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 determined by the principal quantum number (n) and then the azimuthal quantum number (l). Lower n and l values correspond to lower energy.
For the same n, lower l has lower energy. If n and l are same, then energy depends on spin, lower energy for -1/2.
Let’s analyze the electrons:
I. (n=4, l=2)
II. (n=3, l=2)
III. (n=4, l=1)
IV. (n=3, l=1)
Thus, I > III and II > IV based on n and l. Also, I > II and III > IV as n is higher for I and III respectively. Thus, I>III>II>IV