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Question

When a piece of metal is illuminated by monochromatic light of wavelength $ \lambda $, the stopping potential is $3V_s$. When the same surface is illuminated by light of wavelength $2\lambda$, the stopping potential becomes $V_s$. The value of the threshold wavelength for photoelectric emission will be

$4\lambda$

$8\lambda$

$\frac{4}{3}\lambda$

$6\lambda$

Solution

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