Question

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 _______

• A. $n \,= \,10 \,to\, n \,=\, 1$
• B. $n\, = \,3 \,to\, n \,= \,1$
• C. $n \,= \,2 \,to\, n\, =\, 1$
• D. $n \,= \,9 \,to\, n\, =\, 1$

Answer 1

Answer: (B)

$n\, = \,3 \,to\, n \,= \,1$

Answer 2

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 E (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 $n_{1}=1$