JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Modern Physics

A metal surface is illuminated by a radiation of wavelength

4500 \overset{˚}{A}

. The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of

9 0^{\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 :

Answer: (A)

Solution

The kinetic energy of the photoelectron can be found from its circular motion in the magnetic field:

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

. Substituting values:

K = \frac{( 1.6 \times 1 0 ^{- 19} \times 2 \times 1 0 ^{- 3} \times 2 \times 1 0 ^{- 3} ) ^{2}}{2 \times 9.1 \times 1 0 ^{- 31}} \approx 1.40 eV

. The photon energy is

E = \frac{12400}{4500} \approx 2.76 eV

. The work function is

ϕ = E - K = 2.76 - 1.40 = 1.36 eV

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Single correct option