Question

An atomic transition line with a wavelength of 350 nm is observed to be split into three components in a spectrum of light from a sunspot. Adjacent components are separated by 1.7 pm. The strength of the magnetic field in the sunspot is:

• A. \(3\ \text{T}\)
• B. \(0.03\ \text{T}\)
• C. \(3.3\ \text{T}\)
• D. \(0.3\ \text{T}\)

Hint:

Use the normal Zeeman shift \(\Delta\lambda = \dfrac{\lambda^2}{c}\dfrac{eB}{4\pi m_e}\) and solve for \(B\).

Answer 1

The Correct Option is D

Step 1: Splitting into exactly three components is the normal Zeeman effect. The two shifted lines are displaced from the central line by the Larmor frequency \(\Delta\nu = \dfrac{eB}{4\pi m_e}\), so adjacent components are separated by this same \(\Delta\nu\).

Step 2: Convert the frequency shift to a wavelength shift using \(|\Delta\lambda| = \dfrac{\lambda^2}{c}\,\Delta\nu\). Combining, \[\Delta\lambda = \frac{\lambda^2}{c}\cdot\frac{eB}{4\pi m_e}.\]

Step 3: Solve for \(B\): \[B = \frac{4\pi m_e c\,\Delta\lambda}{e\,\lambda^2}.\]

Step 4: Insert \(m_e = 9.11\times10^{-31}\,\text{kg}\), \(c = 3\times10^{8}\,\text{m/s}\), \(e = 1.6\times10^{-19}\,\text{C}\), \(\Delta\lambda = 1.7\times10^{-12}\,\text{m}\), and \(\lambda = 350\times10^{-9}\,\text{m}\) so \(\lambda^2 = 1.225\times10^{-13}\,\text{m}^2\).

Step 5: Numerator: \(4\pi \times 9.11\times10^{-31}\times 3\times10^{8}\times 1.7\times10^{-12} = 5.84\times10^{-33}\). Denominator: \(1.6\times10^{-19}\times 1.225\times10^{-13} = 1.96\times10^{-32}\). Thus \[B = \frac{5.84\times10^{-33}}{1.96\times10^{-32}} \approx 0.30\ \text{T}.\] \[\boxed{B \approx 0.3\ \text{T}}\]