Question

Calculate energy, frequency and wavelength of the radiation which is corresponding to the spectral line of the lowest frequency in Lyman series in the hydrogen atom spectrum . Also calculate the energy for the corresponding line in the spectrum of $L{i}^{2+}$
($R_H=1.09677c{m}^{-1},c=3x{10}^{8}m{s}^{-1},Z=3$)

Answer 1

$\frac{1}{\lambda }=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]z^2$ $R=1.097\times {10}^7;n_1=1;n_2=2$

$\Rightarrow \frac{1}{\lambda }=1.097\times {10}^7[\frac{1}{1}-\frac{1}{2^2}]1$

$\Rightarrow \frac{1}{\lambda }=1.097\times {10}^7\left(\frac{3}{4}\right)$

$\therefore \lambda =121.6nm$

Frequency $\gamma =\frac{C}{\lambda }$

$=\frac{3\times {10}^8}{121.6\times {10}^{-9}}=0.0246\times {10}^{17}$

$=2.46\times {10}^{15}{S}^{-1}$

Energy $E=h\gamma$

$=6.626\times {10}^{-34}\times 2.46\times {10}^{15}$

$=16.3\times {10}^{-19}J$

Energy for the corresponding line in the spectrum of $L{i}^{2+}$

$\frac{1}{\lambda }=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]z^2,z=3$ for $L{i}^{2+}$

$\Rightarrow \frac{1}{\lambda }=1.09677\times {10}^7[\frac{1}{1}-\frac{1}{2^2}]3^2$

$=1.096\times {10}^7\left[\frac{3}{4}\right]9$

$\lambda =\frac{4}{1.096\times {10}^7\times 3\times 9}$

$=\frac{4}{29.592\times {10}^7}=0.1355\times {10}^{-7}$

$=13.55nm$

$E=\frac{hC}{\lambda }=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{0.1355\times {10}^{-7}}$

$=146.7\times {10}^{-19}J$ .