Question

A 12.3 eV electron beam is used to bombard gaseous hydrogen at room temperature. Upto which energy level the hydrogen atoms would be excited? Calculate the wavelength of the second member of Lyman series and second member of Balmer series.

Answer 1

The hydrogen atoms should be excited to $n=3$ as energy diff. between $n=1$ and $n=3$ is $12.08eV$.

$hc=1240eVnm$

For second wavelength of Balmer series:

$\frac{hc}{{\lambda }_2}=\left(13.6eV\right)\{\frac{1}{2^2}-\frac{1}{4^2}\}$

$\frac{1240eV.nm}{{\lambda }_2}=13.6\times \frac{3}{16}$

$\Rightarrow {\lambda }_2=\frac{1240\times 16}{3\times 13.6}nm=486.2nm=4862\dot{A}$

For second wavelength of Lyman series:

$\frac{hc}{{{\lambda }^{\prime }}_2}=\left(13.6eV\right)\{\frac{1}{1^2}-\frac{1}{3^2}\}\frac{1240eVnm}{{{\lambda }^{\prime }}_2}=13.6\times \frac{8}{9}\Rightarrow {\lambda }_2^{\prime }=\frac{1240\times 9}{8\times 13.6}nm=102.5nm=1025\dot{A}$