De-Broglie wavelength of an electron orbiting in the n = 2 state of hydrogen atom is close to (Given Bohr radius = 0.052 nm)

1. 0.067 nm

2. 0.67 nm

3. 1.67 nm

4. 2.67 nm

Doubtly Asked 30 Oct 2026 NEET UG 2025 Practice test neet Save WhatsApp Report

1 answer

Best answer

Answer: 2. 0.67 nm

Key idea: Bohr's quantization $2\pi r_n=n\lambda$ (electron wave fits the orbit).

Radius of the $n=2$ orbit: $r_2=n^2a_0=4\times0.052=0.208$ nm.

$$\lambda=\frac{2\pi r_2}{n}=\frac{2\pi\times0.208}{2}\approx0.65\ \mathrm{nm}\approx0.67\ \mathrm{nm}$$

The nearest option is 0.67 nm.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

Was this answer helpful?