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(D) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = \Phi + K_{max}$,where $\Phi$ is the work function and $K_{max} = \frac{1}{2}

Vedclass · Accepted Answer

$\frac{hc}{3 \lambda_1 \lambda_2}(4 \lambda_2 - \lambda_1)$

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(D) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = \Phi + K_{max}$,where $\Phi$ is the work function and $K_{max} = \frac{1}{2}mv^2$.For wavelength $\lambda_1$ with velocity $V$: $\frac{hc}{\lambda_1} = \Phi + \frac{1}{2}mV^2$  --- $(1)$For wavelength $\lambda_2$ with velocity $2V$: $\frac{hc}{\lambda_2} = \Phi + \frac{1}{2}m(2V)^2 = \Phi + 2mV^2$  --- $(2)$From $(1)$,$\frac{1}{2}mV^2 = \frac{hc}{\lambda_1} - \Phi$. Multiplying by $4$,we get $2mV^2 = \frac{4hc}{\lambda_1} - 4\Phi$.Substitute this into $(2)$: $\frac{hc}{\lambda_2} = \Phi + \frac{4hc}{\lambda_1} - 4\Phi$.$\frac{hc}{\lambda_2} = \frac{4hc}{\lambda_1} - 3\Phi$.$3\Phi = \frac{4hc}{\lambda_1} - \frac{hc}{\lambda_2} = hc \left( \frac{4\lambda_2 - \lambda_1}{\lambda_1 \lambda_2} \right)$.$\Phi = \frac{hc}{3 \lambda_1 \lambda_2} (4 \lambda_2 - \lambda_1)$.

When a metal surface is illuminated by light of wavelength $\lambda_1$ and $\lambda_2$,the maximum velocities of photoelectrons ejected are $V$ and $2V$ respectively. The work function of the metal is ($h=$ Planck's constant,$c=$ velocity of light,$\lambda_1 > \lambda_2$).

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