Question
Ionisation energy of $$H{e^ + }$$ is $$19.6 \times {10^{ - 18}}J\,ato{m^{ - 1}}.$$ The energy of the first stationary state $$( n = 1 )$$ of $$L{i^{2 + }}$$ is
A.
$$4.41 \times {10^{ - 16}}J\,ato{m^{ - 1}}$$
B.
$$ - 4.41 \times {10^{ - 17}}J\,ato{m^{ - 1}}$$
C.
$$ - 2.2 \times {10^{ - 15\,}}J\,ato{m^{ - 1}}$$
D.
$$8.82 \times {10^{ - 17}}J\,ato{m^{ - 1}}$$
Answer :
$$ - 4.41 \times {10^{ - 17}}J\,ato{m^{ - 1}}$$
Solution :
$$\eqalign{
& I.E = \frac{{{Z^2}}}{{{n^2}}} \times 13.6\,eV\,...\left( {\text{i}} \right) \cr
& {\text{or}}\,\frac{{{I_1}}}{{{I_2}}} = \frac{{Z_1^2}}{{n_1^2}} \times \frac{{n_2^2}}{{Z_2^2}}\,...\left( {{\text{ii}}} \right) \cr
& {\text{Given,}} \cr
& {I_1} = - 19.6 \times {10^{ - 18}},\, \cr
& {Z_1} = 2,{n_1} = 1,{Z_2} = 3\,\,{\text{and}}\,\,{n_2} = 1 \cr
& {\text{Substituting these values in equation (ii)}}{\text{.}} \cr
& - \frac{{19.6 \times {{10}^{ - 18}}}}{{{I_2}}} = \frac{4}{1} \times \frac{1}{9} \cr
& {\text{or}}\,{I_2} = - 19.6 \times {10^{ - 18}} \times \frac{9}{4} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = - 4.41 \times {10^{ - 17}}J/atom \cr} $$