The maximum number of electrons in an atom that can have given quantum numbers
Q. Given,

(A) \( n = 5,\; m_l = -1 \)

(B) \( n = 3,\; l = 2,\; m_l = -1,\; m_s = +\frac{1}{2} \)

The maximum number of electron(s) in an atom that can have the quantum numbers as given in (A) and (B) respectively are :
A. 4 and 1
B. 2 and 4
C. 26 and 1
D. 8 and 1
Correct Answer: 8 and 1

Explanation

Case (A): \( n = 5 \) and \( m_l = -1 \) are given.

For a given principal quantum number \( n = 5 \), the possible values of azimuthal quantum number are:

$$ l = 0, 1, 2, 3, 4 $$

For a fixed value \( m_l = -1 \), only those values of \( l \) are allowed for which:

$$ -l \le m_l \le +l $$

So, \( m_l = -1 \) is possible for:

$$ l = 1, 2, 3, 4 $$

Thus, there are 4 different orbitals corresponding to \( m_l = -1 \).

Each orbital can accommodate 2 electrons due to two possible spin states:

$$ m_s = +\frac{1}{2},\; -\frac{1}{2} $$

Hence, maximum number of electrons for case (A) is:

$$ 4 \times 2 = 8 $$

Case (B): All four quantum numbers \( n = 3,\; l = 2,\; m_l = -1,\; m_s = +\frac{1}{2} \) are specified.

A unique set of four quantum numbers corresponds to only one electron.

Therefore, maximum number of electrons for case (B) is:

$$ 1 $$

Hence, the correct answer is:

$$ \boxed{8 \text{ and } 1} $$

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