Question

De-Broglie wavelength of photoelectrons is $1A$ what is stopping potential?

• A. 151V
• B. 15.3V
• C. 12.3V
• D. 13.6V

Answer 1

Answer: (A)

Stopping potential $(V_0)$ is the minimum negative (retarding) potential given to the anode for which the photoelectric current becomes zero.
Photoelectric current becomes zero when the stopping potential equals the maximum kinetic energy $\left(K_{\max }\right)$ the photoelectrons.
That is, $K_{\max }=e{V}_0\dots$ (i)
Now, deBroglie wavelength of electron is given by
${\lambda }_e=\frac{h}{\sqrt{2m{K}_{\max }}}\dots$ (ii)
Equating Eqs. (i) and (ii), we arrive at
${\lambda }_e=\frac{h}{\sqrt{2me{V}_0}}$
$\Rightarrow V_0=\frac{h^2}{2me{\lambda }_e^2}?$ (ii)
Given, ${\lambda }_e=1A^o=1\times 10^{-10}m$ ,
$h=6.6\times 10^{-34}J-s$.
$m=9.1\times 10^{-31}kg$,
$e=1.6\times 10^{-19}C$
$\therefore V_0=\frac{{\left(6.6\times 10^{-34}\right)}^2}{2\times \left(9.1\times 10^{-31}\right)\times \left(1.6\times 10^{-19}\right)}\times {\left(1\times 10^{-10}\right)}^2=151$ volt.
Note: For a given frequency of the incident radiation, the stopping potential is independent of its intensity.