Question

If maximum velocity with which an electron can be emitted from a photo cell is $4\times 10^{8}cm/\sec$, the stopping potential is (mass of electron $=9\times 10^{-31}kg$ )

• A. 30 volt
• B. 45 volt
• C. 59 volt
• D. Information is insufficient

Answer 1

Answer: (B)

$v_{\max }=4\times 10^{8}cm/\sec =4\times 10^{6}m/\sec$
$\therefore K_{\max }=\frac{1}{2}m{v}_{\max }^2$
$=\frac{1}{2}\times 9\times 10^{-31}\times {\left(4\times 10^6\right)}^2$
$=7.2\times 10^{-18}J=45eV$
Hence, stopping potential
$\left|V_0\right|=\frac{K_{\max }}{e}$
$=\frac{45eV}{e}=45$ volt .