The angular momentum of an electron in a Bohr's orbit of H- atom is 4.2178× 10- 34kg-m2/s . Number of spectral line emitted when electrons falls from this level to next lower level :-

Question

The angular momentum of an electron in a Bohr's orbit of H- atom is 4.2178× 10- 34kg-m2/s . Number of spectral line emitted when electrons falls from this level to next lower level :- (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Angular momentum mvr=(nh/2 π )=4.2178× 10- 34 ∴ n=(4 . 2178 × 10- 34 × 2 × 3 . 14/6 . 62 × 10- 34)=4 when e ominus falls from n=4 to next lower level n=3. Then only one spectral line will be obtained.

Q. The angular momentum of an electron in a Bohr's orbit of $H -$ atom is $4.2178 \times 1 0^{- 34} k g - m^{2} / s$ . Number of spectral line emitted when electrons falls from this level to next lower level :-

Angular momentum
$m v r = \frac{nh}{2 π} = 4.2178 \times 1 0^{- 34}$
$∴ n = \frac{4.2178 \times 1 0 ^{- 34} \times 2 \times 3.14}{6.62 \times 1 0 ^{- 34}} = 4$
when $e^{⊖}$ falls from $n = 4$ to next lower level
$n = 3.$ Then only one spectral line will be obtained.

Q. The angular momentum of an electron in a Bohr's orbit of H− atom is 4.2178×10−34kg−m2/s . Number of spectral line emitted when electrons falls from this level to next lower level :-

Angular momentum mvr=2πnh​=4.2178×10−34 ∴n=6.62×10−344.2178×10−34×2×3.14​=4 when e⊖ falls from n=4 to next lower level n=3. Then only one spectral line will be obtained.

Q. The angular momentum of an electron in a Bohr's orbit of $H -$ atom is $4.2178 \times 1 0^{- 34} k g - m^{2} / s$ . Number of spectral line emitted when electrons falls from this level to next lower level :-

Angular momentum
$m v r = \frac{nh}{2 π} = 4.2178 \times 1 0^{- 34}$
$∴ n = \frac{4.2178 \times 1 0 ^{- 34} \times 2 \times 3.14}{6.62 \times 1 0 ^{- 34}} = 4$
when $e^{⊖}$ falls from $n = 4$ to next lower level
$n = 3.$ Then only one spectral line will be obtained.