In the Bohr's model of hydrogen-like atom the force between the nucleus and the electron is modified as F = (e²)/(4πε₀)(1/r² + (β)/(r³)), where β is a constant. For this atom, the radius of the nth orbit in terms of the Bohr radius (a₀ = (ε₀ h²)/(mπ e²)) is: (a) rn = a₀ n - β (b) rn = a₀ n² + β (c) rn = a₀ n² - β (d) rn = a₀ n + β

Answer: (c). mv²/r = (e²)/(4πε₀)·(r + β)/(r³) and Bohr condition v = (nh)/(2π m r). Substitute: (n²h²)/(4π² m r³) = (e²)/(4πε₀)·(r + β)/(r³). r + β = (ε₀ n² h²)/(π m e²) = a₀ n² ⇒ rn = a₀ n² - β So the answer is (c) rn = a₀ n² - β.

In the Bohr's model of hydrogen-like atom the force between the… | JEE Main 2013 (23 Apr, Shift 1)

Q115JEE Main 2013 · 23 APR · SHIFT 1Difficult

Bohr Model: Orbits, Speed & Angular Momentum

In the Bohr's model of hydrogen-like atom the force between the nucleus and the electron is modified as $F = \frac{e^2}{4\pi\epsilon_0}\left(\frac{1}{r^2} + \frac{\beta}{r^3}\right)$ , where $\beta$ is a constant. For this atom, the radius of the $n^{th}$ orbit in terms of the Bohr radius $\left(a_0 = \frac{\epsilon_0 h^2}{m\pi e^2}\right)$ is:

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Answer: (c)

$\frac{mv^2}{r} = \frac{e^2}{4\pi\epsilon_0}\cdot\frac{r + \beta}{r^3}$ and Bohr condition $v = \frac{nh}{2\pi m r}$ .
Substitute: $\frac{n^2h^2}{4\pi^2 m r^3} = \frac{e^2}{4\pi\epsilon_0}\cdot\frac{r + \beta}{r^3}$ .
$r + \beta = \frac{\epsilon_0 n^2 h^2}{\pi m e^2} = a_0 n^2 \Rightarrow r_n = a_0 n^2 - \beta$
So the answer is (c) $r_n = a_0 n^2 - \beta$ .

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