Question

If '$E$' and '$L$' denote the magnitude of total energy and angular momentum of a revolving electron in the $n^{\text{th}}$ Bohr orbit, then

• A. $E \propto L^{-1}$
• B. $E \propto L$
• C. $E \propto L^{-2}$
• D. $E \propto L^2$

Hint:

Since both total energy and angular momentum depend directly on the principal quantum number $n$, you can treat $n$ as a bridge variable. Squaring the momentum relation gives $L^2 \propto n^2$. Inverting this matches the energy relation ($\frac{1}{n^2}$), leading directly to $E \propto L^{-2}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem asks for the proportional relationship between the total mechanical energy magnitude $E$ and the orbital angular momentum magnitude $L$ of an electron traveling within a stable Bohr orbit.

Step 2: Key Formula or Approach:
According to Bohr's atomic model constraints: 1. The total energy $E$ in the $n^{\text{th}}$ quantum shell is inversely proportional to the square of the principal quantum number: $E \propto \frac{1}{n^2}$. 2. The orbital angular momentum $L$ is quantized in integer multiples of $\frac{h}{2\pi}$: $L = \frac{nh}{2\pi} \implies L \propto n$. By combining these two relationships, we can express $E$ directly in terms of $L$.

Step 3: Detailed Explanation:
Let's write down the proportionalities for both parameters relative to the principal quantum number $n$: $$E = -\frac{13.6 \cdot Z^2}{n^2}\text{ eV} \implies E \propto \frac{1}{n^2}$$ $$L = \frac{nh}{2\pi} \implies L \propto n \implies n \propto L$$ Now, substitute the relationship $n \propto L$ into our energy proportionality expression: $$E \propto \frac{1}{(L)^2} \implies E \propto L^{-2}$$ This matches option (C).

Step 4: Final Answer:
The relationship between energy and angular momentum is $E \propto L^{-2}$, which corresponds to option (C).