Answer :
The energy levels of the hydrogen atom are given by
[tex]E_n = -13.6 \frac{1}{n^2} [eV] [/tex]
where n is the level number. Let's use this formula to calculate the energy of the levels n=1 (ground state), n=5 and n=6:
- ground state:
[tex]E_1 = -13.6 \frac{1}{1^2} eV=-13.6 eV = -2.18 \cdot 10^{-18} J[/tex]
- level n=5:
[tex]E_5 = -13.6 \frac{1}{5^2} eV = -13.6 \frac{1}{25} eV=-0.54 eV =-8.64 \cdot 10^{-20} J [/tex]
- level n=6:
[tex]E_6 = -13.6 \frac{1}{6^2}eV= -13.6 \frac{1}{36} eV = -0.38 eV = -6.08 \cdot 10^{-20} J[/tex]
Now that we have the energy for all the levels we are interested in, we can calculate the energy of the emitted photons.
a) In the first transition, the atom goes from n=6 to n=5. The energy of the emitted photon is equal to the energy difference between these two levels:
[tex]E=E_6-E_5 =-6.09 \cdot 10^{-20} J-(-8.65 \cdot 10^{-20}J)=2.56 \cdot 10^{-20}J[/tex]
The energy of a photon is also equal to
[tex]E=hf=h \frac{c}{\lambda} [/tex]
where
h is the Planck constant
c is the speed of light
f is the photon frequency
[tex]\lambda[/tex] is its wavelength
Re-arranging this relationship and using the photon's energy, we find its wavelength:
[tex]\lambda= \frac{hc}{E}= \frac{(6.63 \cdot 10^{-34}Js)(3 \cdot 10^8 m/s)}{2.56 \cdot 10^{-20} J}=7.77 \cdot 10^{-6} m = 7770 nm[/tex]
b) in the second transition, from n=5 to n=1, the energy of the emitted photon is equal to the difference in energy between the two levels:
[tex]E=E_5 - E_1 = -8.65 \cdot 10^{-20} J -(-2.18 \cdot 10^{-18} J) =2.09 \cdot 10^{-18} J[/tex]
Similarly to part a), the wavelength of the photon is given by:
[tex]\lambda= \frac{hc}{E}= \frac{(6.6 \cdot 10^{-34}Js)(3 \cdot 10^8 m/s)}{2.09 \cdot 10^{-18} J}=9.52 \cdot 10^{-8} m = 95.2 nm[/tex]