MCQ
What is the stopping potential when the metal with work function $0.6 eV$ is illuminated with the light of $2 eV$ ............ $V$
- A$2.6 $
- B$3.6 $
- C$0.8 $
- ✓$1.4 $
✓
Answer
Correct option: D.
$1.4 $
d
(d) ${V_0} = \frac{{(E - {W_0})}}{e} = \frac{{(2eV - 0.6\,eV)}}{e} = 1.4V$
(d) ${V_0} = \frac{{(E - {W_0})}}{e} = \frac{{(2eV - 0.6\,eV)}}{e} = 1.4V$
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