JEE Main · 2014 · 06 Apr 2014 · Offline · Modern Physics

The radiation corresponding to

3 \to 2

transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of

3 \times 1 0^{- 4} T

. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to:

Answer: (D)

Solution

The fastest photoelectrons curve on the largest circle, so for that radius

m v = q B R

, and their kinetic energy is

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

. Putting in

B = 3 \times 1 0^{- 4}

T and

R = 10.0

mm gives about

0.8

eV. The incoming photon comes from the

3 \to 2

line of hydrogen, with energy

13.6 (\frac{1}{4} - \frac{1}{9}) = 13.6 \times \frac{5}{36} \approx 1.9

eV. The work function is what remains after the electron's kinetic energy,

W = 1.9 - 0.8 \approx 1.1

eV.

JEE Main · 2014 · 06 Apr 2014 · Offline · Single correct option