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Q.
De-Broglie wavelength of photoelectrons is $1 \,A$ what is stopping potential?
Delhi UMET/DPMTDelhi UMET/DPMT 2006
Solution:
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, $K_{\max }=e V_{0} \ldots$ (i)
Now, deBroglie wavelength of electron is given by
$\lambda_{e}=\frac{h}{\sqrt{2 m K_{\max }}} \ldots$ (ii)
Equating Eqs. (i) and (ii), we arrive at
$\lambda_{e}=\frac{h}{\sqrt{2 m e V_{0}}} $
$\Rightarrow V_{0}=\frac{h^{2}}{2 m e \lambda_{e}^{2}} ?$ (ii)
Given, $\lambda_{e}=1 A^{o}=1 \times 10^{-10} m$ ,
$h=6.6 \times 10^{-34} J -s$.
$m=9.1 \times 10^{-31} kg$,
$e=1.6 \times 10^{-19} C$
$ \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.