The angular momentum of an electron in a particular orbit of H-atom is 5-27 - 10-34 kgm2-s- Identify the orbit-

The correct option is B 5 Angular momentum of an electron is given as; L = nh2π = n× 6.6× 10^ 342× 3.14 Therefore n = 2× 3.14× 5.27× 10^ 346.6× ...

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Standard XII

Physics

Hydrogen Spectra

The angular m...

Question

The angular momentum of an electron in a particular orbit of H-atom is $5.27 \times 10^{- 34} k g m^{2} / s$ . Identify the orbit.

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Solution

The correct option is B $5$
Angular momentum of an electron is given as; $L = \frac{n h}{2 π} = \frac{n \times 6.6 \times 10^{- 34}}{2 \times 3.14}$

Therefore $n = \frac{2 \times 3.14 \times 5.27 \times 10^{- 34}}{6.6 \times 10^{- 34}}$

Therefore, $n = 5$

Hence, option $B$ is correct.

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The angular momentum of an electron in a particular orbit of H-atom is 5.27×10−34kgm2/s. Identify the orbit.

The correct option is B 5Angular momentum of an electron is given as; L=nh2π=n×6.6×10−342×3.14Therefore n=2×3.14×5.27×10−346.6×10−34Therefore, n=5Hence, option B is correct.

The angular momentum of an electron in a particular orbit of H-atom is $5.27 \times 10^{- 34} k g m^{2} / s$ . Identify the orbit.

The correct option is B $5$
Angular momentum of an electron is given as; $L = \frac{n h}{2 π} = \frac{n \times 6.6 \times 10^{- 34}}{2 \times 3.14}$

Therefore $n = \frac{2 \times 3.14 \times 5.27 \times 10^{- 34}}{6.6 \times 10^{- 34}}$

Therefore, $n = 5$

Hence, option $B$ is correct.