Question:medium

A relation connecting the isotopic mass \(M\) of a superconductor with its critical temperature \(T_c\) is given by:

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Isotope effect: \(T_c \propto M^{-1/2}\), so \(M^{1/2}T_c\) stays constant across isotopes.
Updated On: Jul 2, 2026
  • \(M = kT_c\)
  • \(M^{1/2}\,T_c = \text{a constant}\)
  • \(M_c^{1/2} = \text{a constant}\)
  • \(M_c^{2} = \text{a constant}\)
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The Correct Option is B

Solution and Explanation

Step 1: Experiments on different isotopes of the same superconducting element (for example mercury) show that heavier isotopes have a lower critical temperature. This is the isotope effect, a key clue that lattice vibrations drive conventional superconductivity.

Step 2: The empirical law is written $T_c M^{\alpha} = \text{constant}$, where $\alpha$ is the isotope exponent. For the ideal BCS case $\alpha = 1/2$.

Step 3: Substituting $\alpha = 1/2$ gives $T_c M^{1/2} = \text{constant}$, equivalently $M^{1/2}T_c = \text{constant}$. Rearranged, $T_c \propto 1/\sqrt{M}$.

Step 4: The square-root dependence comes directly from the Debye frequency $\omega_D \propto M^{-1/2}$ of the ion lattice; since the pairing energy scale follows the phonon frequency, $T_c$ inherits the same mass dependence.\[\boxed{M^{1/2}\,T_c = \text{constant}}\]
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