Question

A photo-electron is moving with a maximum velocity of ${10}^6$ m/s. [Given $e=1.6\times {10}^{-19}c$, and $m=9.1\times {10}^{-31}kg$], the stopping potential is

• A. 2.5 V
• B. 2.8 V
• C. 2.0 V
• D. 1.4 V

Answer 1

The correct option is B 2.8 V

$K_{max}=e{V}_0$
$e{V}_0=\frac{1}{2}\times 9.1\times {10}^{-31}\times ({10}^6{)}^2$
$V_0=2.8V$

So, the answer is option (B).