In the Bohr model of a hydrogen-like atom,the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The centripetal force is provided by the electrostatic force: $\frac{mv^2}{r} = \frac{e^2}{4\pi \varepsilon_0} \left( \frac{1}{r^2} + \frac{\beta}

Vedclass · Accepted Answer

$r_n = a_0 n^2 - \beta$

Vedclass · Answer

(C) The centripetal force is provided by the electrostatic force: $\frac{mv^2}{r} = \frac{e^2}{4\pi \varepsilon_0} \left( \frac{1}{r^2} + \frac{\beta}{r^3} \right)$.From Bohr's quantization condition,$mvr = \frac{nh}{2\pi}$,so $v = \frac{nh}{2\pi mr}$.Substituting $v$ into the force equation: $\frac{m}{r} \left( \frac{nh}{2\pi mr} \right)^2 = \frac{e^2}{4\pi \varepsilon_0} \left( \frac{r + \beta}{r^3} \right)$.Simplifying: $\frac{n^2 h^2}{4\pi^2 m r^3} = \frac{e^2}{4\pi \varepsilon_0} \left( \frac{r + \beta}{r^3} \right)$.Rearranging terms: $\frac{n^2 h^2 \varepsilon_0}{\pi m e^2} = r + \beta$.Given $a_0 = \frac{\varepsilon_0 h^2}{m \pi e^2}$,we get $a_0 n^2 = r + \beta$.Therefore,$r_n = a_0 n^2 - \beta$.

Similar Questions

The angular momentum of an electron in Bohr's hydrogen atom having energy $(-0.544) \text{ eV}$ is ($h$ = Planck's constant)

In a hydrogen atom,the radius of the $n^{th}$ Bohr orbit is $r_n$. The graph between $\log \left( \frac{r_n}{r_1} \right)$ and $\log n$ will be:

If the radius of the first Bohr orbit is $r$, then the radius of the second Bohr orbit will be

Find the ratio of the area of the orbit of the first excited state to the ground state in a hydrogen atom.

Vedclass Test Series

Exam Paper Generator

Online Exam Module