Question

question 25. calculate the number of orbitals in the 3rd principal energy level (n = 3). show the steps.

Explanation:

Step1: Determine the possible values of \( l \)

For a given principal quantum number \( n \), the angular - momentum quantum number \( l \) can take values \( l = 0,1,\cdots,n - 1 \).
When \( n=3 \), \( l = 0,1,2 \).

Step2: Calculate the number of orbitals for each \( l \)

• When \( l = 0 \) (s - orbital), the magnetic quantum number \( m_l\) has \( 2l + 1=2\times0 + 1 = 1 \) value. So, there is \( 1 \) s - orbital.
• When \( l = 1 \) (p - orbitals), \( m_l=-1,0,1 \). The number of orbitals is \( 2l + 1=2\times1+ 1=3 \).
• When \( l = 2 \) (d - orbitals), \( m_l=-2,-1,0,1,2 \). The number of orbitals is \( 2l + 1=2\times2 + 1=5 \).

Step3: Sum up the number of orbitals

The total number of orbitals in the \( n = 3 \) level is \( 1+3 + 5=9 \).
Another way: Use the formula \( N=n^{2}\). When \( n = 3 \), \( N=3^{2}=9 \)

Answer:

The number of orbitals in the \(3^{rd}\) principal energy level (\(n = 3\)) is \(9\).