Question

The energy of an electron in an orbit of Bohr hydrogen atom is −3.4 eV. Find its angular momentum.

Answer 1

For a hydrogen atom,

E = `-13.6/n^2`

Thus, −3.4 = `-13.6/n^2`

n² = `136/3.4`

n² = 4

n = `sqrt 4`

n = 2

Planks constant (h) = 6.63 × 10⁻³⁴

According to Bohr’s second postulate, the angular momentum L of an electron is quantized and is an integral multiple of `h/(2 pi) `:

L = `(nh)/(2 pi)`

= `(2 xx 6.63 xx 10^-34)/(2 pi)`

= `(6.63 xx 10^-4)/3.14159`

L = 2.11 × 10⁻³⁴ J . s