Question

The angular momentum of an electron in an orbit $X$ of hydrogen atom is $\frac{2h}{\pi}$. Maximum number of orbitals possible in $X$ is

• A. 4
• B. 9
• C. 16
• D. 25

Hint:

Shell number $n$ = (Angular Momentum $\times 2\pi$) $h$. Orbitals = $n^2$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
According to Bohr's model, the angular momentum of an electron is quantized and depends on the principal quantum number $n$. Once $n$ is known, the number of orbitals in that shell can be determined.
Step 2: Key Formula or Approach:
1. Bohr's Angular Momentum: $L = \frac{nh}{2\pi}$.
2. Number of orbitals in a shell = $n^2$.
Step 3: Detailed Explanation:
Given angular momentum $L = \frac{2h}{\pi}$. Set this equal to the formula: \[ \frac{nh}{2\pi} = \frac{2h}{\pi} \] Cancel $h$ and $\pi$ from both sides: \[ \frac{n}{2} = 2 \implies n = 4 \] The principal quantum number is 4. The maximum number of orbitals in the $n^{th}$ shell is $n^2$: \[ \text{Number of orbitals} = 4^2 = 16 \]
Step 4: Final Answer:
The maximum number of orbitals possible in orbit X is 16.