Highest Kinetic Energy of Electron in Bohr Model Orbits

According to Bohr's model of a single-electron atom or ion:

1. Centripetal Force and Electrostatic Attraction: The electrostatic force between the nucleus of atomic number $Z$ and the electron provides the necessary centripetal force for circular motion in the $n$-th orbit: $\frac{1}{4\pi\varepsilon_0} \frac{Z e^2}{r_n^2} = \frac{m v_n^2}{r_n}$

2. Kinetic Energy ($K$): The kinetic energy of the electron is given by: $K = \frac{1}{2} m v_n^2 = \frac{1}{8\pi\varepsilon_0} \frac{Z e^2}{r_n}$

3. Radius of the $n$-th Orbit ($r_n$): The radius $r_n$ of the $n$-th Bohr orbit is proportional to $\frac{n^2}{Z}$: $r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m Z e^2} \propto \frac{n^2}{Z}$

4. Dependence of Kinetic Energy on $Z$ and $n$: Substituting $r_n$ into the expression for kinetic energy gives: $K = \frac{m Z^2 e^4}{8 \varepsilon_0^2 h^2 n^2} \implies K \propto \frac{Z^2}{n^2}$

Now, let us calculate the relative value of kinetic energy proportional to $\frac{Z^2}{n^2}$ for each option:

• (A) First orbit ($n = 1$) of $\text{H}$ atom ($Z = 1$): $\frac{Z^2}{n^2} = \frac{1^2}{1^2} = 1$

• (B) First orbit ($n = 1$) of $\text{He}^+$ ($Z = 2$): $\frac{Z^2}{n^2} = \frac{2^2}{1^2} = 4$

• (C) Second orbit ($n = 2$) of $\text{He}^+$ ($Z = 2$): $\frac{Z^2}{n^2} = \frac{2^2}{2^2} = 1$

• (D) Second orbit ($n = 2$) of $\text{Li}^{2+}$ ($Z = 3$): $\frac{Z^2}{n^2} = \frac{3^2}{2^2} = \frac{9}{4} = 2.25$

Comparing these values, the maximum value of $\frac{Z^2}{n^2}$ is $4$, corresponding to option (B).

Thus, the electron has the highest kinetic energy in the first orbit of $\text{He}^+$.

Correct Answer: (B)