Answer
(a) $3.1$ $keV$
(b) $14.4$ $keV$
Work Step by Step
(a) If an electron moves with a speed $v$ in circular paths of radius $r$ in a region of uniform magnetic field $\vec B$, the magnetic fore acting on the electron provides the necessary centripetal force. Therefore,
$evB=\frac{mv^{2}}{r}$,
where $e$ and $m$ are the charge and mass of the electron respectively.
or, $e\times\frac{v}{r}\times(Br)=\frac{mv^{2}}{r}$,
or, $\frac{v^{2}}{r}\times \frac{r}{v}=\frac{e}{m}\times (Br)$
or, $v=\frac{1.6\times 10^{-19}}{9.1\times 10^{-31}}\times 1.88\times 10^{-4}$ $m/s$
or, $v\approx3.31\times 10^{7}$ $m/s$
$\therefore$ The kinetic energy of the ejected electron is
$K=\frac{1}{2}mv^2$
or, $K=\frac{1}{2}\times 9.1\times 10^{-31} \times (3.31\times 10^{7})^{2}$ $J$
or, $K= 4.985\times 10^{-16}$ $J$
or, $K=\frac{4.985\times 10^{-16}}{1.6\times 10^{-19}}$ $eV$
or, $K\approx 3.1\times 10^{3}$ $eV$
or, $K=3.1$ $keV$
$\therefore$ The maximum kinetic energy of those electrons is $3.1$ $keV$.
(b) 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, the wavelength of the incident x-ray is $\lambda=71$ $pm$ $=71\times 10^{-12}$ $m$
Therefore, the energy of the incident a photon
$E=\frac{3\times 10^{8}\times 6.63\times 10^{-34}}{71\times 10^{-12}}$ $J$
or, $E\approx 2.80\times 10^{-15}$ $J$
or, $E=\frac{2.80\times 10^{-15}}{1.6\times 10^{-19}}$ $eV$
or, $E\approx1.75\times 10^{4}$ $eV$
or, $E=17.5$ $keV$
$\therefore$ The work function $(\phi)$ of the gold foil is given by
$\phi=E-K$
or, $\phi=17.5-3.1$ $keV$
or, $\phi= 14.4$ $keV$
$\therefore$ The work done in removing electron from the gold atoms is $14.4$ $keV$
