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?

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:$\\$ $P = 100 W$$\\ \lambda = 589 nm$$\\$ Energy per photon $= hc / \lambda $$\\ = 3.37 * 10 ^{- 19} J$$\\$ b) No. of photons per = P/E$\\$ $=3 *10 ^{20}$ Photons/ second

8   The threshold frequency for a certain metal is $3.3 × 10^{14} Hz$. If light of frequency $8.2 × 10^ 14 Hz$ is incident on the metal, predict the cutoff-voltage for the photoelectric emission.

Solution :

Given:$\\$ $v_ 0 = 3.3 * 10^{14 }Hz\\ v = 8.2 * 10 ^{14} Hz$$\\ Cut=off voltage \\ eV_ 0 = h ( v - v_ 0 )$$\\$ Substituting the values $V_ 0 = 2.03 eV$

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