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Question

(C) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K_{\max} = eV = \frac{hc}{\lambda

Vedclass · Accepted Answer

$4\lambda$

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(C) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K_{\max} = eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$,where $\lambda_0$ is the threshold wavelength.For the first case: $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$  --- $(i)$For the second case: $e(\frac{V}{3}) = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}$ --- (ii)Dividing equation $(i)$ by equation (ii):$\frac{eV}{eV/3} = \frac{\frac{hc}{\lambda} - \frac{hc}{\lambda_0}}{\frac{hc}{2\lambda} - \frac{hc}{\lambda_0}}$$3 = \frac{\frac{1}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{2\lambda} - \frac{1}{\lambda_0}}$$3(\frac{1}{2\lambda} - \frac{1}{\lambda_0}) = \frac{1}{\lambda} - \frac{1}{\lambda_0}$$\frac{3}{2\lambda} - \frac{3}{\lambda_0} = \frac{1}{\lambda} - \frac{1}{\lambda_0}$$\frac{3}{2\lambda} - \frac{1}{\lambda} = \frac{3}{\lambda_0} - \frac{1}{\lambda_0}$$\frac{1}{2\lambda} = \frac{2}{\lambda_0}$$\lambda_0 = 4\lambda$.

When a certain metal surface is illuminated with light of wavelength $\lambda$,the stopping potential is $V$. When the same surface is illuminated by light of wavelength $2\lambda$,the stopping potential is $\frac{V}{3}$. The threshold wavelength for the surface is:

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