The radiation corresponding to 3 $\to$ 2 transition of a hydrogen atom falls on a gold surface to generate photoelectrons. The 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 : (Mass of electron = 9.1 $\times$ 10^$-$31 kg)

Energy of photon can be given as<br/><br/>${E_p} = 13.6\left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]$ eV<br/><br/>where, n₁ = lower energy level and n₂ = higher energy level.<br/><br/>As per question, n₁ = 2, n₂ = 3<br/><br/>$\therefore$ ${E_p} = 13.6\left[ {{1 \over {{{(2)}^2}}} - {1 \over {{{(3)}^2}}}} \right]$<br/><br/>$$ = 13.6\left[ {{1 \over 4} - {1 \over 9}} \right] = 13.6\left[ {{{9 - 4} \over {36}}} \right] = 1.89$$ eV<br/><br/>We know that work-function is the minimum energy required to eject photoelectrons from metal surface.<br/><br/>For gold plate, it will be<br/><br/>$\phi$ = E_P $-$ KE_max .... (i)<br/><br/>[Given, B = 5 $\times$ 10^$-$4 T, r = 7 mm = 7 $\times$ 10^$-$3 m, q = 1.6 $\times$ 10^$-$19 C and m = 9.1 $\times$ 10^$-$31 kg]<br/><br/>Therefore, velocity of photoelectrons will be<br/><br/>$$v = {{Bqr} \over m} = {{5 \times {{10}^{ - 4}} \times 1.6 \times {{10}^{ - 19}} \times 7 \times {{10}^{ - 3}}} \over {9.1 \times {{10}^{ - 31}}}} = 6.15 \times {10^5}$$ ms^$-$1<br/><br/>Kinetic energy will be<br/><br/>$\therefore$ $$KE = {1 \over 2}m{v^2} = {{1 \times 9.1 \times {{10}^{ - 31}} \times {{(6.15 \times {{10}^5})}^2}} \over {2 \times 1.6 \times {{10}^{ - 19}}}}$$ eV = 1.075 eV<br/><br/>Now, putting the values, in Eq. (i), we get<br/><br/>$\phi$ = (1.89 $-$ 1.075) eV = 0.82 eV