Answer:
[tex]\lambda = 102.7 nm[/tex]
Explanation:
As we know by Bohr's Postulates that when electron makes transition from higher excited state to lower excited state then energy difference of two levels is released in the form of photons.
So we can say
[tex]\frac{hc}{\lambda} = E_2 - E_1[/tex]
here we know that energy of nth excited state in any atom is given as
[tex]E_n = -13.6 \frac{z^2}{n^2} eV[/tex]
here for hydrogen atom z = 1
[tex]n_2 = 3[/tex]
[tex]n_1 = 1[/tex]
also we know that
[tex]hc = 1242 eV -nm[/tex]
now from above equation we have
[tex]\frac{1242 eV-nm}{\lambda} = (-13.6 \frac{1^2}{3^2})-(-13.6 \frac{1^2}{1^2})eV[/tex]
[tex]\frac{1242 eV-nm}{\lambda} = 13.6 (1 - \frac{1}{9}) eV[/tex]
[tex]\lambda = 102.7 nm[/tex]
