The electron belongs to the angular momentum of an $e^{-}$ in a Bohr's orbit of H atom is $4.2178 \times 10^{- 34} K g . m^{2} . S e c^{- 1}$ .

K - shell

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L - shell

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N - shell

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M - shell

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Solution

The correct option is C
N - shell

$\frac{n h}{2 π}$ = $4.2178 \times 10^{- 34}$

$n$ = $\frac{2 π \times 4.2178 \times 10^{- 34}}{h}$

= $\frac{2 \times 3.14 \times 4.2178 \times 10^{- 34}}{6.625 \times 10 - 34}$

$n$ = $4$

N-shell

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Similar questions

The angular momentum of an electron in a Bohr's orbit of H-atom is $4.2178 \times 10^{- 34} K g m^{2} / s e c$ . Then electron belongs to

The electron belongs to the angular momentum of an $e^{-}$ in a Bohr's orbit of H atom is $4.2178 \times 10^{- 34} K g . m^{2} . S e c^{- 1}$ .

4.2178 \times 10^{- 34} {k g m}^{2} / s e c

H

4.2178 \times 10^{- 34} k g m^{2} / s e c

4 \times 10^{- 34} k g m^{2} / s e c .

x \times 10^{- 4}

x

h = 6 \times 10^{- 34} J s, π = 3

R_{H} = 1.08 \times 10^{7} m^{- 1}

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