JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Modern Physics

A light beam of wavelength 500 nm is incident on a metal having work function of 1.25 eV, placed in a magnetic field of intensity

B

. The electrons emitted perpendicular to the magnetic field

B

, with maximum kinetic energy are bent into circular arc of radius 30 cm. The value of

B

is _____

\times 1 0^{- 7}

T.

Given hc

= 20 \times 1 0^{- 26}

J-m, mass of electron

= 9 \times 1 0^{- 31}

Answer: 125

Solution

The photon energy is

E = \frac{h c}{λ} = \frac{20 \times 1 0 ^{- 26}}{500 \times 1 0 ^{- 9}} = 4 \times 1 0^{- 19} J

and the work function is

ϕ = 1.25 \times 1.6 \times 1 0^{- 19} = 2 \times 1 0^{- 19}

J, so the maximum kinetic energy is

K = 2 \times 1 0^{- 19}

J. The corresponding speed is

v = \frac{2 K}{m} = \frac{4 \times 1 0 ^{- 19}}{9 \times 1 0 ^{- 31}} = \frac{2}{3} \times 1 0^{6} m/s

An electron moving perpendicular to the field follows a circle of radius

r = \frac{m v}{e B}

, so

B = \frac{m v}{er} = \frac{9 \times 1 0 ^{- 31} \times \frac{2}{3} \times 1 0 ^{6}}{1.6 \times 1 0 ^{- 19} \times 0.3} = \frac{6 \times 1 0 ^{- 25}}{4.8 \times 1 0 ^{- 20}} = 1.25 \times 1 0^{- 5} T

The magnetic field is

125 \times 1 0^{- 7}

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Integer answer