The radiation corresponding to 3 → 2 transition of a hydrogen atom falls on a gold surface to generate photoelectrons. These electrons are passed through a magnetic field of 5 × 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 × 10 - 31 kg )

Step-by-step solution

3 → 2 ⇒ 1 . 89   eV 5 × 10 - 4   T    r = 7   mm r = m v q B ⇒ m v = q r B ⇒ E = P 2 2   m = ( q R B ) 2 2   m = 1 . 6 × 10 - 19 × 7 × 10 - 3 × 5 × 10 - 4 2 2 × 9 . 1 × 10 - 31 Joule = 3136 × 10 - 52 18 . 2 × 10 - 31 × 1 . 6 × 10 - 19   eV = 1 . 077   eV We know work function = energy incident - ( KE ) electron  ϕ = 1 . 89 - 1 . 077 = 0 . 813   eV