Question:
In the Bohr's model of hydrogen-like atom the force between the nucleus and the electron is modified as \(F=\dfrac{e^2}{4\pi\varepsilon_0}\left(\dfrac{1}{r^2}+\dfrac{\beta}{r^3}\right)\), where \(\beta\) is a constant. For this atom, the radius of the \(n^{th}\) orbit in terms of the Bohr radius \(\left(a_0=\dfrac{\varepsilon_0h^2}{m\pi e^2}\right)\) is :
In the Bohr's model of hydrogen-like atom the force between the nucleus and the electron is modified as \(F=\dfrac{e^2}{4\pi\varepsilon_0}\left(\dfrac{1}{r^2}+\dfrac{\beta}{r^3}\right)\), where \(\beta\) is a constant. For this atom, the radius of the \(n^{th}\) orbit in terms of the Bohr radius \(\left(a_0=\dfrac{\varepsilon_0h^2}{m\pi e^2}\right)\) is :
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Combine the centripetal condition \(mv^2/r=F\) with the quantisation \(mvr=nh/2\pi\).
Updated On: Oct 31, 2026
- \(r_n=a_0n-\beta\)
- \(r_n=a_0n^2+\beta\)
- \(r_n=a_0n^2-\beta\)
- \(r_n=a_0n+\beta\)
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The Correct Option is C
Solution and Explanation
Let \(k=\dfrac{e^2}{4\pi\varepsilon_0}\). Centripetal force: \(\dfrac{mv^2}{r}=k\left(\dfrac1{r^2}+\dfrac{\beta}{r^3}\right)\).
Quantisation: \(mvr=\dfrac{nh}{2\pi}\Rightarrow mv^2=\dfrac{n^2h^2}{4\pi^2mr^2}\).
Substituting: \(\dfrac{n^2h^2}{4\pi^2mr^3}=\dfrac{k(r+\beta)}{r^3}\Rightarrow r+\beta=\dfrac{n^2h^2}{4\pi^2mk}\).
Now \(4\pi^2mk=\dfrac{\pi me^2}{\varepsilon_0}\), so \(\dfrac{h^2}{4\pi^2mk}=\dfrac{\varepsilon_0h^2}{\pi me^2}=a_0\).
Therefore \(r_n=a_0n^2-\beta\).
Answer: (3) \(r_n=a_0n^2-\beta\)
Quantisation: \(mvr=\dfrac{nh}{2\pi}\Rightarrow mv^2=\dfrac{n^2h^2}{4\pi^2mr^2}\).
Substituting: \(\dfrac{n^2h^2}{4\pi^2mr^3}=\dfrac{k(r+\beta)}{r^3}\Rightarrow r+\beta=\dfrac{n^2h^2}{4\pi^2mk}\).
Now \(4\pi^2mk=\dfrac{\pi me^2}{\varepsilon_0}\), so \(\dfrac{h^2}{4\pi^2mk}=\dfrac{\varepsilon_0h^2}{\pi me^2}=a_0\).
Therefore \(r_n=a_0n^2-\beta\).
Answer: (3) \(r_n=a_0n^2-\beta\)
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