Solveeit Logo
Login

Question

Chemistry Question on Bohr’s Model for Hydrogen Atom

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a0a_0 is Bohr radius]

A

h24π2ma02\frac{h^2}{4 \pi^2 ma_0^2}

B

h216π2ma02\frac{h^2}{16 \pi^2 ma_0^2}

C

h232π2ma02\frac{h^2}{32 \pi^2 ma_0^2}

D

h264π2ma02\frac{h^2}{64 \pi^2 ma_0^2}

Answer

h232π2ma02\frac{h^2}{32 \pi^2 ma_0^2}

Explanation

Solution

According to Bohr's model, \hspace15mm mvr=\frac{nh}{2 \pi} \, \, \, \Rightarrow \, \, (mv)^2=\frac{n^2 h^2}{4 \pi^2 r^2}
\hspace15mm KE=\frac{1}{2} mv^2 =\frac{n^2 h^2}{8\pi^2 r^2 m} \hspace25mm ...(i)
Also, Bohr's radius for H-atom is, r =n2 a0n^2 \, a_0 Substituting 'r' in E (i) gives
      KE=h28π2 n2 a02m  when n=2,KE=h232 π2 a02 m\, \, \, \, \, \, KE =\frac{h^2}{8 \pi^2 \, n^2 \, a_0^2 m} \, \, when \, n=2, KE=\frac{h^2}{32 \, \pi^2 \, a_0^2 \, m}