Chapter 27 - Quantum Physics - Learning Path Questions and Exercises - Exercises - Page 936: 22

(a) speed $v=6.02356\times10^{5}m/s$ (b) speed $v=0$,elecrron will come out with zero speed. (c) electrons will not come out of the surface.

From equation 27.8 we know that Cutoff frequency $f_{0}= \frac{\phi_{0}}{h}$ so work function $\phi_{0}= hf_{0}=\frac{hc}{\lambda_{0}}$ If the wavelength of incident light is $\lambda$ and frequency is $f$ the energy of incident light will be $E=hf=\frac{hc}{\lambda}$ From equation 27.7 $hf= K_{max} + \phi_{0}$ $K_{max}=hf- \phi_{0}=\frac{hc}{\lambda}-\frac{hc}{\lambda_{0}}$........equation(1) Also maximum kinetic energy of photoelectron is given by $K_{max}=\frac{1}{2}mv^{2}$ so $v=\sqrt \frac{2K_{max}}{m}$.................equation(2) (a) light of wavelength $\lambda= 300nm=300\times10^{-9}m=3\times10^{-7}m$ $h=6.63\times10^{-34} J.s$, $c=3\times10^{8}m/s$, $\lambda_{0}= 400nm=400\times10^{-9}m=4\times10^{-7}m$ putting values in equation (1) $K_{max}=\frac{hc}{\lambda}-\frac{hc}{\lambda_{0}}=\frac{6.63\times10^{-34} J.s\times3\times10^{8}m/s}{3\times10^{-7}m}-\frac{6.63\times10^{-34} J.s\times3\times10^{8}m/s}{4\times10^{-7}m}$ $K_{max}=6.63\times10^{-19} J-4.9725\times10^{-19} J=1.6575\times10^{-19} J$ mass of electron $m=9.1\times10^{-31}kg $ putting this value in equation(2) $v=\sqrt \frac{2K_{max}}{m}=\sqrt \frac{2\times1.6575\times10^{-19} J}{9.1\times10^{-31}kg }=6.02356\times10^{5}m/s$ (b) light of wavelength $\lambda= 400nm=400\times10^{-9}m=3\times10^{-7}m$ $h=6.63\times10^{-34} J.s$, $c=3\times10^{8}m/s$, $\lambda_{0}= 400nm=400\times10^{-9}m=4\times10^{-7}m$ putting values in equation (1) $K_{max}=\frac{hc}{\lambda}-\frac{hc}{\lambda_{0}}=\frac{6.63\times10^{-34} J.s\times3\times10^{8}m/s}{4\times10^{-7}m}-\frac{6.63\times10^{-34} J.s\times3\times10^{8}m/s}{4\times10^{-7}m}$ $K_{max}=0$, so speed $v=\sqrt \frac{2K_{max}}{m}=0$ (c) light of wavelength $\lambda= 500nm=500\times10^{-9}m=5\times10^{-7}m$ $h=6.63\times10^{-34} J.s$, $c=3\times10^{8}m/s$, $\lambda_{0}= 400nm=400\times10^{-9}m=4\times10^{-7}m$ putting values in equation (1) $K_{max}=\frac{hc}{\lambda}-\frac{hc}{\lambda_{0}}=\frac{6.63\times10^{-34} J.s\times3\times10^{8}m/s}{5\times10^{-7}m}-\frac{6.63\times10^{-34} J.s\times3\times10^{8}m/s}{4\times10^{-7}m}$ $K_{max}=3.978\times10^{-19} J-4.9725\times10^{-19} J=-0.9945\times10^{-19} J$ since $K_{max}$ is negative quantity electrons will not come out of the surface.