Question

Find out the electronic state from which an electron drops to emit radiation with a wavelength of $926\stackrel{o}{A}$ in ultraviolet region (lyman series) of spectrum of hydrogen atom.
Rydberg constant = $1.097\times {10}^7{m}^{-1}$

• A. n = 4
• B. n = 5
• C. n = 6
• D. n = 8

Answer 1

The correct option is D n = 8

Given: wavelength of the radiation = $\lambda =926A^{\circ }=926\times {10}^{-10}m$
We know,
$\overline{v}=\frac{1}{\lambda }=R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
where, $\overline{v}$ = wave number of the radiation.
Since the radiation is in the U.V region and belongs to Lyman series therefore $n_1=1$
$\therefore \overline{v}=\frac{1}{926\times {10}^{-10}m}=1.097\times {10}^7{m}^{-1}(\frac{1}{1^2}-\frac{1}{n_2^2})1-\frac{1}{n_2^2}=\frac{1}{9.26\times {10}^{-10}\times 1.097\times {10}^7}$
On solving, $n_2$ = 8