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When the electron in hydrogen atom is excited from the 4^{\text {th }} stationary orbit to the 5^{\text {th }} stationary orbit, the change in the angular momentum of the electron is (Planck's constant, \mathrm{h}=6.63 \times 10^{-34} \mathrm{Js} )

Option: 1

4.16 \times 10^{-34} \mathrm{Js}

Option: 2

3.32 \times 10^{-34} \mathrm{Js}

Option: 3

1.05 \times 10^{-34} \mathrm{Js}

Option: 4

2.08 \times 10^{-34} \mathrm{Js}

Answers (1)


According to Bohr's quantisation condition \mathrm{L}_{\mathrm{n}}=\frac{\mathrm{nh}}{2 \pi}

\therefore \quad \text { For } \mathrm{n}=4, \mathrm{~L}_4=\frac{4 \mathrm{~h}}{2 \pi} \text{ and for } \mathrm{n}=5, \mathrm{~L}_5=\frac{5 \mathrm{~h}}{2 \pi}
\therefore  Change in angular momentum when an electron is excited from n=4 \text{ to } n=5 is

\Delta \mathrm{L}=\mathrm{L}_5-\mathrm{L}_4=\frac{5 \mathrm{~h}}{2 \pi}-\frac{4 \mathrm{~h}}{2 \pi}=\frac{\mathrm{h}}{2 \pi}=\frac{6.63 \times 10^{-34}}{2 \times 3.14} \mathrm{Js}=1.05 \times 10^{-34} \mathrm{Js}

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Shailly goel

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