Question

The radiation corresponding to the 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, and the work function of the metal is ______.

(Mass of electron = 9.1 × 10^-31 kg)

Options

• 0.82 eV

• 0.16 eV

• 1.88 eV

• 1.36 eV

Answer 1

The radiation corresponding to the 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, and the work function of the metal is 0.82 eV.

Explanation:

Magnetic field = 5 × 10^-4 T

Radius = 7 mm = 7 × 10^-3 m

Applying the relation for energy transition,

`E = 13.6[1/4 - 1/9]`

= `5/36(13.6)` eV = 1.89 eV

The charged particle enters an 'r'-radius magnetic field.

`r = (mv)/(qB)`

p = mυ = rqB

p = `7 xx 10^-3 xx 1.6 xx 10^-19 xx 5 xx 10^-4`

p = `3136 xx 10^-52`

KE = `p^2/(2m)`

⇒ KE = `(3136 xx 10^-52)/(2 xx 9.1 xx 10^-31)`

= `(3136 xx 10^-52)/(2 xx 9.1 xx 10^-31 xx 1.6 xx 10^-19)`eV

= 107.69 × 10^-2 eV

= 1.077 eV

Apply the photoelectric equation,

Work function = E - KE

= [1.89 - 1.077] eV

= 0.813 eV