Question

In hydrogen atom, the product of the angular momentum and the linear momentum of the electron is proportional to \( (n = \) principal quantum number)

• A. \( n^1 \)
• B. \( n^0 \)
• C. \( n^3 \)
• D. \( n^2 \)

Hint:

In the hydrogen atom, the angular and linear momentum are quantized and their product is independent of the principal quantum number \( n \).

Answer 1

The Correct Option is B

Step 1: Understanding the relationship.
In the hydrogen atom, the angular momentum \( L \) of the electron is quantized and is given by: \[ L = n \hbar \] where \( \hbar \) is the reduced Planck constant, and \( n \) is the principal quantum number. The linear momentum \( p \) of the electron in the hydrogen atom is related to its velocity \( v \) and radius \( r \) by: \[ p = mv = \frac{nh}{2\pi r} \] where \( m \) is the mass of the electron, and \( v \) is its velocity. Step 2: Product of angular and linear momentum.
The product of angular momentum \( L \) and linear momentum \( p \) is: \[ L \times p = (n \hbar) \times \left( \frac{nh}{2\pi r} \right) \] The product \( L \times p \) is proportional to \( n^0 \), meaning it is independent of the principal quantum number \( n \). Therefore, the correct answer is \( n^0 \). Step 3: Conclusion.
Thus, the product of the angular and linear momentum is proportional to \( n^0 \), which corresponds to option (B).