The graph shows variation of stopping potential $V_o$ with the frequency $\nu$ of the incident radiation for three photosensitive metals $X_1$, $X_2$ and $X_3$. Which metal will give out electrons with greater kinetic energy, for the same wavelength of incident radiation?

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: $X_1$