Answer:
initial energy state is n = 4
Explanation:
For hydrogen atom when electron makes transition from higher energy state to lower energy state then the photon is released of energy which equal to the energy difference of two states
so here we can say
[tex]h\nu = E_2 - E_1[/tex]
so here we have energy at nth excited state of electron is given as
[tex]E_n = -13.6 \frac{z^2}{n^2} eV[/tex]
here z = 1 for hydrogen
so now we have
[tex]\frac{hc}{\lambda} = -13.6 \frac{1^2}{n^2} + 13.6 \frac{1^2}{2^2}[/tex]
here we know that
[tex]hc = 1242 eV-nm[/tex]
now plug in all values in it
[tex]\frac{1242 eV-nm}{487 nm} = 13.6 eV(\frac{1}{4} - \frac{1}{n^2})[/tex]
[tex]0.187 = 0.25 - \frac{1}{n^2}[/tex]
by solving above equation we got
n = 4
