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Q.

The frequency of one of the lines is Paschen series of hydrogen atom is $2 .340 \times 10^{11} Hz .$ The quantum number $n_{2}$ which produces this transition is

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a

6

b

3

c

4

d

5

answer is B.

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Detailed Solution

$\begin{array}{l} \bar{v} = \frac{1}{λ} R_{H} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] \\ = \frac{1}{λ} R_{H} [\frac{1}{3^{2}} - \frac{1}{n_{2}^{2}}] Since n_{1} = 3 \\ for Paschen series . If n_{2} = 5 then v = 2 .340 \times 10^{14} Hz \end{array}$

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The frequency of one of the lines is Paschen series of hydrogen atom is 2.340 × 1011Hz. The quantum number n2 which produces this transition is