The number of orbitals associated with quantum numbers $$n = 5$$, $$m_s = +\frac{1}{2}$$ is:

We have been told that the principal quantum number is $$n = 5$$. The principal quantum number determines the main shell of the electron and, very importantly, fixes the possible values of the azimuthal quantum number $$\ell$$.

First, recall the rule for $$\ell$$: for any fixed $$n$$, $$\ell$$ can take all integral values from $$0$$ up to $$n-1$$. Therefore, when $$n = 5$$ we write

$$\ell = 0, 1, 2, 3, 4.$$

Now, for each specific value of $$\ell$$, the magnetic quantum number $$m_\ell$$ is allowed to vary from $$-\ell$$ to $$+\ell$$ in steps of one. Thus the total number of distinct $$m_\ell$$ values for a given $$\ell$$ is

$$2\ell + 1.$$

If we want the total number of orbitals for a fixed $$n$$, we must add up all the $$m_\ell$$ values for every possible $$\ell$$. Symbolically, that total is

$$\sum_{\ell = 0}^{n-1} (2\ell + 1) = n^{2}.$$

We now substitute $$n = 5$$ to get

$$\text{Number of orbitals for } n = 5 = 5^{2} = 25.$$

Each of these orbitals may be occupied by electrons having either of the two possible spin quantum numbers $$m_s = +\tfrac{1}{2}$$ or $$m_s = -\tfrac{1}{2}$$. The question explicitly fixes $$m_s = +\tfrac{1}{2}$$. When we restrict ourselves to a single spin orientation, we do not change the count of orbitals; we merely say that, out of the two spin possibilities for every orbital, we are picking just one.

Therefore, the number of orbitals that possess the principal quantum number $$n = 5$$ and the spin quantum number $$m_s = +\tfrac{1}{2}$$ remains

$$25.$$

Hence, the correct answer is Option B.