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

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$4\lambda$

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(B) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by: $eV_s = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function of the metal.For wavelength $\lambda$,$e(4.8) = \frac{hc}{\lambda} - \phi$  --- $(1)$For wavelength $2\lambda$,$e(1.6) = \frac{hc}{2\lambda} - \phi$ --- $(2)$Subtracting equation $(2)$ from $(1)$:$e(4.8 - 1.6) = \frac{hc}{\lambda} - \frac{hc}{2\lambda}$$3.2e = \frac{hc}{2\lambda}$ --- $(3)$From $(1)$,$\phi = \frac{hc}{\lambda} - 4.8e$. Since $\frac{hc}{\lambda} = 2 	imes 3.2e = 6.4e$ (from eq $3$),$\phi = 6.4e - 4.8e = 1.6e$.The threshold wavelength $\lambda_0$ is given by $\phi = \frac{hc}{\lambda_0}$.Since $\phi = 1.6e$ and $\frac{hc}{\lambda} = 6.4e$,we have $\frac{hc}{\lambda_0} = \frac{1}{4} 	imes \frac{hc}{\lambda}$.Therefore,$\lambda_0 = 4\lambda$.

When radiation of wavelength $\lambda$ is incident on a metallic surface,the stopping potential is $4.8 \ V$. If the same surface is irradiated with radiation of wavelength $2\lambda$,the stopping potential is $1.6 \ V$. The threshold wavelength for the metal is:

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