Answer to Question #164204 in Quantum Mechanics for Shivang Gupta

Question

Electrons are emitted with zero velocity from a certain metal surface when it is exposed to radiations of lamda=6800A. Calculate threshold frequency and work function of metal.

Answer 1

Let's use the Einstein equation for the photoelectric effect

$h\nu=A+E_k$

Where

$A-$ Electron work function

$E_k-$ Kinetic energy of an electron

By the condition of the problem

$E_k=0$

Then we write

$A=h\nu=\frac{hc}{\lambda}=\frac{6.63 \cdot 10^{-34}J \cdot s \cdot 3 \cdot 10^{8} \cdot m/s}{6.8 \cdot 10^{-7} m}=2.925 \cdot 10^{-19}J$

$\nu_{min}=\frac{A}{h}=\frac{2.925 \cdot 10^{-19}}{6.63 \cdot 10^{-34}}=4.41 \cdot 10^{14} \cdot \frac {1}{s}$