Question

The ionization energy of hydrogen atom is $13.6$ eV. The longest wavelength of hydrogen spectrum in the ultraviolet region is expected to be:

• A. $121.8$ nm
• B. $102.8$ nm
• C. $97.44$ nm
• D. $95.11$ nm

Answer 1

Answer: (A)

$121.8$ nm

The Lyman series is the only set of hydrogen spectrum in the UV region. So, the line with the longest wavelength of this series, i.e., the one with the shortest transition will be from $n=2$ to $n=1$.
If the wavelength is calculated using the Bohr's equation for electronic transitions using the Rydberg's constant, then

$\frac{1}{\lambda }=-R(\frac{1}{n_2^2}-\frac{1}{n_1^2})$$=-R(\frac{1}{2^2}-\frac{1}{1^2})$

$\Delta E=\frac{hc}{\lambda }=-13.6hc(\frac{1}{n_2^2}-\frac{1}{n_1^2})$

or, $\Delta E=-13.6hc(\frac{1}{4}-\frac{1}{1})$

or, $\Delta E=10.2eV$

or, $\frac{hc}{\lambda }=10.2eV$

$\therefore \lambda =\frac{hc}{10.2\times 1.6\times {10}^{-19}J}$ $=121.8nm.$