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JEE Mains · Physics · STD 12 - 12. atoms
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 \(n^{th}\) orbit in terms of the Bohr radius \(\left( {{a_0} = \,\frac{{{\varepsilon _0}{h^2}}}{{m\pi {e^2}}}} \right)\) is
- A \({r_n} = {a_0}n - \beta \)
- B \({r_n} = {a_0}{n^2} + \beta \)
- C \({r_n} = {a_0}{n^2} - \beta \)
- D \({r_n} = {a_0}n + \beta \)
Answer & Solution
Correct Answer
(C) \({r_n} = {a_0}{n^2} - \beta \)
Step-by-step Solution
Detailed explanation
As \(F=\frac{m v^{2}}{r}=\frac{e^{2}}{4 \pi \epsilon_{0}}\left(\frac{1}{r^{2}}+\frac{B}{r^{3}}\right)\) and \(\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi} \Rightarrow \mathrm{v}=\frac{\mathrm{nh}}{2 \pi \mathrm{m} \mathrm{r}}\)…
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