Question

An hydrogen atom ionisation energy 13.6 ev jumps from third excited state to first excited state the energy of photon emitted in process is ?

Answer 1

According to Bohr's model, the energy of n=n quantum state in hydrogen atom is given by

En= -13.6 eV/n^2. Here, n=1,2,3,….. n is called in general principal quantum number.

n=1 gives ground state, n=2 gives first excited state and so on.

n=4 state is called third excited state and in this state energy of electron is:

E4=-13.6/4^2=-0.85 eV. …….(1)

First excited state corresponds to n=2. Energy in this state is:

E2=-13.6/2^2=-3.4 eV…..(2)

Energy of photon resulting from translation from n=4 to n=2 state is:

E=E4-E2=-0.85+3.4=2.55 eV =2.55x1.6x10^-19J=4.08x10^-19 J.