MCQ
When an electron in hydrogen atom is excited, from its $4^{th}$ to $5^{th}$ stationary orbit, the change in angular momentum of electron is (Planck’s constant: $h = 6.6 \times {10^{ - 34}}J{\rm{ - s}}$)
  • A
    $4.16 \times {10^{ - 34}}\,J{\rm{ - }}s$
  • B
    $3.32 \times {10^{ - 34}}\,J{\rm{ - }}s$
  • $1.05 \times {10^{ - 34}}\,J{\rm{ - }}s$
  • D
    $2.08 \times {10^{ - 34}}\,J{\rm{ - }}s$

Answer

Correct option: C.
$1.05 \times {10^{ - 34}}\,J{\rm{ - }}s$
c
(c) Change in the angular momentum 

$\Delta L = {L_2} - {L_1} = \frac{{{n_2}h}}{{2\pi }} - \frac{{{n_1}h}}{{2\pi }}$$ \Rightarrow \Delta L = \frac{h}{{2\pi }}({n_2} - {n_1})$ 

$ = \frac{{6.6 \times {{10}^{ - 34}}}}{{2 \times 3.14}}(5 - 4)$$ = 1.05 \times {10^{ - 34}}J{\rm{ - }}S$

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