Question
The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [ $${{a_0}}$$ is Bohr radius ] :
A.
$$\frac{{{h^2}}}{{4{\pi ^2}ma_0^2}}$$
B.
$$\frac{{{h^2}}}{{16{\pi ^2}ma_0^2}}$$
C.
$$\frac{{{h^2}}}{{32{\pi ^2}ma_0^2}}$$
D.
$$\frac{{{h^2}}}{{64{\pi ^2}ma_0^2}}$$
Answer :
$$\frac{{{h^2}}}{{32{\pi ^2}ma_0^2}}$$
Solution :
$$\eqalign{
& {\text{As per Bohr's}}\,{\text{postulate,}} \cr
& mvr = \frac{{nh}}{{2\pi }}\,\,{\text{So}},\,\,\upsilon = \frac{{nh}}{{2\pi mr}} \cr
& KE = \frac{1}{2}m{\upsilon ^2}\,{\text{So}},\,\,KE = \frac{1}{2}m{\left( {\frac{{nh}}{{2\pi mr}}} \right)^2} \cr
& {\text{Since}},\,\,r = \frac{{{a_0} \times {n^2}}}{z} \cr
& {\text{So,}}\,{\text{for}}\,{{\text{2}}^{{\text{nd}}}}\,{\text{Bohr}}\,{\text{orbit}} \cr
& r = \frac{{{a_0} \times {2^2}}}{1} = 4{a_0} \cr
& KE = \frac{1}{2}m\left( {\frac{{{2^2}{h^2}}}{{4{\pi ^2}{m^2} \times {{\left( {4{a_0}} \right)}^2}}}} \right) \cr
& = \frac{{{h^2}}}{{32{\pi ^2}ma_0^2}} \cr} $$