Radiation \(p\) with wavelength \(6561\;\mathring A\) falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of \(3 \times 10^{-4}\; T\). If the radius of the largest circular path followed by the electrons is \(10 \;\mathrm{mm},\) the work function of the metal is close to ............... \(eV\)

\(\therefore \quad \mathrm{KE}_{\max }=\frac{\mathrm{hc}}{\lambda}-\phi\)

Again, \(\mathrm{R}_{\max }=\frac{\sqrt{2 \mathrm{mKE}_{\max }}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{m}\left(\frac{\mathrm{hc}}{\lambda}-\phi\right)}}{\mathrm{qB}}\)

\(\therefore \quad \frac{\mathrm{R}_{\max }^{2} \mathrm{q}^{2} \mathrm{B}^{2}}{2 \mathrm{m}}=\frac{\mathrm{hc}}{\lambda}-\phi\)

\(\therefore \quad \phi=\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{R}_{\mathrm{max}}^{2} \mathrm{q}^{2} \mathrm{B}^{2}}{2 \mathrm{m}}=1.0899 \mathrm{eV} \approx 1.1 \mathrm{eV}\)