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What is the De-Broglie wavelength of an electron orbiting in the n = 2 state of a hydrogen atom?
(Given: Bohr radius = 0.052 nm)

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  1. 0.067 nm
  2. 0.67 nm
  3. 1.67 nm
  4. 2.67 nm

Answer (Detailed Solution Below)

Option 2 : 0.67 nm
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Detailed Solution

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Correct option is: (2) 0.67 nm

Given n = 2, Z = 1

2πr = nλ

2π × (0.052 × n2 / Z) = nλ

On solving, λ = 0.67 nm

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