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(A) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by $eV_s = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{hc}

Vedclass · Accepted Answer

$4\lambda$

Vedclass · Answer

(A) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by $eV_s = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{hc}{\lambda_0}$ is the work function and $\lambda_0$ is the threshold wavelength.For the first case: $3eV_0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$  --- $(1)$For the second case: $eV_0 = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}$  --- $(2)$Dividing equation $(1)$ by equation $(2)$:$3 = \frac{\frac{hc}{\lambda} - \frac{hc}{\lambda_0}}{\frac{hc}{2\lambda} - \frac{hc}{\lambda_0}}$$3 \left( \frac{1}{2\lambda} - \frac{1}{\lambda_0} \right) = \frac{1}{\lambda} - \frac{1}{\lambda_0}$$\frac{3}{2\lambda} - \frac{3}{\lambda_0} = \frac{1}{\lambda} - \frac{1}{\lambda_0}$$\frac{3}{2\lambda} - \frac{1}{\lambda} = \frac{3}{\lambda_0} - \frac{1}{\lambda_0}$$\frac{1}{2\lambda} = \frac{2}{\lambda_0}$$\lambda_0 = 4\lambda$.

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