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(C) According to Einstein's photoelectric equation,the kinetic energy $E$ of an emitted photoelectron is given by $E = \frac{hc}{\lambda} - W_0$,where

Vedclass · Accepted Answer

$\frac{(E_1 \lambda_1 - E_2 \lambda_2)}{(\lambda_2 - \lambda_1)}$

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(C) According to Einstein's photoelectric equation,the kinetic energy $E$ of an emitted photoelectron is given by $E = \frac{hc}{\lambda} - W_0$,where $W_0$ is the work function.For wavelength $\lambda_1$,$E_1 = \frac{hc}{\lambda_1} - W_0 \implies E_1 \lambda_1 = hc - W_0 \lambda_1 \implies hc = E_1 \lambda_1 + W_0 \lambda_1$ ... $(i)$For wavelength $\lambda_2$,$E_2 = \frac{hc}{\lambda_2} - W_0 \implies E_2 \lambda_2 = hc - W_0 \lambda_2 \implies hc = E_2 \lambda_2 + W_0 \lambda_2$ ... (ii)Equating $(i)$ and (ii):$E_1 \lambda_1 + W_0 \lambda_1 = E_2 \lambda_2 + W_0 \lambda_2$$E_1 \lambda_1 - E_2 \lambda_2 = W_0 \lambda_2 - W_0 \lambda_1$$E_1 \lambda_1 - E_2 \lambda_2 = W_0 (\lambda_2 - \lambda_1)$$W_0 = \frac{E_1 \lambda_1 - E_2 \lambda_2}{\lambda_2 - \lambda_1}$

When a photosensitive surface is irradiated by light of wavelengths $\lambda_1$ and $\lambda_2$,the kinetic energies of the emitted photoelectrons are $E_1$ and $E_2$ respectively. The work function of the photosensitive surface is:

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