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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 \times 10^{-34} \mathrm{~J}-\mathrm{s})

Option: 1

4.16 \times 10^{-34} \mathrm{~J}-\mathrm{s}


Option: 2

3.32 \times 10^{-34} \mathrm{~J}-\mathrm{s}


Option: 3

1.05 \times 10^{-34} \mathrm{~J}-\mathrm{s}


Option: 4

2.08 \times 10^{-34} \mathrm{~J}-\mathrm{s}


Answers (1)

best_answer

Change in the angular momentum

\mathrm{\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}\left(n_2-n_1\right)}
        \mathrm{=\frac{6.6 \times 10^{-34}}{2 \times 3.14}(5-4)=1.05 \times 10^{-34} J-S}

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seema garhwal

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