Question

The electric field associated with a light wave is $E=E_{0}sin[1.57\times {10}^7(x-ct\left)\right]$ where x is in meters & t in seconds. If this is incident on a metal surface of work function 1.1 eV, the stopping potential is approximately

• A. 1.2 V
• B. 1.5 V
• C. 1.75 V
• D. 2 V

Answer 1

The correct option is D

2 V

$\frac{2\pi }{\lambda }=1.57\times {10}^7=\frac{\pi }{2}\times {10}^7$ $\therefore \lambda =400nm$

$E_{ph}=\frac{1242}{400}ev=3.105eV$

$K_{max}=E_{ph}-W=3.105-1.1\cong 2eV$

$\therefore$ stopping potential=2V