Consider an electron in a hydrogen atom, revolving in its second excited state (having radius 4.65  …

Consider an electron in a hydrogen atom, revolving in its second excited state (having radius 4.65 Å ). The de-Broglie wavelength of this electron is:
  1. 12.9 Å
  2. 6.6 Å
  3. 9.7 Å
  4. 3.5 Å

Solution

According to Bohr’s atomic model
mvr=nh2π
And the de-Broglie wavelength
λ=hmv
From the above two equations we can write, λ=2πrn
In the 2nd excited state
λ=2πr3
∴  λ=23π×4.65  Å=9.7 Å

Asked in: JEE Main 2019 (12 Apr Shift 2)

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