Question

A hydrogen atom in a state having a binding energy of 0.85 e V makes transition to a state with excitation energy 10.2 e V.

(a) Identify the quantum numbers n of the upper and the lower energy states involved in the transition.

(b) Find the wavelength of the emitted radiation.

Answer 1

(a) From the energy data, we see that the 'H' atom transists from binding energy of 0.85 e V to excitation energy of 102 e V.

= Binding Energy of - 3.4 e V

So, n = 4 to n =2

(b) We know

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

$\Rightarrow \lambda =\frac{16}{1.097\times 3\times {10}^7}$

= $4.8617\times {10}^{-7}$

= 487 nm