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Question: When an electron in hydrogen atom is excited, from its 4<sup>th</sup> to 5<sup>th</sup> stationary o...

When an electron in hydrogen atom is excited, from its 4th to 5th stationary orbit, the change in angular momentum of electron is (Planck’s constant: h=6.6×10−34J⥂−⥂sh = 6.6 \times 10^{- 34}J ⥂ - ⥂ s)

A

4.16×10−34J⥂−s4.16 \times 10^{- 34}J ⥂ - s

B

3.32×10−34J⥂−⥂s3.32 \times 10^{- 34}J ⥂ - ⥂ s

C

1.05×10−34J⥂−⥂s1.05 \times 10^{- 34}J ⥂ - ⥂ s

D

2.08×10−34J⥂−⥂s2.08 \times 10^{- 34}J ⥂ - ⥂ s

Answer

1.05×10−34J⥂−⥂s1.05 \times 10^{- 34}J ⥂ - ⥂ s

Explanation

Solution

Change in angular momentum

ΔL=L2−L1=n2h2π−n1h2π\Delta L = L_{2} - L_{1} = \frac{n_{2}h}{2\pi} - \frac{n_{1}h}{2\pi} ⇒

ΔL=h2π(n2−n1)=6.6×10−342×3.14(5−4)=1.05×10−34\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-*s