Explanation:
It is known that formula for energy in the 1st exited state is as follows.
E = [tex]\frac{-13.6 eV}{n^{2}}[/tex]
And, for 1st exited state n = 2.
So, E = [tex]\frac{-13.6 eV}{4}[/tex]
= -3.4eV
It is given that energy added E' = 2.86 eV.
So, energy of the electron in nth state is as follows.
E" = E + E'
= -0.54 eV
we know E" = [tex]\frac{-13.6 eV}{n'^{2}}[/tex]
-0.54 eV = [tex]\frac{-13.6 eV}{n'^{2}}[/tex]
n' = 5
i.e., the electron exited from n = 1 to n = 5.
Thus, we can conclude that the quantum number n of the state into which the electron moves is from n = 1 to n = 5.
