Question

Calculate energy of the radiation in kcal/mol corresponding to the spectral line of lowest frequency in Lyman series in the spectrum of $H-$ atom

Answer 1

For Lyman series $n_1=1,n_2=2$

$\Rightarrow \frac{1}{\lambda }=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]z^2$ $z=1$

$=1.09\times {10}^7[\frac{1}{1}-\frac{1}{2^2}]$

$=1.09\times {10}^7\left[\frac{3}{4}\right]$

$\Rightarrow \lambda =1216\stackrel{o}{A}$

$\therefore E=\frac{hc}{\lambda }=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{1216\times {10}^{-10}}=0.0163\times {10}^{-16}J$

$\Rightarrow 1J=0.000239Kcal$

$\Rightarrow E=0.0163\times {10}^{-16}\times 0.000239$

$=3.89\times {10}^{-22}Kcal$