Question

If the ionisation energy of hydrogen atom is 13.6 eV, the energy required to excite it from ground state to the next higher state is approximately:

• A. 3.4 eV
• B. 10.2 eV
• C. 17.2 eV
• D. 13.6 eV

Answer 1

The correct option is A 3.4 eV

For ionization process: Trnasition is from $n_i=1$ to $n_f=\infty$,

given:$IE=13.6eV$

Transition energy for ionization process is given as:

$\Delta E=E_{\infty }-E_1=IE$

as $E_{\infty }=0$

thus $E_1=-13.6eV$

First excited state is n=2

as $E_n=\frac{E_1}{n^2}$

$E_2=\frac{-13.6}{2^2}=-3.4eV$

Transition energy from ground state i.e. $n=1$ to first excited state i.e. $n=2$ in Hydrogen atom is given as:

$\Delta E=E_2-E_1=-3.4-(-13.6)=10.2eV$