Question

which of the following sets of quantum numbers represents an electron with the highest energy in a multi - electron atom?
a
n = 4,
ℓ = 0,
m_ℓ = 0
b
n = 3,
ℓ = 2,
m_ℓ=-1
c
n = 3,
ℓ = 1,
m_ℓ = 0
d
n = 2,
ℓ = 1,
m_ℓ=-1
e
n = 4,
ℓ = 1,
m_ℓ = 1

Explanation:

In a multi - electron atom, the energy of an electron is mainly determined by the principal quantum number \(n\) and the azimuthal quantum number \(l\). The higher the value of \(n + l\), the higher the energy of the electron. If two electrons have the same \(n + l\) value, the one with the higher \(n\) value has higher energy.

For option A: \(n = 4\), \(l=0\), so \(n + l=4\)
For option B: \(n = 3\), \(l = 2\), so \(n + l=3 + 2=5\)
For option C: \(n = 3\), \(l = 1\), so \(n + l=3+1 = 4\)
For option D: \(n = 2\), \(l = 1\), so \(n + l=2 + 1=3\)
For option E: \(n = 4\), \(l = 1\), so \(n + l=4+1 = 5\)

Since \(n + l\) for B and E is 5. Comparing B (\(n = 3\)) and E (\(n = 4\)), the electron with \(n = 4\) (option E) has higher energy.

Answer:

E. \(n = 4\), \(l = 1\), \(m_{l}=1\)