Dual-Nature of Radiation and Matter

Physics

NCERT

1   Find the$\\$ a) maximum frequency, and$\\$ b) Minimum wavelength of X-rays produced by $30 kV$ electrons.

Solution :

a) We know,$\\$ $K_{max}=hv-\Phi_0\\ \text{Or} eV_0=hv-\Phi_0\\ v_{max}=eV_0/h$$\\$ Substituting the required values$\\$ $v_{max}=(1.6*10^{-19}).(30000)/(6.63*10^{-34})\\ =7.24*10^{18}Hz$ (b)By the relation$\lambda =c/v_{max}$$\\$ $=0.041nm$

2   The work function of caesium metal is $2.14 eV$, When light of frequency $6 * 10^{14} Hz$ is incident on the metal surface, photoemission of electrons occurs.$\\$ a) What is the maximum kinetic energy of the emitted electrons,$\\$ b) Stopping potential, and$\\$ c) Maximum speed of the emitted photoelectrons?

Solution :

b) Stopping potential can be deduced by maximum kinetic energy, by the expression$\\$ $eV_0=K_{max}\\ \text{Or} V_0=K_{max}/e$$\\$ Putting the required values$\\$ $V_0=5.54*10^{-20}/1.6*10^{-19}=0.34V$$\\$ c) Maximum speed of photoelectron is given by the maximum kinetic energy it possesses,$\\$ $K_{max}=5.54*10^{-20}\\ Or mv^2/2=5.54*10^{-20}\\ Or v_{max}=344 km/s$

Given:$\\$ $\Phi_0=2.14V\\ v=6*10^{-14}Hz$$\\$ a) Maximum kinetic energy is the difference between the photon energy and the work function of the metal. For caesium it is given to be $2.14 eV$ Using the expression$\\$ $K_{max}=hv-\Phi_0$$\\$ Substitution yields $\\$ $K_{max}=[(6.63*10^{-34}).(6*10^{14})]-[(2.114).(1.6*10^{19})]\\ =5.54 * 10^{-20} J\\ =0.346eV$$\\$

3   The photoelectric cut-off voltage in a certain experiment is $1.5 V$. What is the maximum kinetic energy of photo electrons emitted?

Solution :

Maximum kinetic energy,$\\$ $K_{max }= ^eV_o$$\\$ Putting values and solving$\\$ $=2.4*10^{-19}J$

4   Monochromatic light of wavelength $632.8 nm$ is produced by a helium-neon laser. The power emitted is $9.42 m W.$$\\$ 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 :

Given:$\\$ Wavelength of monochromatic light$\\$ $\lambda =632.8 nm$$\\$ Power of He-be laser$\\$ $P=9.42 mW$$\\$ (a)Energy of a photon is given by$\\$ $E=hv\\ \text{Or} E=hc/\lambda$$\\$ Which gives$\\$ $E=3.14 * 10^{-19} J$$\\$ Now momentum of a photon$\\$ $p = h / \lambda \\ p = 1.05 *10^{-27} kg m / s$ $\\$ b) For a beam of uniform cross-section having cross-sectional area less than target area$\\$ $P=E*N$$\\$ where,$P=$power emitted$\\$ $E=$ energy of photon$\\$ $N=$ number of photons$\\$ Therefore$\\$ $N=P/E$$\\$ Substitution gives $\\$ $N=3*10^{16} \text{photons/second}$$\\$ (c) Momentum of He-Ne laser $\\$ $=1.05*10^{-27}kg m/s$$\\$ For this much momentum of a hydrogen atom $mv=1.05*10^{-27}\\ \text{Or} v=1.05*10^{-27}/1.6*10^{-27}\\ \text{Or} v=0.63m/s$$\\$ The required speed for hydrogen atom is $0.63 m / s.$

5   The energy flux of sunlight reaching the surface of the earth is $1.388 *10^3 W/m ^2 $. How many photons (nearly) per square meter are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of $ 550 nm$.

Solution :

Given:$\\$ Enrgy flux$=1388 W/sq.m$$\\$ wavelength $=550 nm$$\\$ Energy of photon $=hc/\lambda$$\\$ $= 3.61*10^{-19} J$$\\$ So no. of photons =$ P / E$$\\$ $= 4 * 10 ^{21}\text{ Photons/ $m ^2$ s}$