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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 $\mathrm{R}$ by a perpendicular magnetic field having a magnitude $\mathrm{B}$. The work function of the metal is (Where symbols have their usual meanings) -
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$\frac{\mathrm{hc}}{\lambda}-2 \mathrm{~m}_{\mathrm{c}}\left(\frac{\mathrm{eBR}}{2 \mathrm{~m}_{\mathrm{e}}}\right)^{2}$
$\begin{aligned} \mathrm{R} &=\frac{\mathrm{mv}}{\mathrm{qB}} \\ \mathrm{V} &=\frac{\mathrm{qBR}}{\mathrm{m}}=\frac{\mathrm{eBR}}{\mathrm{m}_{\mathrm{e}}} \\ \frac{\mathrm{hc}}{\lambda}-\phi=\mathrm{KE}_{\max } \text { (Einstein photo electric equation) } \\ \phi &=\frac{\mathrm{hc}}{\lambda}-\mathrm{KE}_{\max } \\ &=\frac{\mathrm{hc}}{\lambda}-\frac{1}{2} \mathrm{~m}_{\mathrm{e}}\left(\frac{\mathrm{eBR}}{\mathrm{m}_{\mathrm{e}}}\right)^{2} \\ &=\frac{\mathrm{hc}}{\lambda}-2 \mathrm{~m}_{\mathrm{e}}\left(\frac{\mathrm{eBR}}{2 \mathrm{~m}_{\mathrm{e}}}\right)^{2} \end{aligned}$
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