The velocity of the photoelectrons is found by the relation :

\(e v B=m \frac{v^2}{R} \text { or } v=\frac{e}{m} B R\)

The kinetic energy of the hotoelectrons is

\(K  =\frac{1}{2} m v^2=\frac{1}{2} \frac{e^2 B^2 R^2}{m}\)

\(=\frac{1}{2} \frac{\left(1.6 \times 10^{-19}\right)^2\left(2 \times 101^{-2}\right)^2\left(23 \times 10^{-3}\right)^2}{\left(9.1 \times 10^{-31}\right)}\)

\(=2.97 \times 10^{-15}\,J\)

\(=\left(2.97 \times 10^{-15}\right) \frac{1}{1.6 \times 10^{-19}}=18.36\,KeV\)

The energy of the incident photon is \(E_v=\frac{h c}{\lambda}=\frac{12.4}{0.50}=24.8\,KeV\)

The binding energy is the difference between these two values:

\(B E=E_v-K=24.8 - 18.6- 6.2\,keV\)