Start Learning Free

Before You Begin

Question #43668

Difficulty: 8/10 | 🟢 Free

The Question

A gas discharge tube contains a mixture of Hydrogen (1H^1\text{H}) and Deuterium (2H^2\text{H}) atoms. The atoms are excited and emit light corresponding to the first line of the Balmer series (n=3→n=2n=3 \to n=2). This emitted light is passed through a Michelson interferometer. As the movable mirror is displaced, the interference fringes periodically wash out (reach minimum visibility) due to the superposition of the slightly different wavelengths from the two isotopes. Assuming the mass of the proton is MM, the mass of the electron is mm (with M≫mM \gg m), and the mass of the Deuterium nucleus is 2M2M, what is the minimum non-zero path difference Δx\Delta x introduced by the interferometer at which the fringes completely wash out? (Let R∞R_\infty be the Rydberg constant for an infinitely massive nucleus).

Options

  • A.

    36M5mR∞\frac{36 M}{5 m R_\infty}

  • B.

    72M5mR∞\frac{72 M}{5 m R_\infty}

  • C.

    18M5mR∞\frac{18 M}{5 m R_\infty}

  • D.

    36m5MR∞\frac{36 m}{5 M R_\infty}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Emission vs Absorption|Reduced Mass Effect|Phase Relationship

Why This Question Is Tricky

Common Mistakes

This question tests the integration of the Bohr model's reduced mass effect with wave optics.

A common trap is treating the Rydberg constant as an immutable scalar, ignoring the isotope-dependent frequency shift.

Furthermore, students often confuse the condition for minimum visibility (fringes washing out, Δϕ=π\Delta \phi = \pi) with the spatial beat period (fringes repeating, Δϕ=2π\Delta \phi = 2\pi), which incorrectly leads to a path difference twice as large.

By requiring the calculation of the phase shift directly from the wavenumber difference Δ(1/λ)\Delta(1/\lambda), the problem elegantly bypasses cumbersome wavelength approximations while strictly penalizing conceptual gaps in both modern physics and interferometry.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterAtoms - Class 12
ConceptEmission vs Absorption
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree