Chapter 27 - Quantum Physics - Learning Path Questions and Exercises - Exercises - Page 936: 25

(a) stopping potential $V_{0} =0.625V$ (b) work function $\phi_{0}=2.3348eV$ (c) Energy of red light $E=1.77589eV$ is lower than the work function $\phi_{0}=2.3348eV$ of metal. So electron will not come out of the surface so there is no question of stopping potential.

wavelength of the incident blue light $\lambda= 420nm=420\times10^{-9}m=4.2\times10^{-7}m$ $E=hf=\frac{hc}{\lambda}$ putting $h=6.63\times10^{-34} J.s$, $c=3\times10^{8}m/s$,$\lambda=4.2\times10^{-7}m$ $E=hf=\frac{6.63\times10^{-34} J.s\times3\times10^{8}m/s}{4.2\times10^{-7}m}=4.7357\times10^{-19} J$ $1.6\times10^{-19} J$ is equal to $1eV$ so $4.7357\times10^{-19} J$ is equal to $\frac{4.7357\times10^{-19}}{1.6\times10^{-19}}eV=2.9598eV$ $E=hf=4.7357\times10^{-19} J=2.9598eV$ .....equation(1) (a) Calculation of stopping voltage given that Maximum kinetic energy of electron is $K_{max}=1.0\times10^{-19}J$ $1.6\times10^{-19} J$ is equal to $1eV$ so $1.0\times10^{-19} J$ is equal to $\frac{1.0\times10^{-19}}{1.6\times10^{-19}}eV=0.625eV$.......equation(2) from equation 27.5 $K_{max}=eV_{0}$ so stopping potential $V_{0} =\frac{K_{max}}{e}=\frac{0.625eV}{e}=0.625V$ (b) Calculation of work function From equation 27.7 $hf= K_{max} + \phi_{0}$ $\phi_{0}=hf- K_{max}$ putting the values from equation (1) and (2) $\phi_{0}=2.9598eV -0.625eV$ $\phi_{0}=2.3348eV$ (c) If red light is used instead. wavelength of the incident Red light $\lambda= 700nm=700\times10^{-9}m=7\times10^{-7}m$ $E=hf=\frac{hc}{\lambda}$ putting $h=6.63\times10^{-34} J.s$, $c=3\times10^{8}m/s$,$\lambda=7\times10^{-7}m$ $E=hf=\frac{6.63\times10^{-34} J.s\times3\times10^{8}m/s}{7\times10^{-7}m}=2.84143\times10^{-19} J$ $1.6\times10^{-19} J$ is equal to $1eV$ so $2.84143\times10^{-19} J$ is equal to $\frac{2.84143\times10^{-19}}{1.6\times10^{-19}}eV=1.77589eV$ $E=hf=2.84143\times10^{-19} J=1.77589eV$ Energy of red light $E=1.77589eV$ is lower than the work function $\phi_{0}=2.3348eV$ of metal. So electron will not come out of the surface so there is no question of stopping potential. From equation 27.7 $hf= K_{max} + \phi_{0}$ $K_{max}=hf-\phi_{0}$ $K_{max}=1.77589eV-2.3348eV$=$-0.5589eV$ since kinetic energy is negative so there is no question of stopping potential.