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Question: The angular momentum of an electron in a Bohr's orbit of He<sup>+</sup> is 3.1652 × 10<sup>–34</sup>...

The angular momentum of an electron in a Bohr's orbit of He+ is 3.1652 × 10–34 kg-m2/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.

[Use h = 6.626 × 10–34 J.s]

A

3R

B

5R9\frac{5R}{9}

C

3R4\frac{3R}{4}

D

8R9\frac{8R}{9}

Answer

5R9\frac{5R}{9}

Explanation

Solution

Angular momentum =nh2π\frac{nh}{2\pi}

3.1652 × 10–34 =n×6.626×10−342π\frac{n \times 6.626 \times 10^{- 34}}{2\pi};

n = 3

\ vˉ\bar{v}= R.Z2.(1n12−1n22)\left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right);

vˉ\bar{v}= R.22 (122−132)\left( \frac{1}{2^{2}} - \frac{1}{3^{2}} \right)Ž 5R9\frac{5R}{9}