Question:
The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength $491 \mathrm{~nm}$ is $0.710 \mathrm{~V}$. When the incident wavelength is changed to a new value, the stopping potential is $1.43 \mathrm{~V}$. The new wavelength is :
Correct Option: , 3
Solution:
$\frac{\mathrm{hc}}{\lambda}=\phi+\mathrm{eV}_{\mathrm{s}}$
$\frac{1240}{491}=\phi+0.71$......(1)
$\frac{1240}{\lambda}=\phi+1.43$........(2)
$\therefore \lambda=382 \mathrm{~nm}$ Ans.
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