Question

The angular momentum and the energy of the electron in the second Bohr’s orbit are respectively

• A. \( \frac{h}{\pi} \) and -13.6 eV
• B. \( \frac{2h}{\pi} \) and -1.5 eV
• C. \( \frac{h}{\pi} \) and -3.4eV
• D. \( \frac{h}{2\pi} \) and - 3.4 eV
• E. \( \frac{2h}{\pi} \) and -3.4 Ev

Hint:

Memorize the energy levels of Hydrogen for quick calculations:
\(n=1 \to -13.6 \text{ eV} \)
\(n=2 \to -3.4 \text{ eV} \)
\(n=3 \to -1.51 \text{ eV} \)
\(n=4 \to -0.85 \text{ eV} \)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Bohr’s model of the atom quantizes both the angular momentum and the energy levels of electrons in hydrogen-like atoms.

Step 2: Key Formula or Approach:
1. Angular momentum (\(L\)): \( L = n \frac{h}{2\pi} \)
2. Energy (\(E\)): \( E_n = -\frac{13.6}{n^2} \text{ eV} \)

Step 3: Detailed Explanation:
For the second Bohr orbit, the principal quantum number \(n = 2\).

Calculating Angular Momentum (\(L\)):
\[ L = 2 \times \frac{h}{2\pi} = \frac{h}{\pi} \]

Calculating Energy (\(E\)):
\[ E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \text{ eV} \]

Comparing with the options, Option (C) provides both values correctly.

Step 4: Final Answer:
The angular momentum is \(h/\pi\) and the energy is -3.4 eV.