Question

A particle with mass m is held in circular orbit around the origin by an attractive force F(r)=-Dr; where D is a positive constant. Assume that the Bohr model idea that ,the angular momentum is quantized i.e. it takes only integral multiples of $\frac{h}{2\pi }$ holds.

The value of energy in the $n^{th}$ orbit (Let the potential energy be zero when r = 0)

• A. $\frac{nh}{\pi }\sqrt{\frac{D}{m}}$
  
• B. ${\left(\frac{n^2{h}^{2}D}{{\pi }^{2}m}\right)}^{\frac{1}{4}}$
  
• C. $\frac{nh}{2\pi }\sqrt{\frac{D}{m}}$
  
• D. ${\left(\frac{n^2{h}^{2}D}{2{\pi }^{2}m}\right)}^{\frac{1}{4}}$

Answer 1

The correct option is C

$\frac{nh}{2\pi }\sqrt{\frac{D}{m}}$

${\int }_0^{r}dU={\int }_0^r\overrightarrow{f}.\overrightarrow{dr}U\left(r\right)=\frac{D(r_n{)}^2}{2}$ and $K=\frac{1}{2}m(v_n{)}^2=\frac{1}{2}D(r_n{)}^2\therefore T.E=U+K=D(r_n{)}^2=\frac{nh}{2\pi }\sqrt{\frac{D}{M}}$