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(B) The Einstein's photoelectric equation is given by $eV_0 = \frac{hc}{\lambda} - \phi_0$,where $\phi_0 = \frac{hc}{\lambda_0}$ and $\lambda_0$ is th

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$4 \lambda$

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(B) The Einstein's photoelectric equation is given by $eV_0 = \frac{hc}{\lambda} - \phi_0$,where $\phi_0 = \frac{hc}{\lambda_0}$ and $\lambda_0$ is the threshold wavelength.For the first case: $e(4.8) = \frac{hc}{\lambda} - \phi_0 \quad ... (i)$For the second case: $e(1.6) = \frac{hc}{2\lambda} - \phi_0 \quad ... (ii)$Dividing equation $(i)$ by equation $(ii)$:$\frac{4.8}{1.6} = \frac{\frac{hc}{\lambda} - \phi_0}{\frac{hc}{2\lambda} - \phi_0}$$3 = \frac{\frac{hc}{\lambda} - \phi_0}{\frac{hc}{2\lambda} - \phi_0}$$3 \left( \frac{hc}{2\lambda} - \phi_0 \right) = \frac{hc}{\lambda} - \phi_0$$\frac{3hc}{2\lambda} - 3\phi_0 = \frac{hc}{\lambda} - \phi_0$$\frac{3hc}{2\lambda} - \frac{hc}{\lambda} = 2\phi_0$$\frac{hc}{2\lambda} = 2\phi_0 \implies \phi_0 = \frac{hc}{4\lambda}$Since $\phi_0 = \frac{hc}{\lambda_0}$,we have $\frac{hc}{\lambda_0} = \frac{hc}{4\lambda}$.Therefore,the threshold wavelength $\lambda_0 = 4\lambda$.

$X = x_i$	$-2$	$-1$	$0$	$1$	$2$
$P(X = x_i)$	$\frac{1}{6}$	$k$	$\frac{1}{4}$	$k$	$\frac{1}{6}$

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

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$A$ curve passes through the point $\left(1, \frac{\pi}{6}\right)$. Let the slope of the curve at each point $(x, y)$ be $\frac{y}{x}+\sec \left(\frac{y}{x}\right)$,where $x>0$. Then,the equation of the curve is:

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