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(D) Given: $\lambda_1 = 300 \ nm$,$\lambda_2 = 500 \ nm$,and the ratio of maximum velocities $\frac{v_1}{v_2} = 3$.Using Einstein's photoelectric equa

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$2.28$

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(D) Given: $\lambda_1 = 300 \ nm$,$\lambda_2 = 500 \ nm$,and the ratio of maximum velocities $\frac{v_1}{v_2} = 3$.Using Einstein's photoelectric equation: $E = \phi_0 + K_{max}$,where $K_{max} = \frac{1}{2}mv^2$.Thus,$\frac{1}{2}mv^2 = \frac{hc}{\lambda} - \phi_0$.Taking the ratio for the two cases:$\left(\frac{v_1}{v_2}\right)^2 = \frac{E_1 - \phi_0}{E_2 - \phi_0} = \frac{\frac{1240}{\lambda_1} - \phi_0}{\frac{1240}{\lambda_2} - \phi_0}$.Substituting the values: $3^2 = \frac{\frac{1240}{300} - \phi_0}{\frac{1240}{500} - \phi_0}$.$9 = \frac{4.133 - \phi_0}{2.48 - \phi_0}$.$9(2.48 - \phi_0) = 4.133 - \phi_0$.$22.32 - 9\phi_0 = 4.133 - \phi_0$.$8\phi_0 = 18.187$.$\phi_0 \approx 2.273 \ eV \approx 2.28 \ eV$.

The surface of a metal is first illuminated with a light of wavelength $300 \ nm$ and later illuminated by another light of wavelength $500 \ nm$. It is observed that the ratio of maximum velocities of photoelectrons in the two cases is $3$. The work function of the metal is close to: (in $eV$)

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