When a light beam of frequency is incident on a metal surface,...

Using Einstein's photoelectric equation:

K_\max = h\upsilon - W

When an electron moves in a circular path in a uniform magnetic field, the necessary centripetal force is provided by the magnetic force:

$\dfrac{m_e v^2}{r} = e v B$

$v = \dfrac{e B r}{m_e}$

The maximum kinetic energy of the photoelectron is:

K_\max = \dfrac{1}{2} m_e v^2 = \dfrac{1}{2} m_e \left( \dfrac{e B r}{m_e} \right)^2 = \dfrac{(B e r)^2}{2 m_e}

Substituting K_\max into the photoelectric equation:

$\dfrac{(B e r)^2}{2 m_e} = h\upsilon - W$

$W = h\upsilon - \dfrac{(B e r)^2}{2 m_e}$

Answer: $h\upsilon - \dfrac{(Ber)^2}{2m_e}$