Answer:
a) [tex]E_photon =0.306 eV[/tex]
b) [tex]E_photon =0.166 eV[/tex]
Explanation:
The energy of the photon (E) for [tex]n^th[/tex] orbit of the hydrogen atom is given as:
[tex]E_photon = E_o(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}})[/tex]
where,
[tex]E_o[/tex] = 13.6 eV
n = orbit
a) Now for the transition from n = 4 to n = 5
[tex]E_photon =13.6(\frac{1}{4^{2}}-\frac{1}{5^{2}})[/tex]
[tex]E_photon =0.306 eV[/tex]
b) Now for the transition from n = 5 to n = 6
[tex]E_photon =13.6(\frac{1}{5^{2}}-\frac{1}{6^{2}})[/tex]
[tex]E_photon =0.166 eV[/tex]
