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 photoelectron. The electrons from the metal surface with maximum kinetic energy are made to move perpendicular to a magnetic field of `1/320`T in a radius of 10^-3m. Find the 320 work function of metal:

पर्याय

• 1.03 eV

• 1.89 eV

• 0.86 eV

• 2.03 eV

Answer 1

1.03 eV

Explanation:

E₃ – E₂ = 13.6 `[1/2^2-1/3^2]`

= `(13.6xx5)/36`

= 1.89 eV

Photoelectrons with kE_max are moving on circular path.

r = `"mv"/"qB"`

mv = qBr

P = qBr

= `1.6xx10^(–19) 1/3200xx10^-3`

`1/2xx10^-24 = 5xx10^-25`kg m/s

Energy of photoelectron = KE_max = `"P"^2/(2"m")`

= `(25xx10^-50)/(2xx9.1xx10^-31xx1.6xx10^-19)`

= 0.86 eV

Now use Einstein equation

hn = Φ + kE_max

1.89 = 0.56 + Φ;

Φ = 1.03 eV