Question

If the work function of a metal is \(1.2\,eV\) and the stopping potential is \(1.8\,V\), find the frequency of incident light on the metal surface.

• A. \(4.84\times10^{14}\,Hz\)
• B. \(7.25\times10^{14}\,Hz\)
• C. \(1.45\times10^{14}\,Hz\)
• D. \(9.67\times10^{14}\,Hz\)

Hint:

For photoelectric effect problems, remember Einstein's equation: \[ \boxed{hf=\phi+eV_s} \] If the work function is given in electron volts, \[ \boxed{\text{Photon Energy (eV)}=\phi+V_s.} \] Then convert the energy into joules and use \[ \boxed{f=\frac{E}{h}} \] to calculate the frequency.

Answer 1

The Correct Option is B

The photoelectric equation gives $hf = \phi + eV_s = 1.2 + 1.8 = 3.0$ eV. Converting to SI: $f = 3.0 \times 1.6 \times 10^{-19}/(6.63 \times 10^{-34}) \approx 7.24 \times 10^{14}$ Hz.