Question

A hydrogen atom in is ground state absorbs 10.2 eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of ______.

(Given, Planck's constant = 6.6 × 10^-34 Js)

Options

• 2.10 × 10^-34 Js

• 1.05 × 10^-34 Js

• 3.15 × 10^-34 Js

• 4.2 × 10^-34 Js

Answer 1

A hydrogen atom in is ground state absorbs 10.2 eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of 1.05 × 10^-34 Js.

Explanation:

`Delta"E"=-13.6(1/"n"^2)-(-13.6/1^2)`

⇒ 10.2 = -13.6 `(1-1/"n"^2)`

⇒ 10.2 = 13.6`(1-1/"n"^2)`

⇒ n = 2

So, `"L"_("n"=2)=(2"h")/(2pi)-"h"/pi`

Then, `Delta"L"="L"_(("n"=2))-"L"_(("n"=1))=(2"h")/(2pi)-"h"/(2pi)`

= `"h"/(2pi)`

= 1.05 × 10^-34 Js