Question

The quantum numbers of four electrons are given below:
I. \( n = 4; l = 2; m_l = -2; s = -\frac{1}{2} \)
II. \( n = 3; l = 2; m_l = 1; s = +\frac{1}{2} \)
III. \( n = 4; l = 1; m_l = 0; s = +\frac{1}{2} \)
IV. \( n = 3; l = 1; m_l = -1; s = +\frac{1}{2} \)
The correct decreasing order of energy of these electrons is:

• A. IV > II > III > I
• B. I > III > II > IV
• C. III > I > II > IV
• D. I > II > III > IV

Answer 1

The Correct Option is B

The energy of an electron is primarily dictated by its principal quantum number (n) and secondarily by its azimuthal quantum number (l). Lower values of n and l result in lower energy. When n is constant, a lower l yields lower energy. If both n and l are identical, energy is influenced by spin, with -1/2 having lower energy. Analyzing the given electron states:
I. (n=4, l=2)
II. (n=3, l=2)
III. (n=4, l=1)
IV. (n=3, l=1)
Based on n and l, the energy order is I>III and II>IV. Considering n alone, I>II and III>IV due to higher n values for I and III. Therefore, the overall energy ranking is I>III>II>IV.