Question

A hydrogen atom in a state having a binding energy of 0.85 eV 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

The wavelength of the emitted radiation 487 nm

Explanation:

The binding energy of hydrogen is given by  

E = $\frac {13.6}{n^6} eV$

For binding energy of 0.85 eV,

$n_{2}^{2} = \frac {13.6}{0.85} = 16$

$n_2 = 4$

For binding Energy of 10.2 eV,

$n_{1}^{2} = \frac {13.6}{10.2}$

$n_1 = 1.15\ or\ n_1 = 2$

The quantum number of the upper and the lower energy state are 4 and 2, respectively.

(b) Wavelength of the emitted radiation (\lambda) is given by

$\frac {1}{\lambda} = R(n_{1}^{2} - n_{2}^{2})$

Here,

R = Rydberg constant  

$n_1\ and\ n_2$ are quantum numbers.

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

So, $\lambda = \frac {16}{1.097 \times 3 \times 10^-^7}$

= $4.8617 \times 10^-^7$

= 487 nm