Question

Calculate the energy of radiation corresponding to spectrum line of lowest frequency in Lyman series in the spectra of of $L{i}^{2+}$.

• A. $14.7\times {10}^{-18}J$
• B. $16.35\times {10}^{-19}J$
• C. $16.35\times {10}^{-18}J$
• D. $4.9\times {10}^{-18}J$

Answer 1

The correct option is A $14.7\times {10}^{-18}J$

For electronic transition of $H$-atom
$\frac{1}{\lambda }=R_H[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
For Lyman series, $n_1=1$
Also, for lowest frequency,i.e., minimum energy photon, $n_2=2$
$\frac{1}{\lambda }=R_H[\frac{1}{1^2}-\frac{1}{2^2}]=109678\times \frac{3}{4}$
$\lambda =1215.6\times {10}^{-8}cm$
$\therefore \lambda =1215.6\mathring{A}$
Also, $v=\frac{c}{\lambda }=\frac{3.0\times {10}^8}{1215.6\times {10}^{-10}}=2.468\times {10}^{15}Hz$
and $E=hv=6.626\times {10}^{-27}\times 2.468\times {10}^{15}$
$=16.35\times {10}^{-12}erg$
$=16.35\times {10}^{-19}J$
Furthermore energy of corresponding line of $L{i}^{2+}=E_n\times Z^2=16.35\times {10}^{-19}\times 9$
$14.7\times {10}^{-18}J$