 10^{-19}J$$
$$E=\frac{6.018\times 10^{-34} }{1.6 \times 10^{-19}}eV$$
$$E= 3.76eV$$
$$E < W_{0}$$
Since, the energy of the incident radiation is less than the work function of the metal. Thus, there will be no photo electric emission.
11.8 .Light of frequency $7.21 \times 10^{14}Hz$ is incident on a metal surface.Electrons with a maximum speed of $6.0 \times 10^{5}m/s$ are ejected from the surface. What is the threshold frequency for photoemission of electrons?
Solution :
$$f = 7.21 \times 10^{14}Hz$$
$$v_{max} = 6.0 \times 10^{5}m/s$$
$$f_{0} =?$$
Now
$$E =W_{0}+ K.E_{max}$$
$$hf = hf_{o} +K.E_{max}$$
$$hf = hf_{0} + \frac{1}{2}mv_{max}^{2}$$
$$f_{0} = f - \frac{mv_{max}^{2}}{2h}$$
$$f_{0} = 7.21 \times 10^{14} - \frac{9.1 \times 10^{-31} \times (6.0 \times 10^{5})^{2}}{2 \times 6.62 \times 10^{-34}}$$
$$f_{0} = 7.21 \times 10^{14} - 2.47 \times 10^{14}$$
$$f_{0} = 4.74 \times 10^{14}Hz$$
11.9 .Light of wavelength 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 0.38 V. Find the work function of the material from which the emitter is made.
Solution :
Here, $$\lambda = 488 nm = 488 \times 10^{-9}m$$
$$V_{0} = 0.38 V$$
$$W_{0} =?$$
Now,
$$E =W_{0}+ K.E_{max}$$
$$\frac{hc}{\lambda} = W_{0} +K.E_{max}$$
$$\because \ K.E_{max} = eV_{0}$$
$$\frac{hc}{\lambda} = W_{0} + eV_{0}$$
$$W_{0} = \frac{hc}{\lambda} - eV_{0}$$
$$W_{0} = \frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{488 \times 10^{-9}} - 1.6 \times 10^{-19} \times 0.38$$
$$W_{0}= 4.076 \times 10^{-19} - 0.608 \times 10^{-19}$$
$$W_{0}= 3.468 \times 10^{-19}J$$
$$W_{0}= \frac{3.468 \times 10^{-19}}{1.6 \times 10^{-19}}eV$$
$$W_{0}= 2.17eV$$
11.10 What is the de Broglie wavelength of
(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,
(b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
(c) a dust particle of mass 1.0 × 10–9 kg drifting with a speed of 2.2 m/s ?
Solution :
de-Broglie wavelength,
$$\lambda = \frac{h}{mv}$$
(a) For bullet, here
$$m = 0.040 kg$$
$$v = 1.0 km/s$$
$$v= 10^{3} m/s$$
$$\therefore \lambda = \frac{6.62 \times 10^{-34}}{0.040 \times 10^{3}}$$
$$\lambda = 1.7 \times 10^{-35}m$$
(b) For ball, here
$$m = 0.060 kg$$
$$v = 1.0 m/s$$
$$\therefore \lambda = \frac{6.62 \times 10^{-34}}{0.060 \times 1}$$
$$\lambda = 1.1 \times 10^{-32}m$$
(c) For dust particles, here
$$m = 1.0 \times 10^{-9} kg$$
$$v = 2.2 m/s$$
$$\therefore \lambda = \frac{6.62 \times 10^{-34}}{1.0 \times 10^{-9} \times 2.2}$$
$$\lambda = 3.0 \times 10^{-25}m$$
11.11 Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
Solution :
Consider an electromagnetic radiation of wavelength $\lambda$ and frequency f the momentum of its Photon is given by :
$$p = \frac{E}{c}$$
$$p = \frac{hf}{c}$$
$$p = \frac{h}{\lambda}$$
$$\lambda = \frac{h}{p}$$
De-Broglie wavelength of photons
$$\lambda = \frac{h}{mv} = \frac{h}{p}$$
The wavelength of electromagnetic radiation is equal to the de Broglie wavelength associated with it's quantum Photon.