Question

Maximum velocity of photo electron emitted is $4.8m{s}^{-1}$ . If $\left[\frac{e}{m}\right]$ ratio of electron is $1.76\times {10}^{11}\left(Ck{g}^{-1}\right)$ then stopping potential is given by

• A. 5×10^-10V
• B. 3×10^-7V
• C. 6.5×10^-11V
• D. 2.5×10²V

Answer 1

The correct option is C

6.5×10^-11V

$\frac{1}{2}m{v}^2=e{V}_0{V}_0=\frac{v^2}{2\left(\frac{e}{m}\right)}=\frac{4.8\times 4.8}{2\times 1.76\times {10}^{1}1}=6.54\times {10}^{-11}V$