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(C) According to Einstein's photoelectric equation,the energy of the incident photon is equal to the sum of the work function $(W_0)$ and the maximum

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 energy of the incident photon is equal to the sum of the work function $(W_0)$ and the maximum kinetic energy $(K_{max})$ of the emitted photoelectrons: $E = W_0 + K_{max}$.For wavelength $\lambda_1$: $\frac{hc}{\lambda_1} = W_0 + E_1$  --- $(1)$For wavelength $\lambda_2$: $\frac{hc}{\lambda_2} = W_0 + E_2$  --- $(2)$From $(1)$,$hc = W_0 \lambda_1 + E_1 \lambda_1$. From $(2)$,$hc = W_0 \lambda_2 + E_2 \lambda_2$.Equating the two expressions for $hc$: $W_0 \lambda_1 + E_1 \lambda_1 = W_0 \lambda_2 + E_2 \lambda_2$.Rearranging to solve for $W_0$: $W_0 (\lambda_1 - \lambda_2) = E_2 \lambda_2 - E_1 \lambda_1$.$W_0 = \frac{E_2 \lambda_2 - E_1 \lambda_1}{\lambda_1 - \lambda_2} = \frac{E_1 \lambda_1 - E_2 \lambda_2}{\lambda_2 - \lambda_1}$.

If light of wavelength $\lambda_1$ is allowed to fall on a metal,then the kinetic energy of the photoelectrons emitted is $E_1$. If the wavelength of light changes to $\lambda_2$,then the kinetic energy of the electrons changes to $E_2$. Then the work function of the metal is:

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