For a photo cell stopping potential is found to be 9V. The maximum speed (× 10⁵m/s)of the electron is equal to _____

Given Data :

photo cell stopping potential is found to be $9v$

Formula Used :

Energy given $=$ Work function + Kinetic Energy

Detailed Solution :

We know from photoelectric equation, Energy given $=$ Work function + Kinetic Energy
The stopping potential must be equal to kinetic energy of the particle given, $v_0 =9eV$ (stopping potentia)
So, $\frac{1}{2}m{v}_{\text{max }}^2=9eV$

$$\begin{gathered} V=\sqrt{\frac{2\times 9\times 1.6\times {10}^{-19}}{9.11\times {10}^{-31}}} \\ =\sqrt{\frac{18\times 1.6\times {10}^{12}}{9.11}} \\ =18\times {10}^5 \mathrm{m}/\mathrm{s}. \end{gathered}$$

Hence, the correct answer is 18