Question

For which one of the following sets of four quantum numbers, an electron will have the highest energy? 

• A. \( 3 | 2| 1 |  +\frac{1}{2} \)
• B. \(| 4 | 2 | 1 |  +\frac{1}{2}\)
• C. \(| 4 | 1 | 0 |  +\frac{1}{2} \)
• D. \(| 5 | 0 | 0 |  +\frac{1}{2} \)

Hint:

The energy of an electron in an atom is primarily determined by the sum of the principal quantum number (\( n \)) and the azimuthal quantum number (\( l \)). The higher the value of \( n + l \), the higher the energy.

Answer 1

The Correct Option is B

Step 1: Quantum Numbers Defined
The four quantum numbers are: 
\( n \) (Principal): dictates energy level. 
\( l \) (Azimuthal): dictates subshell. 
\( m \) (Magnetic): dictates orbital orientation. 
\( s \) (Spin): represents electron spin. 

Step 2: Electron Energy Determination 
An electron's energy is primarily governed by \( n \) and \( l \). 
Energy increases with increasing \( n + l \). 
For equal \( n + l \) values, higher \( n \) indicates higher energy. 

Step 3: \( n + l \) Calculation 

The maximum \( n + l \) value is 6, corresponding to option (B). 

Step 4: Highest Energy Identification 
The highest energy is associated with the largest \( n + l \) value. 
In case of a tie, the greater \( n \) value determines the highest energy. 
Therefore, the electron with the highest energy is in option (B), with \( n = 4, l = 2 \). 

Final Answer: The electron with the highest energy resides in (B), characterized by \( n = 4, \, l = 2, \, m = 1, \, s = +\frac{1}{2} \).