The angular momentum of an electron in a Bohr's orbit of H e+ is 3.1652× 10- 34 kg m2 / s e c . What is the wave number in terms of Rydberg constant (R) of the spectral line emitted when an electron falls from this level to the first excited state. [Use h = 6.626 × 10- 34 J . s]

Question

The angular momentum of an electron in a Bohr's orbit of H e+ is 3.1652× 10- 34 kg m2 / s e c . What is the wave number in terms of Rydberg constant (R) of the spectral line emitted when an electron falls from this level to the first excited state. [Use h = 6.626 × 10- 34 J . s] (A) 3R (B) (5 R/9) (C) (3 R/4) (D) (8 R/9)

Tardigrade Admin · Accepted Answer

Angular momentum =( nh /2 π) 3.1652 × 10-34=(n × 6.626 × 10-34/2 π) n =3 ∵ barv=R ⋅ Z2 ((1/n12)-(1/n22)) overline v = R ⋅ 22 (1/22)-(1/32) ⇒ (5 R /9)

$3 R$

$\frac{5 R}{9}$

$\frac{3 R}{4}$

$\frac{8 R}{9}$

Angular momentum $= \frac{nh}{2 π}$
$3.1652 \times 1 0^{- 34} = \frac{n \times 6.626 \times 1 0 ^{- 34}}{2 π}$
$n = 3$
$∵ \overset{v}{ˉ} = R \cdot Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$
$\overline{v} = R \cdot 2^{2} \frac{1}{2 ^{2}} - \frac{1}{3 ^{2}} \Rightarrow \frac{5 R}{9}$

Q. The angular momentum of an electron in a Bohr's orbit of He+ is 3.1652×10−34 kgm2/sec . What is the wave number in terms of Rydberg constant (R) of the spectral line emitted when an electron falls from this level to the first excited state. [Useh=6.626×10−34J.s]

3R

95R​

43R​

98R​

Angular momentum =2πnh​ 3.1652×10−34=2πn×6.626×10−34​ n=3 ∵vˉ=R⋅Z2(n12​1​−n22​1​) v=R⋅22221​−321​⇒95R​

$3 R$

$\frac{5 R}{9}$

$\frac{3 R}{4}$

$\frac{8 R}{9}$

Angular momentum $= \frac{nh}{2 π}$
$3.1652 \times 1 0^{- 34} = \frac{n \times 6.626 \times 1 0 ^{- 34}}{2 π}$
$n = 3$
$∵ \overset{v}{ˉ} = R \cdot Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$
$\overline{v} = R \cdot 2^{2} \frac{1}{2 ^{2}} - \frac{1}{3 ^{2}} \Rightarrow \frac{5 R}{9}$