Answer
$7.41\times 10^{11}$ $m/s$
Work Step by Step
Energy of a photon, $E=\frac{ch}{\lambda}$,
in which $c$ is the speed of light in vacuum, $\lambda$ is the wavelength of the light and $h$ is the Planck's constant
In our case, $\lambda=590$ $nm$ $=590\times 10^{-9}$ $m$
Therefore, $E=\frac{3\times 10^{8}\times 6.63\times 10^{-34}}{590\times 10^{-9}}$ $J$
or, $E\approx3.37\times 10^{-19}$ $J$
If $m$ be the mass of electron which is moving with a speed $v$, the kinetic energy of the electron is given by
$K=\frac{1}{2}mv^{2}$
According to the condition,
$\frac{1}{2}mv^{2}=3.37\times 10^{-19}$
or, $v =\sqrt {\frac{2\times 3.37\times 10^{-19}}{9.1\times 10^{-31}}}$ $m/s$
or, $v =7.41\times 10^{11}$ $m/s$
$\therefore$ The electron has to move with a speed of $7.41\times 10^{11}$ $m/s$
