Question

The electron in a hydrogen atom is in the first excited state, when the electron acquires an additional 2.86 eV of energy. What is the quantum number n of the state into which the electron moves?

Answer 1

Explanation:

It is known that formula for energy in the 1st exited state is as follows.

              E = $\frac{-13.6 eV}{n^{2}}$

And, for 1st exited state n = 2.

So,        E = $\frac{-13.6 eV}{4}$

                = -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" = $\frac{-13.6 eV}{n'^{2}}$

                  -0.54 eV = $\frac{-13.6 eV}{n'^{2}}$

                           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.