In the Bohr model of hydrogen,the force on th | vedclass

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

Question

(B) The electrostatic force $F$ between the nucleus and the electron is given by Coulomb's law: $F = \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r^2}$,which

Vedclass · Accepted Answer

$F \propto \frac{1}{n^4}$

Vedclass · Answer

(B) The electrostatic force $F$ between the nucleus and the electron is given by Coulomb's law: $F = \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r^2}$,which implies $F \propto \frac{1}{r^2}$.According to the Bohr model,the radius of the $n^{th}$ orbit is given by $r_n = a_0 n^2$,which implies $r \propto n^2$.Substituting the relation $r \propto n^2$ into the force equation,we get $F \propto \frac{1}{(n^2)^2} = \frac{1}{n^4}$.Therefore,the force on the electron is inversely proportional to the fourth power of the principal quantum number $n$.

In the Bohr model of hydrogen,the force on the electron depends on the principal quantum number $n$ as:

Similar Questions

The wavelength for an electron in an orbit of a hydrogen atom is $10^{-9} \ m$. The principal quantum number for this electron is:

The frequency of revolution of an electron in the $n^{th}$ Bohr orbit is $v_n$. The graph between $\log n$ and $\log (v_n / v_1)$ may be

Write the quantum condition suggested by Bohr for the angular momentum of the electron.

The radius of the first (lowest) orbit of the hydrogen atom is $a_0$. The radius of the second (next higher) orbit will be (in $a_0$)

Kinetic energy of an electron in one of the orbits of a hydrogen atom is $x$. Then, its total energy is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module