Subtopic: Hydrogen Atom Spectrum & Radial Wavefunctions
Question Statement:
An atomic state of hydrogen is represented by the following wavefunction:
$\psi(r, \theta, \phi) = \frac{1}{4\sqrt{2\pi}} \left(\frac{1}{a_0}\right)^{3/2} \left(2 - \frac{r}{a_0}\right) e^{-r/(2a_0)} \cos\theta$
where $a_0$ is the Bohr radius. The quantum numbers of the state are
Options:
(A) $l = 0, \; m = 0, \; n = 1$
(B) $l = 1, \; m = 1, \; n = 2$
(C) $l = 1, \; m = 0, \; n = 2$
(D) $l = 2, \; m = 0, \; n = 3$
Official Answer Key:
C
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2007 Q.40)**
**1. Theoretical Formulation & Hydrogen Atom Wavefunctions**:
- **Field & Concept**: *Section 4: Quantum Mechanics* (`Hydrogen Atom Solutions & Radial Wavefunctions`)
- The general solution for the hydrogen atom state in spherical coordinates is $\psi_{nlm}(r, \theta, \phi) = R_{nl}(r) Y_l^m(\theta, \phi)$.
**2. Detailed Step-by-Step Derivation**:
- **Angular Part**:
- The angular dependence is $\cos\theta$, which is proportional to the spherical harmonic $Y_1^0(\theta, \phi) = \sqrt{\frac{3}{4\pi}} \cos\theta$.
- Therefore, the orbital angular momentum quantum number is $l = 1$ and the magnetic quantum number is $m = 0$.
- **Radial Part & Principal Quantum Number $n$**:
- The exponential decay factor is $e^{-r / (n a_0)} = e^{-r / (2 a_0)}$, which immediately yields $n = 2$.
- Furthermore, $n = 2, l = 1, m = 0$ corresponds to the standard $2p_z$ state.
**3. Official Verification & Conclusion**:
- **Official Master Answer**: **Option (C)**: $l = 1, \; m = 0, \; n = 2$.
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