Physics · Atomic Physics

NEET (UG) 2025 — Question 32

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

Answer: B

Given $\mathrm{n}=2, \mathrm{Z}=1$

$2 \pi \mathrm{r}=\mathrm{n} \lambda$

$2 \pi \times\left(0.052 \frac{n^{2}}{Z}\right)=n \lambda$

On solving $\lambda=0.67 \mathrm{~nm}$

Answer key and solution verified before publishing.

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