Question

A hydrogen atom in its 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).

• A. \(2.10 × 10^{–34}\) Js
• B. \(1.05 × 10^{–34}\) Js
• C. \(3.15 × 10^{–34}\) Js
• D. \(4.2 × 10^{–34}\) Js

Answer 1

The Correct Option is B

\(-13.6+10.2=\frac{-13.6}{n^2}\)

\(⇒ \frac{13.6}{n^2} = 3.4\)

\(⇒ n = 2\)

\(⇒ Δ L = 2 \times \frac{h}{2λ} - 1 \times \frac{h}{2λ}\)

\(⇒ \frac{h}{2λ}\)

\(⇒ Δ L ≅ 1.05 \times 10^{-34} \;J\;s\)

Hence, the correct option is (B): \(1.05 × 10^{–34}\) \(J s\)