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

<p>The correct set of four quantum numbers for the valence electron of a rubidium atom ($Z = 37$) is:</p> <p><p>Principal quantum number ($n$): This indicates the main energy level or shell of the electron. For rubidium, the valence electron is in the 5th energy level, so $n = 5$.</p></p> <p><p>Azimuthal quantum number ($l$): This defines the subshell or orbital type. The possible values for $l$ range from 0 to $n-1$. For rubidium, the valence electron is in the $5s$ orbital, so $l = 0$.</p></p> <p><p>Magnetic quantum number ($m_l$): This indicates the orientation of the orbital in space. For $l = 0$, the only possible value is $m_l = 0$.</p></p> <p><p>Spin quantum number ($m_s$): This represents the spin of the electron, which can be either $+\frac{1}{2}$ or $-\frac{1}{2}$. Generally, when an electron is added to an unoccupied orbital, it starts with a spin of $+\frac{1}{2}$.</p></p> <p>Therefore, the correct set of quantum numbers for the valence electron of rubidium is:</p> <p>Option B: $5,0,0,+\frac{1}{2}$</p>