Question

Hydrogen atoms are excited by a laser to the n = 4 state and then allowed to emit. What is the maximum number of distinct emission spectral lines (lines of different wavelengths) that can be observed from this system? Calculate the wavelength of the 4 -> 2 transition.

Answer 1

An atom of hydrogen emits electromagnetic waves when the initial state is greater than the final state (), so the changes of state that will emits an electromagnetic wave are:

• $n_i=4\rightarrow n_f=3$
• $n_i=4\rightarrow n_f=2$
• $n_i=4\rightarrow n_f=1$

In our systme we need to have to take in count to the emission when the initial state is lower than n_i =4.That is:

• When n_i =3:
  • $n_i=3\rightarrow n_f=2$
  • $n_i=3\rightarrow n_f=1$
• When n_i =2:
  • $n_i=2\rightarrow n_f=1$
• When n_i =1:
  • This is the lowest state of an atom, at this state the atom cannot emits energy.

So the maximum number of distinct emission lines will be 6. This is regarless of the atomic numbers and , because they are not mentioned in the header of the problem.

When the atom change from one state to another smaller state, the formula to calculate the wavelength is

$\frac{1}{\lambda}=R \left(\frac{1}{(n_f)^2}-\frac{1}{(n_i)^2} \right)$

where

$R = 1.0973732 \times 10^{7}m^{-1}$

so from the change of state 4 $\rightarrow$ 2

$\frac{1}{\lambda}= 1.0973732 \times 10^{7}m^{-1}\left(\frac{1}{(2)^2}-\frac{1}{(4)^2} \right)$

$\frac{1}{\lambda}= 1.0973732 \times 10^{7}m^{-1 }\left(\frac{1}{(2)^2}-\frac{1}{(4)^2} \right)$

$\frac{1}{\lambda } = 2057574.8m^{-1}$

$\lambda= 4.86 \times 10^{-7}m$