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

Question

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. A. 0.067 nm
2. B. 0.67 nm (Correct answer)
3. C. 1.67 nm
4. D. 2.67 nm

Correct Answer

Option B — 0.67 nm

Detailed Solution & Explanation

The correct answer is 0.67 nm.

Key Points

• Bohr radius of the $n$-th orbit: $r=0.052\,n^2$ nm. For $n=2$: $r=0.052\times4=0.208$ nm.
• Angular momentum quantization $mvr=\dfrac{nh}{2\pi}$ gives $\lambda=\dfrac{h}{mv}=\dfrac{2\pi r}{n}$.
• For $n=2$: $\lambda=\dfrac{2\pi\times0.208}{2}=\pi\times0.208\approx 0.65$ nm $\approx$ 0.67 nm.