The correct set of four quantum numbers for the valence electron of rubidium atom (Z = 37) is:

🧠 The Alkali Location Rubidium ($Z=37$) is located in Group 1, right below Potassium. As the first member of the $5^{\text{th}}$ period, its valence electron marks the beginning of a fresh principal shell.

🗺️ The Quantum Address

1. Electronic Configuration: The noble gas before Rubidium is Krypton ($Z=36$). Config: $[\text{Kr}] 5s^1$.
2. Assigning the Numbers:
   • $n = 5$ (Floor number)
   • $l = 0$ (For all 's' orbitals)
   • $m_l = 0$ (For $l=0$, this is the only option)
   • $s = +\frac{1}{2}$ (Standard starting spin)

⚡ The "Group 1, Period 5" Hook If you know Rubidium is in the $5^{\text{th}}$ period, $n$ must be 5. If it's in Group 1, it must be an $s^1$ electron ($l=0$). This eliminates options b, c, and d instantly.

⚠️ Common Traps Don't be tempted by $l=1$. That would be the $5p$ subshell, which belongs to elements like Indium. Rubidium is very much an s-block metal.

$\boxed{\text{Answer: (a)}}$