The photoelectric threshold wavelength for silver is $\lambda_0$. The energy of the electron ejected from the surface of silver by an incident wavelength $\lambda$ will be...

Question

$hc(\lambda_0 - \lambda)$

$\frac{hc}{\lambda_0 - \lambda}$

$\frac{h}{c}\left(\frac{\lambda_0 - \lambda}{\lambda_0 \lambda}\right)$

$hc\left(\frac{\lambda_0 - \lambda}{\lambda_0 \lambda}\right)$

Solution

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