The radiation corresponding to \(3 \rightarrow 2\) transition of a hydrogen atom falls on a gold surface to generate photoelectrons. These electrons are passed through a magnetic field of \(5 \times 10^{-4} \,{T}\). Assume that the radius of the largest circular path followed by these electrons is \(7\, {mm}\), the work function of the metal is \(.....\,{eV}\) (Mass of electron \(=9.1 \times 10^{-31} \,{kg}\) )

\(3.4 \longrightarrow {N}=2\)

\(13.6 \longrightarrow {N}=1\)

\(3 \rightarrow 2 \Rightarrow 1.89\, {eV}\)

\(5 \times 10^{-4}\, {T}\)

\({r}=7 \,{mm}\)

\(r=\frac{m v}{q B} \Rightarrow m v=q r B\)

\(\Rightarrow E=\frac{p^{2}}{2 m}=\frac{(q R B)^{2}}{2 m}\)

\(=\frac{\left(1.6 \times 10^{-19} \times 7 \times 10^{-3} \times 5 \times 10^{-1}\right)^{2}}{2 \times 9.1 \times 10^{-31} \text { Joule }}\)

\(=\frac{3136 \times 10^{-52}}{18.2 \times 10^{-31} \times 1.6 \times 10^{-19}} \,{eV}\)

\(=1.077\, {eV}\)

We know work function = energy incident \(-(K E)_{\text {electron }}\)

\(\Phi=1.89-1.077=0.813\, {eV}\)