Question

A particle of mass m is moving in a circular orbit under the influence of the central force F(r) = -kr, corresponding to the potential energy \(V(r)=k r^{2} / 2\), where k is a positive force constant and r is the radial distance from the origin.  According to the Bohr's quantization rule, the angular momentum of the particle is given by L = nh, where  \(h=h /(2 \pi), h\) is the Planck's constant, and n a positive integer. If v and E are the speed and total energy of the  particle, respectively, then which of the following expression(s) is(are) correct?

(A) \(r^{2}=n h \sqrt{\frac{1}{m k}}\)

(B) \(v^{2}=n h \sqrt{\frac{k}{m^{3}}}\)

(C) \(\frac{L}{m r^{2}}=\sqrt{\frac{k}{m}}\)

(D) \(E=\frac{n h}{2} \sqrt{\frac{k}{m}}\)

Answer 1

Correct option is A,B,C

\(L=m v r=n h\), also \(\frac{m v^{2}}{r}=k r\)

\( m v^{2}=k r^{2} \)

\(m^{2} v^{2}=m k r^{2}\)

\(m v=\sqrt{m k r^{2}}\)

\(m v r=r^{2} \sqrt{m k}\)

\( n h=r^{2} \sqrt{m k}\)

\( \frac{n h}{\sqrt{m k}}=r^{2} \)

Option (A) is correct

Also, \(v^{2}=\frac{k r^{2}}{m}\)

\(=\frac{n h}{\sqrt{m k}} \cdot \frac{k}{m}\)

\(v^{2}=n h \sqrt{\frac{k}{m^{3}}} \)

Option (B) is correct

Now,

\(\mathrm{T} . \mathrm{E}=E=\frac{1}{2} k r^{2}+\frac{1}{2} k r^{2}\)

\(E=k r^{2}\)

\(=k \frac{n h}{\sqrt{m k}}\)

\(=n h \sqrt{\frac{k}{m}}\)

Option (D) is incorrect

\(\Rightarrow \frac{L}{m r^{2}}=\frac{h \sqrt{m k}}{m n h}\)

\(\frac{L}{m r^{2}}=\sqrt{\frac{k}{m}}\)

Option (C) is correct