Question

Calculate the change in angular momentum of the electron when it jumps from third orbit to first orbit in hydrogen atom.
(Take h = 6.33 × 10⁻³⁴ Js)

Answer 1

According to Bohr's second postulate,

Angular momentum = ${\displaystyle \frac{nh}{2\pi }}$

For the first orbit, n₁ = 1

∴ ${\displaystyle L_1=\frac{\left(1\right)h}{2\pi }=\frac{h}{2\pi }}$    ...(i)

For the third orbit, n₃ = 3

∴ ${\displaystyle L_3=\frac{\left(3\right)h}{2\pi }=\frac{3h}{2\pi }}$    ...(ii)

When an electron jumps from 3rd orbit to 1st orbit, the change in angular momentum is

${\displaystyle L_3-L_1=\frac{3h}{2\pi }-\frac{h}{2\pi }}$

= ${\displaystyle \frac{2h}{2\pi }}$

= ${\displaystyle \frac{h}{\pi }}$

Putting h and π values, we get

Change in angular momentum = ${\displaystyle \frac{6.33\times {10}^{-34}}{3.142}}$

= 2.11 × 10⁻³⁴ kg m²/s

The change in angular momentum of an electron when it jumps from the 3^rd orbit to the 1^st orbit in a hydrogen atom is 2.11 × 10⁻³⁴ kg m²/s.