The graph shows variation of stopping potential V_o with the...

From Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by:

$K_{\text{max}} = \dfrac{hc}{\lambda} - \Phi$

where $\lambda$ is the wavelength of incident radiation and $\Phi$ is the work function of the metal.

The work function is related to the threshold frequency $\nu_0$ by $\Phi = h\nu_0$. From the given graph, the x-intercept represents the threshold frequency. We can observe that:

$\nu_0(X_1) < \nu_0(X_2) < \nu_0(X_3)$

Therefore, the work functions of the metals follow the order:

$\Phi(X_1) < \Phi(X_2) < \Phi(X_3)$

For the same wavelength $\lambda$ of incident radiation, the term $\dfrac{hc}{\lambda}$ is constant. Thus, the kinetic energy $K_{\text{max}}$ will be greatest for the metal with the smallest work function.

Since metal $X_1$ has the minimum work function, it will emit electrons with the greatest kinetic energy.

Answer: $X_1$