SATHEE: Atomic Structure Question 41

Atomic Structure Question 41

Question: The frequency corresponding to transition $n = 2$ to $n = 1$ in hydrogen atom is

Options:

A) $15.66 \times 10^{10} H z$

B) $24.66 \times 10^{14} H z$

C) $30.57 \times 10^{14} H z$

D) $40.57 \times 10^{24} H z$

Show Answer

Answer:

Correct Answer: B

Solution:

$v = \frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] = 109678 [\frac{1}{1} - \frac{1}{4}] = 82258.5$

$λ = 1.21567 \times 10^{- 5} c m o r λ = 12.1567 \times 10^{- 6} c m$

$= 12.1567 \times 10^{- 8}, m$

$v = \frac{c}{λ} = \frac{3 \times 10^{8}}{12.567 \times 10^{- 8}} = 24.66 \times 10^{14} H z$ .

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