Question

The longest wavelength in Lyman series of hydrogen atom spectrum corresponding to $H_{\alpha }$ line is approximately.

• A. $1218\stackrel{\circ }{A}$
• B. $216\stackrel{\circ }{A}$
• C. $1016\stackrel{\circ }{A}$
• D. $6578\stackrel{\circ }{A}$

Answer 1

The correct option is A $1218\stackrel{\circ }{A}$

The longest wavelength would correspond to the shortest energy gap. Lyman series corresponds to all the transitions from energy level 2 to $\infty$ to energy level 1.
Hence the minimum energy gap is that from 2 to 1.

Now $\frac{1}{\lambda }=R\times Z^2\times (\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2})$
where $R=1.0968\times {10}^7{m}^{-1}$
Here $n_1=1$ and $n_2=2$
Hence
$\frac{1}{\lambda }=1.0968\times {10}^7\times 1^2\times (\frac{1}{1^2}-\frac{1}{2^2})$
This gives
$\lambda \approx 1218\stackrel{\circ }{A}$