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(B) The Einstein's photoelectric equation is given by $eV_0 = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{hc}{\lambda_0}$ is the work function and

Vedclass · Accepted Answer

$4 \lambda$

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(B) The Einstein's photoelectric equation is given by $eV_0 = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{hc}{\lambda_0}$ is the work function and $\lambda_0$ is the threshold wavelength.Thus,$V_0 = \frac{hc}{e} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)$.For the first case: $4.8 = \frac{hc}{e} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) \quad \dots (i)$For the second case: $1.6 = \frac{hc}{e} \left( \frac{1}{2\lambda} - \frac{1}{\lambda_0} \right) \quad \dots (ii)$Dividing equation $(i)$ by $(ii)$:$\frac{4.8}{1.6} = \frac{\frac{1}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{2\lambda} - \frac{1}{\lambda_0}}$$3 = \frac{\frac{1}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{2\lambda} - \frac{1}{\lambda_0}}$$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$.

When radiation of the wavelength $\lambda$ is incident on a metallic surface,the stopping potential is $4.8 \ V$. If the same surface is illuminated with radiation of double the wavelength,then the stopping potential becomes $1.6 \ V$. Then,the threshold wavelength for the surface is:

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