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In hydrogen atom, the de-Broglie wavelength of an electron in the second Bohr orbit is [Given that Bohr radius, a₀ = 52.9 pm]

1. 105.8 pm
2. 211.6 pm
3. 211.6 π pm
4. 52.9 π pm

Correct answer: 211.6 π pm

Solution

The de-Broglie wavelength of an electron in a Bohr orbit is given by λ = 2πr, where r is the radius of the orbit. For the second Bohr orbit, r = 2² × a₀ = 4 × 52.9 pm. Substituting, λ = 2π × 4 × 52.9 pm = 211.6π pm.

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