In the Bohr's model of hydrogen-like atom the force between the nucleus and the electron is modified as…

In the Bohr's model of hydrogen-like atom the force between the nucleus and the electron is modified as $F=\frac{e^2}{4 \pi \varepsilon_0}\left(\frac{1}{r^2}+\frac{\beta}{r^3}\right)$, where $\beta$ is a constant. For this atom, the radius of the $n^{\text {th }}$ orbit in terms of the Bohr radius $\left(a_0=\frac{\varepsilon_0 h^2}{m \pi e^2}\right)$ is :
  1. $r_n=a_0 n-\beta$
  2. $r_n=a_0 n^2+\beta$
  3. $r_n=a_0 n^2-\beta$
  4. $r_n=a_0 n+\beta$

Solution

As $F=\frac{m^2}{r}=\frac{e^2}{4 \pi \epsilon_0}\left(\frac{1}{r^2}+\frac{B}{r^3}\right)$ and $\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi} \Rightarrow \mathrm{v}=\frac{\mathrm{nh}}{2 \pi \mathrm{mr}}$ $\therefore \mathrm{m}\left(\frac{\mathrm{nh}}{2 \pi \mathrm{mr}}\right)^2 \times \frac{1}{\mathrm{r}}=\frac{\mathrm{e}^2}{4 \pi \epsilon_0}\left(\frac{1}{\mathrm{r}^2}+\frac{\mathrm{B}}{\mathrm{r}^3}\right)$ or, $\frac{1}{\mathrm{r}^2}+\frac{\mathrm{B}}{\mathrm{r}^3}=\frac{\mathrm{mn}^2 \mathrm{~h}^2 4 \pi \epsilon_0}{4 \pi^2 \mathrm{~m}^2 \mathrm{e}^2 \mathrm{r}^3}$ $ \begin{aligned} & \text { or, } \frac{\mathrm{a}_0 \mathrm{n}^2}{\mathrm{r}^3}=\frac{1}{\mathrm{r}^2}+\frac{\mathrm{B}}{\mathrm{r}^3} \\ & \left(\because \mathrm{a}_0=\frac{\epsilon_0 \mathrm{~h}^2}{\mathrm{~m} \pi \mathrm{e}^2} \text { Given }\right) \\ & \therefore \mathrm{r}=\mathrm{a}_0 \mathrm{n}^2-\mathrm{B} \end{aligned} $

Asked in: JEE Main 2013 (23 Apr Online)

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