Question

Protons of energy $6eV$ are indicated on a potassium surface of work function $2.1eV$. What is the stopping potential?

• A. $-6V$
• B. $-2.1V$
• C. $-3.9V$
• D. $-8.1V$

Answer 1

The correct option is C $-3.9V$

According to photoelectric effect,
$hv=h{v}_0+KE$

$\frac{1}{2}m{u}^2=E_k^{max}=hv-W=(6-2.1)eV$

$=3.9eV$

and stopping potential is the potential requires to stop this particle.

So $V_0=3.9eV$.

Stopping potential $=-3.9eV$.