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Chemistry · PYQ NEET UG 2019(2)
हिन्दीIn hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is ________. [Given that Bohr radius, \( a_0 = 52.9 \text{ pm} \)]
Correct answer · Option 3
\( 211.6 \pi \text{ pm} \)
Explanation
For Bohr orbit, \( 2\pi r = n\lambda \). For \( n = 2 \), \( r = n^2 a_0 = 4 a_0 \). Thus, \( 2\pi (4 a_0) = 2\lambda \implies \lambda = 4\pi a_0 = 4\pi (52.9) = 211.6\pi \text{ pm} \).
