Question
The energy of second Bohr orbit of the hydrogen atom is $$ - 328\,kJ\,mo{l^{ - 1}},$$ hence the energy of fourth Bohr orbit would be
A.
$$ - 41\,kJ\,mo{l^{ - 1}}$$
B.
$$ - 1312\,kJ\,mo{l^{ - 1}}$$
C.
$$ - 164\,kJ\,mo{l^{ - 1}}$$
D.
$$ - 82\,kJ\,mo{l^{ - 1}}$$
Answer :
$$ - 82\,kJ\,mo{l^{ - 1}}$$
Solution :
The energy of second Bohr orbit of hydrogen atom $$\left( {{E_2}} \right)$$ is $$ - 328\,kJ\,mo{l^{ - 1}}$$
$$\eqalign{
& \,\,\,\,\,\,\,{E_n} = - \frac{{1312}}{{{n^2}}}kJ\,mo{l^{ - 1}} \cr
& \therefore \,\,{E_2} = - \frac{{1312}}{{{2^2}}}kJ\,mo{l^{ - 1}} \cr
& {\text{If}}\,\,n = 4 \cr
& \therefore \,\,{E_4} = - \frac{{1312}}{{{4^2}}}kJ\,mo{l^{ - 1}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 82\,kJ\,mo{l^{ - 1}} \cr} $$