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  4. The wave number of the spectral line in the emission spectrum of hydrogen will be equal to (8/9) times the Rydberg's constant if the electron jumps from

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Q. The wave number of the spectral line in the emission spectrum of hydrogen will be equal to $\frac {8}{9}$ times the Rydberg's constant if the electron jumps from _______

KCETKCET 2010Structure of Atom

Solution:

Wave number of spectral line in emission spectrum of hydrogen,

$\bar{v}=R_{H}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$...(i)

Given, $\bar{v}=\frac{8}{9} R_{H}$

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

$\frac{8}{9} R_{ H } =R_{ H }\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$

$\frac{8}{9} =\frac{1}{(1)^{2}}-\frac{1}{n_{2}^{2}}$

$\frac{8}{9}-1 =-\frac{1}{n_{2}^{2}}$

$\frac{1}{3} =\frac{1}{n_{2}}$

$\therefore n_{2}=3$

Hence, electron jumps from $n_{2}=3$ to $n_{1}=1$