According to Bohr's atomic theory: (A) Kinetic energy of electron is $\propto \frac{{Z}^{2}}{{n}^{2}}.$ (B) The product of velocity (v) of electron and principal quantum number $(n),'vn'\propto {Z}^{2}.$ (C) Frequency of revolution of electron in an orbit is $\propto \frac{{Z}^{3}}{{n}^{3}}.$ (D) Coulombic force of attraction on the electron is $\propto \frac{{Z}^{3}}{{n}^{4}}.$ Choose the most appropriate answer from the options given below:

According to the Bohr's atomic theory, the kinetic energy of an electron is directly proportional to the square of the atomic number and inversely proportional to the square of the principal quantum number. (A) $\mathrm{KE}=13.6\frac{{z}^{2}}{{n}^{2}}\frac{\mathrm{eV}}{\mathrm{atom}}\Rightarrow KE\alpha \frac{{z}^{2}}{{n}^{2}}$ (B) speed of ${e}^{-}\alpha \frac{z}{n}$ $\therefore v\times n\alpha z$ (C) Frequency of revolution of ${e}^{-}=\frac{v}{2\pi r}$ $\therefore$ Frequency $\alpha \frac{{z}^{2}}{{n}^{3}}$ (D) Coulombic force of attraction on the electron is directly proportional to the square of the atomic number and inversely proportional to the fourth power of the principal quantum number. $F=\frac{{\mathrm{kq}}_{1}{q}_{2}}{{r}^{2}}=\frac{{\mathrm{kze}}^{2}}{{r}^{2}}{r\alpha \frac{{n}^{2}}{z}$ $\Rightarrow F\alpha \frac{z}{{(\frac{{n}^{2}}{z})}^{2}}$ $\Rightarrow F\alpha \frac{{z}^{3}}{{n}^{4}}$

JEE Main 2021 — Chemistry Physical Chemistry

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According to Bohr's atomic theory:

(A) Kinetic energy of electron is $\propto \frac{{Z}^{2}}{{n}^{2}}.$

(B) The product of velocity (v) of electron and principal quantum number $(n),'vn'\propto {Z}^{2}.$

(D) Coulombic force of attraction on the electron is $\propto \frac{{Z}^{3}}{{n}^{4}}.$

Choose the most appropriate answer from the options given below:

Official previous-year question

Held on 24 Feb 2021 · Verified 6 Jul 2026.

Options

A
(A), (C) and (D) only
B
(A) and (D) only
C
(C) only
D
(A) only

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Solution

According to the Bohr's atomic theory, the kinetic energy of an electron is directly proportional to the square of the atomic number and inversely proportional to the square of the principal quantum number.

(A) $\mathrm{KE}=13.6\frac{{z}^{2}}{{n}^{2}}\frac{\mathrm{eV}}{\mathrm{atom}}\Rightarrow KE\alpha \frac{{z}^{2}}{{n}^{2}}$

(B) speed of ${e}^{-}\alpha \frac{z}{n}$

$\therefore v\times n\alpha z$

$\therefore$ Frequency $\alpha \frac{{z}^{2}}{{n}^{3}}$

(D) Coulombic force of attraction on the electron is directly proportional to the square of the atomic number and inversely proportional to the fourth power of the principal quantum number.

$F=\frac{{\mathrm{kq}}_{1}{q}_{2}}{{r}^{2}}=\frac{{\mathrm{kze}}^{2}}{{r}^{2}}{r\alpha \frac{{n}^{2}}{z}$

$\Rightarrow F\alpha \frac{z}{{(\frac{{n}^{2}}{z})}^{2}}$

$\Rightarrow F\alpha \frac{{z}^{3}}{{n}^{4}}$

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Solution

(A) $\mathrm{KE}=13.6\frac{{z}^{2}}{{n}^{2}}\frac{\mathrm{eV}}{\mathrm{atom}}\Rightarrow KE\alpha \frac{{z}^{2}}{{n}^{2}}$

(B) speed of ${e}^{-}\alpha \frac{z}{n}$

$\therefore v\times n\alpha z$

$\therefore$ Frequency $\alpha \frac{{z}^{2}}{{n}^{3}}$

(D) Coulombic force of attraction on the electron is directly proportional to the square of the atomic number and inversely proportional to the fourth power of the principal quantum number.

$F=\frac{{\mathrm{kq}}_{1}{q}_{2}}{{r}^{2}}=\frac{{\mathrm{kze}}^{2}}{{r}^{2}}{r\alpha \frac{{n}^{2}}{z}$

$\Rightarrow F\alpha \frac{z}{{(\frac{{n}^{2}}{z})}^{2}}$

$\Rightarrow F\alpha \frac{{z}^{3}}{{n}^{4}}$

JEE Main 2021 — Chemistry Physical Chemistry

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JEE Main 2021 — Chemistry Physical Chemistry

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Solution

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JEE Main Physical Chemistry in other years

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Inorganic Chemistry

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