Stopping Potential versus Frequency Graph for Photosensitive Metals

To determine which metal will emit photoelectrons with the greatest kinetic energy for incident radiation of the same wavelength, we use Einstein's photoelectric equation:

$K_{\text{max}} = h\nu - \Phi$

where:

• $K_{\text{max}}$ is the maximum kinetic energy of the emitted photoelectrons,
• $h$ is Planck's constant,
• $\nu$ is the frequency of the incident radiation,
• $\Phi$ is the work function of the photosensitive metal.

The relationship between stopping potential $V_0$ and incident frequency $\nu$ is expressed as:

$e V_0 = h\nu - \Phi \implies V_0 = \left(\frac{h}{e}\right)\nu - \frac{\Phi}{e}$

At the threshold frequency $\nu_0$, the stopping potential $V_0 = 0$, giving:

$\Phi = h\nu_0$

From the provided graph of $V_0$ versus $\nu$:

• Threshold frequency for metal $X_1$: $\nu_{0,1} = 1.0 \times 10^{14}\text{ Hz}$
• Threshold frequency for metal $X_2$: $\nu_{0,2} = 1.5 \times 10^{14}\text{ Hz}$
• Threshold frequency for metal $X_3$: $\nu_{0,3} = 2.0 \times 10^{14}\text{ Hz}$

Since $\nu_{0,1} < \nu_{0,2} < \nu_{0,3}$, the work functions satisfy:

$\Phi_1 < \Phi_2 < \Phi_3$

For a given wavelength $\lambda$ of incident radiation, the photon energy $E = h\nu = \frac{hc}{\lambda}$ is constant for all three metals. The maximum kinetic energy is:

$K_{\text{max}} = E - \Phi$

Since $X_1$ has the smallest work function $\Phi_1$, it yields the maximum kinetic energy:

$K_{\text{max}, 1} > K_{\text{max}, 2} > K_{\text{max}, 3}$

Hence, metal $X_1$ will give out electrons with greater kinetic energy.

Correct Option: A ($X_1$)