Question

The radiation emitted, when an electron jumps from n = 3 to n = 2 orbit is a hydrogen atom, falls on a metal to produce photo electrons. The electrons from the metal surface with maximum kinetic energy are made to move perpendicular to a magnetic field of $\frac{1}{320}$ T in a radius of 10^–3 m. Find the work function of metal –

• A. 1.03 eV
• B. 1.89 eV
• C. 0.86 eV
• D. 2.03 eV

Answer 1

Answer: (A)

1.03 eV

Answer 2

E₃ – E₂ = 13.6 $\left\lbrack \frac{1}{2^{2}} - \frac{1}{3^{2}} \right\rbrack$ = $\frac{13.6 \times 5}{36}$= 1.89 eV

Photoelectron with KE_max are moving on circular path.

r = $\frac{mv}{qB}$

mv = qBr

P = qBr = 1.6 × 10^–19 ×$\frac{1}{3200}$× 10^–3

= $\frac{1}{2}$× 10^–24 = 5 × 10^–25 kg m/s

Energy of photoelectron = KE_max

= $\frac{p^{2}}{2m}$ = $\frac{25 \times 10^{- 50}}{2 \times 9.1 \times 10^{- 31} \times 1.6 \times 10^{- 19}}$

= 0.86 eV

Now use Einstein equation

hn = f + KE_max

1.89 = 0.86 + f Ž f = 1.03 eV