Dual Nature of Radiation

Stopping potential (V₀) is given by

$e{V_0} = {{hc} \over \lambda } - \phi $

Graph between V₀ and $\lambda$ :

$e{V_0} + \phi = {{hc} \over \lambda }$

$(e{V_0} + \phi )\lambda = hc$

$(e{V_0} + \phi )\lambda $ = constant

Here, both e and $\phi$ are also constant. It represents a hyperbola.

For, V₀ = 0, $\lambda = {{constant} \over \phi } = $ constant

So correct option is (a).

Graph between V₀ and ${1 \over \lambda }$ :

${V_0} = \left( {{{hc} \over e}} \right)\left( {{1 \over \lambda }} \right) - \left( {{\phi \over e}} \right)$

It represents a straight line with slope $\left( {{{hc} \over e}} \right)$ and intercept $\left( { - {\phi \over e}} \right)$ on V₀ axis.

The energy of incident photon $\frac{h c}{\lambda}$ is equal to the sum of minimum energy required to emit the photoelectron (work function) and kinetic energy K of the ejected photoelectron.

The stopping potential V converts the kinetic energy K of ejected photoelectron into potential energy eV , there by stopping its motion.

Thus, $\frac{h c}{\lambda}=\phi+\mathrm{K}=\phi+\mathrm{eV}$

$ \Rightarrow \frac{h c}{\lambda}-\phi=\mathrm{eV} . $

i.e., $\mathrm{V}=\frac{h c}{e} \cdot \frac{1}{\lambda}-\frac{\phi}{e} \quad$ Slope, $m=\tan \theta=\frac{h c}{e}$

(on comparing with $y=m x-c$ )

At $\mathrm{V}=0, \phi=\frac{h c}{\lambda}$ which gives

$ \phi_1=0.001 h c, \phi_2=0.002 h c, \phi_3=0.004 h c, $

(For metal 1)           (metal 2)              (metal 3)

Thus, $\phi_1: \phi_2: \phi_3:: 1: 2: 4$.

Also, $\phi=\frac{h c}{\lambda_0}$, where $\lambda_0$ is threshold wavelength,

$\therefore$ The wavelength corresponding to $\phi_1, \phi_2, \lambda_3$ are

$ \begin{aligned} & \lambda_1=\frac{1}{0.001}=1000 \mathrm{~nm} \\ & \lambda_2=\frac{1}{0.002}=500 \mathrm{~nm} \\ & \lambda_3=\frac{1}{0.004}=250 \mathrm{~nm} \end{aligned} $

The wavelength of violet colour light is close to 400 nm , which is more than the threshold wavelength for metal 3 . Thus violet light can eject photoelectrons from from metal 2 but not from metal 3 .