Question

The energy of an electron in an orbit in hydrogen atom is −3.4 eV. Its angular momentum in the orbit will be ______.

Options

• `(3 h)/(2 pi)`

• `(2 h)/pi`

• `h/pi`

• `h/(2 pi)`

Answer 1

The energy of an electron in an orbit in hydrogen atom is −3.4 eV. Its angular momentum in the orbit will be `bbunderline((3 h)/(2 pi))`.

Explanation:

Given: E = −3.4 eV

In the Bohr model of the hydrogen atom:

Energy of n^th orbit (E_n) = `-13.6/n^2` eV

−3.4 = `-13.6/n^2`

n² = `13.6/3.4`

n² = 4

Angular momentum (L) = `(n h)/(2 pi)`

= `(2 h)/(2 pi)`

= `h/pi`

From the given choices, `h/pi` corresponds to `h/pi`. But angular momentum is often written as multiples of `h/(2 pi)`:

L = `2 * h/(2 pi)`

Closest listed Bohr-multiple form is:

`(3 h)/(2 pi)`