Question

Suppose an electron is attracted towards the origin by a force \( K/r \), where K is a constant and r is the distance of the electron from the origin. By applying Bohr model to this system, the radius of the \( n^{th} \) orbit of the electron is found to be \( r_n \), and the kinetic energy of the electron to be \( T_n \), then which of the following is true?

• A. \( T_n \) is independent of n, \( r_n \propto n \)
• B. \( T_n \propto 1/n, r_n \propto n \)
• C. \( T_n \propto 1/n, r_n \propto n^2 \)
• D. \( T_n \propto 1/n^2, r_n \propto n^2 \)

Hint:

In non-Coulombic central force fields, the behavior of orbital radii and energy levels can differ significantly from the standard hydrogen-like atom model.

Answer 1

The Correct Option is A

Step 1: Set up the two Bohr conditions.
Bohr's model needs two ideas. First, the attractive force supplies the centripetal force, so $\dfrac{mv^2}{r} = F(r)$. Second, angular momentum is quantised, $mvr = \dfrac{nh}{2\pi}$. Here the force given is $F(r) = \dfrac{K}{r}$.
Step 2: Find the speed from the force balance.
Put $F(r) = K/r$ into the centripetal condition. \[ \frac{mv^2}{r} = \frac{K}{r} \] The $r$ cancels nicely, leaving $mv^2 = K$.
Step 3: Notice the speed is fixed.
So $v = \sqrt{K/m}$, which contains only constants. The electron moves at the same speed in every orbit, no matter what $n$ is.
Step 4: Get the kinetic energy.
Since $T_n = \tfrac{1}{2} m v^2 = \tfrac{1}{2} K$, the kinetic energy is just a constant. It does not depend on $n$ at all.
Step 5: Get the radius from quantisation.
Use $mvr = \dfrac{nh}{2\pi}$ to write \[ r_n = \frac{nh}{2\pi m v} \] Everything except $n$ is constant, so $r_n \propto n$.
Step 6: Choose the matching option.
We found $T_n$ independent of $n$ and $r_n \propto n$, which is option 1.
\[ \boxed{T_n \text{ is independent of } n, \ r_n \propto n} \]