Zeeman, Paschen-Back and Stark Effects Questions and Answers

Q: An atom is subjected to a weak uniform magnetic field \(\vec{B}\). The number of lines in its Zeeman spectrum for transition from \(n = 2, l = 1\) to \(n = 1, l = 0\) is

The Zeeman spectrum has 10 lines.In a weak field, spin-orbit coupling splits the n = 2, l = 1 level into 2P3/2 and 2P1/2, and the anomalous Zeeman effect splits each level into 2j + 1 sublevels with different Landé g-factors (g = 4/3, 2/3 and 2 for 2P3/2, 2P1/2 and 2S1/2). With the selection rule Δmj = 0, ±1:2P3/2 → 2S1/2: 6 lines (the mj = ±3/2 → &mp;1/2 transitions are forbidden);2P1/2 → 2S1/2: 4 lines.Because the g-factors differ, all 6 + 4 = 10 lines have different energies. 5 or 8 undercount the allowed transitions, and 12 counts all 4 × 2 + 2 × 2 pairs without the selection rule.

Q: For a singlet state of electronic system, the Landé splitting factor will be equal to :

The Landé factor is 1.\[ g = 1 + \frac{J(J+1) + S(S+1) - L(L+1)}{2J(J+1)} \]In a singlet state \( S = 0 \), so \( J = L \) and the fraction vanishes: \( g = 1 \). Such states show the normal Zeeman effect, with the splitting of the classical orbital value.3/2, 5/2 and 1/2 need a non-zero spin; for example the 3P1 level has g = 3/2. None of them is possible when \( S = 0 \).

Q: The ground state of an alkali metal is a 2S1/2 state. The difference in energy levels arising in the presence of a weak magnetic field B, given in terms of Bohr magneton \(\mu_{B}\) is :

The splitting is \(2\mu_{B}B\).For a \(^{2}S_{1/2}\) state, L = 0 and S = J = 1/2, so the Landé factor is \(g_{J} = 2\). In a weak field each level shifts by \(\Delta E = g_{J}m_{J}\mu_{B}B\), with \(m_{J} = \pm\frac{1}{2}\):\[ \begin{aligned} E_{\pm} &= \pm\frac{1}{2} \times 2\mu_{B}B = \pm\mu_{B}B \\ E_{+} - E_{-} &= 2\mu_{B}B \end{aligned} \]\(\mu_{B}B\) is the shift of each sublevel from the original level, not the difference between them. \(4\mu_{B}B\) and \(6\mu_{B}B\) would need larger g or J values.

Q: The \(D_{1}\) and \(D_{2}\) lines of sodium (\({}^{2}P_{1/2} \rightarrow {}^{2}S_{1/2}\) and \({}^{2}P_{3/2} \rightarrow {}^{2}S_{1/2}\)) will split on application of a weak magnetic field into :

In a weak field the D1 line splits into 4 lines and the D2 line into 6.This is the anomalous Zeeman effect. The Lande g factors are 2 for \({}^{2}S_{1/2}\), 2/3 for \({}^{2}P_{1/2}\) and 4/3 for \({}^{2}P_{3/2}\). Because upper and lower g factors differ, every allowed transition (\(\Delta m_{J} = 0, \pm 1\)) has its own frequency.For D1 (\(m_{J} = \pm\frac{1}{2}\) to \(\pm\frac{1}{2}\)) there are 4 allowed transitions. For D2, the upper level has four sublevels (\(m_{J} = \pm\frac{3}{2}, \pm\frac{1}{2}\)), and the allowed transitions to the two lower sublevels number 6, since \(\pm\frac{3}{2} \rightarrow \mp\frac{1}{2}\) would need \(\Delta m_{J} = 2\).Three lines each is the normal Zeeman pattern, which applies only to singlet transitions, not to these doublets.

Q: The expression for the weak field Zeeman effect correction \(E_{z}\) to hydrogen atom energies and its values for the ground state energy are (\(\mu_{B}\) is the Bohr magneton, \(g_{J}\) the Lande g-factor, \(m_{j}\) the total angular momentum magnetic quantum number, B external magnetic field) :

The correct answer is \(E_{z} = \mu_{B}g_{J}Bm_{j}\), for ground state \(E_{z} = \pm\mu_{B}B\).In a weak field, the energy shift of a level with total angular momentum \(j\) is \(E_{z} = \mu_{B}g_{J}Bm_{j}\). The hydrogen ground state is \(1s_{1/2}\): \(l = 0\), \(s = 1/2\), \(j = 1/2\), for which the Lande factor is that of a pure spin, \(g_{J} = 2\). With \(m_{j} = \pm 1/2\):\[ \begin{aligned} E_{z} &= \mu_{B}\cdot 2\cdot B\cdot(\pm\tfrac{1}{2}) \\ &= \pm\mu_{B}B \end{aligned} \]The ground state does split, because the electron's spin carries a magnetic moment even though \(l = 0\), so \(E_{z} = 0\) is wrong. The factor 3/2 in the other options would correspond to a different \(g_{J}\) or \(m_{j}\) from those of the ground state.

An atom is subjected to a weak uniform magnetic field \(\vec{B}\). The number of lines in its Zeeman spectrum for transition from \(n = 2, l = 1\) to \(n = 1, l = 0\) is

A.8
B.10
C.12
D.5

Exam Year(s)

GATE Physics - 2024,

Answer: B

The Zeeman spectrum has 10 lines.

