Question

The energies of three conservative energy levels $L_3,L_2$ and $L_1$ of hydrogen atom are $E_0,\frac{4E_o}{9}$ and $\frac{E_1}{4}$ respectively. A photon of wavelength $\lambda$ is emitted for a transition $L_3$ to $L_1$. What will be the wavelength of emission for transition $L_2$ to $L_1$?

Answer 1

ATQ, $E_{L3}={\epsilon }_{\circ }{E}_{L2}=\frac{4{\epsilon }_{\circ }}{9}{E}_{L1}=\frac{{\epsilon }_{\circ }}{4}$

In case-1, when the transition is from L3 to L1, the wavelength $\lambda$ is given by,

$\frac{hc}{\lambda }={\epsilon }_{\circ }-\frac{{\epsilon }_{\circ }}{4}=\frac{{3\epsilon }_{\circ }}{4}\to \lambda =\frac{4hc}{{3\epsilon }_{\circ }}$

The wavelength of photon when transition is from L2 to L1 is given by,

$\frac{hc}{{\lambda }^{{}^{\prime }}}=E_{L2}-E_{L1}=\frac{4{\epsilon }_{\circ }}{9}-\frac{{\epsilon }_{\circ }}{4}=\frac{16{\epsilon }_{\circ }-9{\epsilon }_{\circ }}{36}=\frac{7{\epsilon }_{\circ }}{36}or{\lambda }^{{}^{\prime }}=\frac{36hc}{7{\epsilon }_{\circ }}{\lambda }^{{}^{\prime }}=\frac{27}{7}\lambda$