Question

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

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

Hint:

|E| goes as 1/n^2 and L goes as n.

Answer 1

The Correct Option is C

Step 1: Use mechanics:
For circular orbit, $E=-\tfrac12mv^2$ in magnitude, and $L=mvr$.

Step 2: Eliminate r:
With $v=\dfrac{L}{mr}$, and using $mv^2=\dfrac{ke^2}{r}$, one gets $r=\dfrac{L^2}{mke^2}$.

Step 3: Energy:
$|E|=\dfrac{ke^2}{2r}=\dfrac{mk^2e^4}{2L^2}\propto L^{-2}$. Option (C).

Final Answer:
Energy magnitude goes as 1/L^2. \[ \boxed{C} \]