Atoms & Nuclei

JEE Main level, single correct, medium

84 of 239

Question 84 of 239

In the Bohr's model of hydrogen-like atom the force between the nucleus and the electron is modified as F=e24πε0(1r2+βr3)F=\frac{e^2}{4 \pi \varepsilon_0}\left(\frac{1}{r^2}+\frac{\beta}{r^3}\right), where β\beta is a constant. For this atom, the radius of the nth n^{\text {th }} orbit in terms of the Bohr radius (a0=ε0h2mπe2)\left(a_0=\frac{\varepsilon_0 h^2}{m \pi e^2}\right) is :

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Answer: (C) rn=a0n2−βr_n=a_0 n^2-\beta

As F=m2r=e24πϵ0(1r2+Br3)F=\frac{m^2}{r}=\frac{e^2}{4 \pi \epsilon_0}\left(\frac{1}{r^2}+\frac{B}{r^3}\right) and mvr=nh2π⇒v=nh2πmr\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi} \Rightarrow \mathrm{v}=\frac{\mathrm{nh}}{2 \pi \mathrm{mr}} ∴m(nh2πmr)2×1r=e24πϵ0(1r2+Br3)\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, 1r2+Br3=mn2 h24πϵ04π2 m2e2r3\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}

 or, a0n2r3=1r2+Br3(∵a0=ϵ0 h2 mπe2 Given )∴r=a0n2−B\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}

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