Question:medium

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:

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Use the normal Zeeman shift \(\Delta\lambda = \dfrac{\lambda^2}{c}\dfrac{eB}{4\pi m_e}\) and solve for \(B\).
Updated On: Jul 2, 2026
  • \(3\ \text{T}\)
  • \(0.03\ \text{T}\)
  • \(3.3\ \text{T}\)
  • \(0.3\ \text{T}\)
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The Correct Option is D

Solution and Explanation

Three equally spaced lines is the signature of the normal (Lorentz) Zeeman triplet. The frequency displacement of each outer line equals the Larmor precession frequency $\nu_L = \dfrac{eB}{4\pi m_e}$.

Work directly in wavelength. Differentiating $\nu = c/\lambda$ gives $|d\nu| = \dfrac{c}{\lambda^2}|d\lambda|$, so the measured wavelength gap of the adjacent lines corresponds to

\[\Delta\nu = \frac{c\,\Delta\lambda}{\lambda^2} = \frac{(3\times10^{8})(1.7\times10^{-12})}{(3.5\times10^{-7})^2}.\]

Evaluate the numerator $= 5.1\times10^{-4}$ and denominator $= 1.225\times10^{-13}$, giving $\Delta\nu = 4.16\times10^{9}\ \text{Hz}$.

Now invert the Larmor relation for the field: $B = \dfrac{4\pi m_e \Delta\nu}{e} = \dfrac{4\pi (9.11\times10^{-31})(4.16\times10^{9})}{1.6\times10^{-19}}$.

The numerator is $4.76\times10^{-20}$, and dividing by $1.6\times10^{-19}$ yields $B \approx 0.30\ \text{T}$, matching the direct estimate. \[\boxed{B \approx 0.3\ \text{T}}\]
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