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(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ of the ejected electrons is given by:$K_{\max} = \frac{h c}{\

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$\frac{h c}{\lambda} - 2 m \left( \frac{q B R}{2 m} \right)^{2}$

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ of the ejected electrons is given by:$K_{\max} = \frac{h c}{\lambda} - \phi_{0}$Therefore,the work function $\phi_{0}$ is:$\phi_{0} = \frac{h c}{\lambda} - K_{\max} \quad \dots(i)$When an electron with charge $q$ and mass $m$ moves with velocity $v$ in a perpendicular magnetic field $B$,it follows a circular path of radius $R$ given by:$R = \frac{m v}{q B} \implies v = \frac{q B R}{m}$The maximum kinetic energy is:$K_{\max} = \frac{1}{2} m v^{2} = \frac{1}{2} m \left( \frac{q B R}{m} \right)^{2} = \frac{q^{2} B^{2} R^{2}}{2 m}$Substituting $K_{\max}$ into equation $(i)$:$\phi_{0} = \frac{h c}{\lambda} - \frac{q^{2} B^{2} R^{2}}{2 m}$Note that option $(d)$ is equivalent to this result:$2 m \left( \frac{q B R}{2 m} \right)^{2} = 2 m \left( \frac{q^{2} B^{2} R^{2}}{4 m^{2}} \right) = \frac{q^{2} B^{2} R^{2}}{2 m}$Thus,option $(d)$ is correct.

Photons of wavelength $\lambda$ are incident on a metal. The most energetic electrons ejected from the metal are bent into a circular arc of radius $R$ by a perpendicular magnetic field having a magnitude $B$. The work function of the metal is (where,symbols have their usual meanings)

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