$(a)\;1.3 KeV\qquad(b)\;4.2 KeV\qquad(c)\;5.8 KeV\qquad(d)\;6.2 KeV$

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Answer : (d) $\;6.2 KeV$

Explanation :

The velocity of photo electrons is found

$F=ma$

$eVB=\large\frac{mV^2}{R} \quad \; V=\large\frac{e}{m}\;BR$

The kinetic energy of photo electrons is then

$K=\large\frac{1}{2}\;mV^2=\large\frac{1}{2}\;\large\frac{e^2B^2R^2}{m}$

$=\large\frac{1}{2}\;\large\frac{(1.6 \times 10^{-19}C)^2\;(2 \times10^{-2}T)^2\;(23 \times10^{-3} m)^2}{(9.1\times10^{-31} kg)}$

$=2.97\times10^{-15} J$

$or\; K=(2.97\times10^{-15} J)\;\large\frac{(1 Kev)}{(1.6\times10^{-16}) J}=18.36eV$

The energy of incident photon is

$\varepsilon \nu =\large\frac{hc}{\lambda}=\large\frac{12.4 KeV A^{0}}{0.5A^{0}}=24.8 eV$

The binding energy is difference between these two valves

$BE=\varepsilon \nu-K=24.8-18.6=6.2 KeV\;.$

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