According to the principle of conservation of mechanical energy, the decrease in potential energy is equal to the increase in kinetic energy . When the mass $m$ slips down the wall of a semispherical surface of radius $R$ to the bottom, its vertical displacement is $h = R$.

Loss in potential energy, $\Delta U = mgh = mgR$ Gain in kinetic energy, $\Delta K = \frac{1}{2}mv^2 - 0 = \frac{1}{2}mv^2$

Equating both: $\frac{1}{2}mv^2 = mgR$ $v^2 = 2gR$ $v = \sqrt{2Rg}$

Thus, the velocity at the bottom of the surface is $\sqrt{2Rg}$.