Question

A hydrogen atom in a state having a binding energy of $0.85eV$ makes transition to a state with excitation energy $10.2eV$
Find the wavelength of the emitted radiation.

Answer 1

From the energy data we see that the $H$ atom transists from binding energy of $0.85$ ev to exitation energy of $10.2$ ev $=$ Binding Energy of $–3.4$ ev.

So, $n=4$ to $n=2$

We know that $\frac{1}{\lambda }=1.097\times {10}^7(\frac{1}{4}-\frac{1}{16})$

Wavelength of the emitted radiation$=\lambda =\frac{16}{1.097\times 3\times {10}^7}=4.8617\times {10}^{-7}=487$nm