A metal surface is illuminated by a radiation of wavelength \(4500 \; \mathring A\). The ejected photo-electron enters a constant magnetic field of \(2 \; mT\) making an angle of \(90^{\circ}\) with the magnetic field. If it starts revolving in a circular path of radius \(2 \; mm\), the work function of the metal is approximately ............. \(eV\)

\(B =2 \; mT , R =2 \; mm\)

\(R =\frac{\sqrt{2 \; Km }}{ qB }\)

\(\frac{(q B R)^{2}}{2 \; m}=K\)

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

\(\frac{(6.4)^{2}}{2 \times 9.1} \times \frac{10^{-50}}{10^{-31}}= K\)

\(K =2.25 \times 10^{-19} \; J\)

\(=\frac{2.25 \times 10^{-19}}{1.6 \times 10^{-19}} eV =1.40 \; eV\)

\(E =\frac{12400}{4500}=2.76 \; eV\)

\(\phi= E - K =(2.76-1.40) eV =1.36 \; eV\)