In a weak field, spin-orbit coupling splits the n = 2, l = 1 level into ²P_3/2 and ²P_1/2, and the anomalous Zeeman effect splits each level into 2j + 1 sublevels with different Landé g-factors (g = 4/3, 2/3 and 2 for ²P_3/2, ²P_1/2 and ²S_1/2). With the selection rule Δm_j = 0, ±1:

²P_3/2 → ²S_1/2: 6 lines (the m_j = ±3/2 → ∓1/2 transitions are forbidden);
²P_1/2 → ²S_1/2: 4 lines.

Because the g-factors differ, all 6 + 4 = 10 lines have different energies. 5 or 8 undercount the allowed transitions, and 12 counts all 4 × 2 + 2 × 2 pairs without the selection rule.

Discuss in Forum

For a singlet state of electronic system, the Landé splitting factor will be equal to :

A.3/2
B.5/2
C.1/2
D.1

Exam Year(s)

MH SET - 2018, MH SET - 2020, SET Exam - 2018, SET Exam - 2020,

Answer: D

The Landé factor is 1.

\[ g = 1 + \frac{J(J+1) + S(S+1) - L(L+1)}{2J(J+1)} \]

In a singlet state \( S = 0 \), so \( J = L \) and the fraction vanishes: \( g = 1 \). Such states show the normal Zeeman effect, with the splitting of the classical orbital value.

3/2, 5/2 and 1/2 need a non-zero spin; for example the ³P₁ level has g = 3/2. None of them is possible when \( S = 0 \).

Discuss in Forum

The ground state of an alkali metal is a ²S_1/2 state. The difference in energy levels arising in the presence of a weak magnetic field B, given in terms of Bohr magneton \(\mu_{B}\) is :

A.\(\mu_{B}B\)
B.\(2\mu_{B}B\)
C.\(4\mu_{B}B\)
D.\(6\mu_{B}B\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The splitting is \(2\mu_{B}B\).

For a \(^{2}S_{1/2}\) state, L = 0 and S = J = 1/2, so the Landé factor is \(g_{J} = 2\). In a weak field each level shifts by \(\Delta E = g_{J}m_{J}\mu_{B}B\), with \(m_{J} = \pm\frac{1}{2}\):

\[ \begin{aligned} E_{\pm} &= \pm\frac{1}{2} \times 2\mu_{B}B = \pm\mu_{B}B \\ E_{+} - E_{-} &= 2\mu_{B}B \end{aligned} \]

\(\mu_{B}B\) is the shift of each sublevel from the original level, not the difference between them. \(4\mu_{B}B\) and \(6\mu_{B}B\) would need larger g or J values.

Discuss in Forum

The \(D_{1}\) and \(D_{2}\) lines of sodium (\({}^{2}P_{1/2} \rightarrow {}^{2}S_{1/2}\) and \({}^{2}P_{3/2} \rightarrow {}^{2}S_{1/2}\)) will split on application of a weak magnetic field into :

A.3 lines each
B.6 lines each
C.4 lines each
D.4 and 6 lines, respectively

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

In a weak field the D₁ line splits into 4 lines and the D₂ line into 6.

This is the anomalous Zeeman effect. The Lande g factors are 2 for \({}^{2}S_{1/2}\), 2/3 for \({}^{2}P_{1/2}\) and 4/3 for \({}^{2}P_{3/2}\). Because upper and lower g factors differ, every allowed transition (\(\Delta m_{J} = 0, \pm 1\)) has its own frequency.

For D₁ (\(m_{J} = \pm\frac{1}{2}\) to \(\pm\frac{1}{2}\)) there are 4 allowed transitions. For D₂, the upper level has four sublevels (\(m_{J} = \pm\frac{3}{2}, \pm\frac{1}{2}\)), and the allowed transitions to the two lower sublevels number 6, since \(\pm\frac{3}{2} \rightarrow \mp\frac{1}{2}\) would need \(\Delta m_{J} = 2\).

Three lines each is the normal Zeeman pattern, which applies only to singlet transitions, not to these doublets.

Discuss in Forum

The expression for the weak field Zeeman effect correction \(E_{z}\) to hydrogen atom energies and its values for the ground state energy are (\(\mu_{B}\) is the Bohr magneton, \(g_{J}\) the Lande g-factor, \(m_{j}\) the total angular momentum magnetic quantum number, B external magnetic field) :

A.\(E_{z} = \mu_{B}g_{J}Bm_{j}\), for ground state \(E_{z} = 0\)
B.\(E_{z} = \mu_{B}g_{J}Bm_{j}\), for ground state \(E_{z} = \pm\mu_{B}B\)
C.\(E_{z} = \frac{3}{2}\mu_{B}g_{J}Bm_{j}\), for ground state \(E_{z} = \pm\frac{3}{2}\mu_{B}B\)
D.\(E_{z} = \mu_{B}g_{J}Bm_{j}\), for ground state \(E_{z} = \pm\frac{3}{2}\mu_{B}B\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The correct answer is \(E_{z} = \mu_{B}g_{J}Bm_{j}\), for ground state \(E_{z} = \pm\mu_{B}B\).

In a weak field, the energy shift of a level with total angular momentum \(j\) is \(E_{z} = \mu_{B}g_{J}Bm_{j}\). The hydrogen ground state is \(1s_{1/2}\): \(l = 0\), \(s = 1/2\), \(j = 1/2\), for which the Lande factor is that of a pure spin, \(g_{J} = 2\). With \(m_{j} = \pm 1/2\):

\[ \begin{aligned} E_{z} &= \mu_{B}\cdot 2\cdot B\cdot(\pm\tfrac{1}{2}) \\ &= \pm\mu_{B}B \end{aligned} \]

The ground state does split, because the electron's spin carries a magnetic moment even though \(l = 0\), so \(E_{z} = 0\) is wrong. The factor 3/2 in the other options would correspond to a different \(g_{J}\) or \(m_{j}\) from those of the ground state.

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