Answer to Question #154704 in Quantum Mechanics for dhanashree

Question

a)If the work function foe a metal is 1.85 ev, what would be the stopping potential for light having a wavelength of 410 nm? b) what would be the maximum speed of the emitted photoelectrons at the metals surface?

Answer 1

By Enstein's photoelectric equation:

$$KE_{max} = hc/\lambda-A$$

where $KE_{max}$ is the maximum kinetic energy of the photoelectrons, $h = 6.63\times 10^{-34}J\cdot s$ is Plank's constant, $c = 3\times 10^{8}m/s$ is the speed of light, $\lambda = 410nm = 4.1\times 10^{-7}m$ is the wavelenght of the light, and $A = 1.85eV = 1.6\times 10^{-19}J$ is the work function. On the other hand, the kinetic energy is:

$$KE_{max} = eV_s$$

where $e = 1.6\times 10^{-19}C$ is the elementary charge, and $V_s$ is the stopping potential. Thus, obtain:

$$eV_s= hc/\lambda-A\\ V_s = \dfrac{ hc/\lambda-A}{e}\\ V_s = \dfrac{ 6.63\times 10^{-34}\cdot 3\times 10^8/(4.1\times 10^{-7})-1.6\times 10^{-19}}{1.6\times 10^{-19}}\\ V_s \approx 2.03V$$

Since $KE_{max} = \dfrac{mv_{max}^2}{2}$ (assuming classical limit), the maximum speed is:

$$v_{max} = \sqrt{\dfrac{2\cdot KE_{max}}{m}} = \sqrt{\dfrac{2 eV_s}{m}}$$

where $m = 9.1\times 10^{-31}kg$ is the mass of the electron. Thus, obtain:

$$v_{max} = \sqrt{\dfrac{2 \cdot 1.6\times 10^{-19}\cdot 2.03}{9.1\times 10^{-31}}} \approx 7.16\times 10^5m/s$$

Answer. a) $2.03V$ , b) $7.16\times 10^5m/s$.