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De-Broglie wavelength of an electron orbiting in the $n = 2$ state of hydrogen atom is close to (Given Bohr radius = $0.052\text{ nm}$):
A
0.067 nm
B
0.67 nm
C
1.67 nm
D
2.67 nm
Explanation
De Broglie condition states $2\pi r_n = n \lambda$. With $r_2 = 2^2 a_0 = 4 \times 0.052 = 0.208\text{ nm}$, $\lambda = \frac{2\pi (0.208)}{2} \approx 0.65\text{ nm} \approx 0.67\text{ nm}$.
Detailed Solution
According to de Broglie's standing wave condition for Bohr orbits: $2\pi r_n = n \lambda \implies \lambda = \frac{2\pi r_n}{n}$. For $n = 2$, orbit radius is $r_2 = n^2 a_0 = 4 \times 0.052\text{ nm} = 0.208\text{ nm}$. Substituting gives: $\lambda = \frac{2\pi \times 0.208\text{ nm}}{2} = \pi \times 0.208 \approx 3.1416 \times 0.208 \approx 0.653\text{ nm}$. This is closest to $0.67\text{ nm}$.