When light of wavelength 'lambda' is incident | vedclass

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(D) According to Einstein's photoelectric equation: $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$, where $\lambda_0$ is the threshold wavelength.Fo

Vedclass · Accepted Answer

$\frac{9}{5}\lambda$

Vedclass · Answer

(D) According to Einstein's photoelectric equation: $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$, where $\lambda_0$ is the threshold wavelength.For the first case: $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$  --- $(1)$For the second case: $e(\frac{V}{4}) = \frac{hc}{1.5\lambda} - \frac{hc}{\lambda_0}$  --- $(2)$Multiply equation $(2)$ by $4$: $eV = \frac{4hc}{1.5\lambda} - \frac{4hc}{\lambda_0} = \frac{8hc}{3\lambda} - \frac{4hc}{\lambda_0}$ --- $(3)$Equating $(1)$ and $(3)$: $\frac{hc}{\lambda} - \frac{hc}{\lambda_0} = \frac{8hc}{3\lambda} - \frac{4hc}{\lambda_0}$$\frac{4hc}{\lambda_0} - \frac{hc}{\lambda_0} = \frac{8hc}{3\lambda} - \frac{hc}{\lambda}$$\frac{3hc}{\lambda_0} = \frac{5hc}{3\lambda}$$\frac{3}{\lambda_0} = \frac{5}{3\lambda}$$\lambda_0 = \frac{9}{5}\lambda$.

When light of wavelength '$\lambda$' is incident on a photosensitive surface, the stopping potential is '$V$'. When a light of wavelength $1.5\lambda$ is incident on the same surface, the stopping potential is '$\frac{V}{4}$'. Threshold wavelength for the surface is

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