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Question

The wave number of the spectral line in the emission spectrum of Hydrogen will be equal to $$\dfrac {8}{9}$$ times the Rydberg's constant if the electron jumps from:


A
n=3 to n=1
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B
n=10 to n=1
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C
n=9 to n=1
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D
n=2 to n=1
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Solution

The correct option is B $$n = 3$$ to $$n = 1$$
Wave number of spectral line in emission spectrum of hydrogen,

$$\bar{v}=R_H\begin{pmatrix}\dfrac{1}{n^2_1}-\dfrac{1}{n^2_2}\end{pmatrix}$$ ...(i)

Given, $$\bar{v}=\dfrac{8}{9}R_H$$

On putting the value of $$\bar{v}$$ in Eq. (i), we get-

$$\dfrac{8}{9}R_H=R_H\begin{pmatrix}\dfrac{1}{n^2_1}-\dfrac{1}{n^2_2}\end{pmatrix}$$

$$\dfrac{8}{9}=\dfrac{1}{(1)^2}-\dfrac{1}{n^2_2}$$

$$\dfrac{8}{9}-1=\dfrac{1}{n^2_2}$$

$$\dfrac{1}{3}=\dfrac{1}{n_2} \therefore n_2=3$$

Hence, electron jumps from $$n_2=3$$ to $$n_1=1$$

Chemistry

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