1. Use Bohr model formulae for a hydrogen-like species

For an electron in the $n^{\text{th}}$ orbit of an atom/ion with atomic number $Z$:

• Radius: $r_n \propto \frac{n^2}{Z}$
• Velocity: $v_n \propto \frac{Z}{n}$
• Kinetic energy: $K \propto \frac{Z^2}{n^2}$
• Total energy: $E \propto -\frac{Z^2}{n^2}$

Also, $\text{frequency of revolution } f = \frac{v}{2\pi r}$

and Coulomb force: $F = \frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r^2}$

2. Check statement (A)

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

From Bohr theory, $K = \frac{1}{2}mv^2 \propto \left(\frac{Z}{n}\right)^2 = \frac{Z^2}{n^2}$

So, (A) is correct.

3. Check statement (B)

Statement (B): $vn \propto Z^2$

But since $v \propto \frac{Z}{n}$ we get $vn \propto Z$

not $Z^2$.

So, (B) is incorrect.

4. Check statement (C)

Statement (C): Frequency of revolution $\propto \frac{Z^3}{n^3}$

Now, $f \propto \frac{v}{r}$ Using $v \propto \frac{Z}{n}, \qquad r \propto \frac{n^2}{Z}$

Therefore, $f \propto \frac{Z/n}{n^2/Z} = \frac{Z^2}{n^3}$

So the correct proportionality is $f \propto \frac{Z^2}{n^3}$ not $\dfrac{Z^3}{n^3}$.

Hence, (C) is incorrect.

5. Check statement (D)

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

Since $F \propto \frac{Z}{r^2}$ and $r \propto \frac{n^2}{Z} \implies r^2 \propto \frac{n^4}{Z^2}$

Thus, $F \propto \frac{Z}{n^4/Z^2} = \frac{Z^3}{n^4}$

So, (D) is correct.

6. Final evaluation of options

• (A) correct
• (B) incorrect
• (C) incorrect
• (D) correct

Hence the correct choice is: $\boxed{\text{D: (A) and (D) only}}$

7. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

So, the derived answer agrees with the stored correct answer.