QJEEM24Phb275: JEE Main 2024 Physics - Class 12 JEEM24Phb275 Solution: The radius of third stationary orbit of electron for Bohr's…

The radius of the nth stationary orbit in the Bohr model of an atom is directly proportional to the square of its principal quantum number n and inversely proportional to the atomic number Z. This relationship is represented by the formula $$r_n=\frac{n^2{h}^2}{4{\pi }^{2}km{e}^{2}Z},$$ where :

• $n$ is the principal quantum number,

• $h$ is Planck's constant,

• $k$ is the Coulomb constant,

• $m$ is the mass of the electron,

• $e$ is the charge of the electron, and

• $Z$ is the atomic number of the nucleus (for hydrogen, Z=1).

To find the radius of the fourth orbit ( $r_4$ ), in comparison to the third orbit ( $r_3$ ), we apply the formula with $n=4$ for the fourth orbit and $n=3$ for the third orbit, and simplify as follows:

$$\begin{array}{rl}& \frac{r_4}{r_3}=\frac{(4^2)h^2}{4{\pi }^{2}km{e}^{2}Z}÷\frac{(3^2)h^2}{4{\pi }^{2}km{e}^{2}Z}=\frac{(4^2)}{(3^2)}\\ \\ & =\frac{16}{9}\end{array}$$

Thus, the radius of the fourth stationary orbit is $\frac{16}{9}$ times the radius of the third stationary orbit, $R$ . Therefore, $$r_4=\frac{16}{9}R$$.