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(D) According to Einstein's photoelectric equation,$K_{max} = \frac{hc}{\lambda} - \phi_0$,where $K_{max} = \frac{1}{2}mv^2$.Let $v_1$ and $v_2$ be th

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$4.0 	imes 10^{-19} \, J$

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(D) According to Einstein's photoelectric equation,$K_{max} = \frac{hc}{\lambda} - \phi_0$,where $K_{max} = \frac{1}{2}mv^2$.Let $v_1$ and $v_2$ be the maximum velocities for $\lambda_1$ and $\lambda_2$ respectively. Given $v_1 = 2v_2$,so $K_1 = 4K_2$.$E_1 = \frac{hc}{\lambda_1} = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{350 	imes 10^{-9}} \approx 5.68 	imes 10^{-19} \, J$.$E_2 = \frac{hc}{\lambda_2} = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{450 	imes 10^{-9}} \approx 4.42 	imes 10^{-19} \, J$.From the equations: $E_1 - \phi_0 = 4(E_2 - \phi_0)$.$E_1 - \phi_0 = 4E_2 - 4\phi_0$.$3\phi_0 = 4E_2 - E_1$.$3\phi_0 = 4(4.42 	imes 10^{-19}) - 5.68 	imes 10^{-19} = 17.68 	imes 10^{-19} - 5.68 	imes 10^{-19} = 12.0 	imes 10^{-19} \, J$.$\phi_0 = 4.0 	imes 10^{-19} \, J$.

If the surface of a metal is successively exposed to radiation of wavelengths $\lambda_1 = 350 \, nm$ and $\lambda_2 = 450 \, nm$,the maximum velocity of photoelectrons differs by a factor of $2$. The work function of this metal is:

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