Question
If ionisation potential for hydrogen atom is $$13.6\,\,eV,$$ then ionisation potential for $$H{e^ + }$$ will be
A.
54.4$$\,\,eV$$
B.
6.8$$\,\,eV$$
C.
13.6$$\,\,eV$$
D.
24.5$$\,\,eV$$
Answer :
54.4$$\,\,eV$$
Solution :
$$\eqalign{
& {\text{For hydrogen atom}}\,\,Z = 1 \cr
& \therefore \,\,{\text{Ionisation energy,}}\,{E_H} = \frac{{2{\pi ^2}m{e^4}}}{{{n^2}{h^2}}}\,\,...{\text{(i)}} \cr
& {\text{For }}H{e^ + }{\text{ }}ion,\left( {H{e^ + } = 1{s^1}} \right) \cr
& {\text{so, }}(H{e^ + } = H)\,{\text{ionisation energy,}} \cr
& {{\text{E}}_{H{e^ + }}} = \frac{{2{\pi ^2}m{e^4}{Z^2}}}{{{n^2}{h^2}}}\,\,...{\text{(ii)}} \cr
& Eq{\text{ (i)}}/Eq{\text{ (ii)}},{\text{ we get}} \cr
& {E_{H{e^ + }}} = {E_H} \times {Z^2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 13.6 \times 4 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 54.4\,\,eV \cr} $$