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Physics · PYQ NEET UG 2025
हिन्दीDe-Broglie wavelength of an electron orbiting in the \( n = 2 \) state of hydrogen atom is close to (Given Bohr radius = \( 0.052\text{ nm} \))
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Correct answer · Option B
\( 0.67\text{ nm} \)
Explanation
For an electron in the \( n^{\text{th}} \) Bohr orbit, the circumference equals \( n\lambda \). Thus, \( \lambda = \frac{2\pi r_n}{n} \). For \( n = 2 \), \( r_2 = 2^2 \times r_0 = 4 \times 0.052\text{ nm} \), giving \( \lambda = 2\pi (0.052) \approx 0.67\text{ nm} \).
