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(D) According to Einstein's photoelectric equation,the stopping potential $V_0$ is given by:$eV_0 = \frac{hc}{\lambda} - \phi_0 = \frac{hc}{\lambda} -

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$3 \lambda$

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(D) According to Einstein's photoelectric equation,the stopping potential $V_0$ is given by:$eV_0 = \frac{hc}{\lambda} - \phi_0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$  --- $(1)$When the wavelength is changed to $2\lambda$,the stopping potential becomes $\frac{V_0}{4}$:$e\left(\frac{V_0}{4}\right) = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}$  --- $(2)$From equation $(1)$,we have $eV_0 = hc(\frac{1}{\lambda} - \frac{1}{\lambda_0})$. Substituting this into equation $(2)$:$\frac{hc}{4}(\frac{1}{\lambda} - \frac{1}{\lambda_0}) = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}$Dividing both sides by $hc$:$\frac{1}{4\lambda} - \frac{1}{4\lambda_0} = \frac{1}{2\lambda} - \frac{1}{\lambda_0}$Rearranging the terms to solve for $\lambda_0$:$\frac{1}{\lambda_0} - \frac{1}{4\lambda_0} = \frac{1}{2\lambda} - \frac{1}{4\lambda}$$\frac{3}{4\lambda_0} = \frac{1}{4\lambda}$Therefore,$\lambda_0 = 3\lambda$.

$A$ metallic surface is illuminated with radiation of wavelength $\lambda$,the stopping potential is $V_0$. If the same surface is illuminated with radiation of wavelength $2 \lambda$,the stopping potential becomes $\frac{V_0}{4}$. The threshold wavelength for this metallic surface will be -

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