Question

The energy of electron in the nth orbit of hydrogen atom is expressed as $E_n=\frac{-13.6}{n^2}eV$. The shortest and longest wavelength of Lyman series will be

• A. $910\stackrel{{}_{\circ }}{A}$, $1213\stackrel{{}_{\circ }}{A}$
• B. $5463\stackrel{{}_{\circ }}{A}$, $7858\stackrel{{}_{\circ }}{A}$
• C. $1315\stackrel{{}_{\circ }}{A}$, $1530\stackrel{{}_{\circ }}{A}$
• D. None of these

Answer 1

The correct option is A $910\stackrel{{}_{\circ }}{A}$, $1213\stackrel{{}_{\circ }}{A}$

The energy difference between $n_1$ and $n_2$ orbit of hydrogen atom is given by: $\Delta E=-13.6(\frac{1}{n_2^2}-\frac{1}{n_1^2})eV$

Also, $\Delta E=\frac{hc}{\lambda }=\frac{1242}{\lambda \left(nm\right)}$ where $\lambda$ is the wavelength of the spectral line due to that transition

$\therefore$ The wavelength of the spectral line observed due to the transition from $n_2$ to $n_1$ state of a hydrogen atom is given by $\frac{1}{\lambda \left(nm\right)}=0.01097(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Let the shortest and longest wavelengths of Lyman series be ${\lambda }_s$ and ${\lambda }_l$ respectively.

Now for shortest wavelength of lyman series, $n_1=1$ and $n_2=\infty$

$\therefore \frac{1}{{\lambda }_s\left(nm\right)}=0.01097(\frac{1}{1^2}-\frac{1}{\infty })⟹{\lambda }_s\approx 91.15nm=911.5A^o$

For longest wavelength of Lyman series, $n_1=1$ and $n_2=2$

$\therefore \frac{1}{{\lambda }_l\left(nm\right)}=0.01097(\frac{1}{1^2}-\frac{1}{2^2})⟹{\lambda }_l\approx 121.54nm=1215.4A^o$