The electron belongs to the angular momentum of an e − in a Bohr's orbit of H atom is 4.2178 × 10 − 34 K g . m 2 . S e c − 1 . Options: A . &nbsp K - shell B . &nbsp L - shell C . &nbsp N - shell D . &nbsp M - shell

Answer: Option C : C n h 2 π = 4.2178 × 10 − 34 n = 2 π × 4.2178 × 10 − 34 h = 2 × 3.14 × 4.2178 × 10 − 34 6.625 × 10 − 34 n = 4 N-shell

The electron belongs to the angular momentum of an e−in a Bohr&apo

Question -

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

Options:

A . K - shell

B . L - shell

C . N - shell

D . M - shell

Answer: Option C
:
C

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

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

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

$n$ = $4$

N-shell

Was this answer helpful ?

Next Question