Question:

A white dwarf star has volume \( V \) and contains \( N \) electrons so that the density of electrons is \( n = \frac{N}{V} \). Taking the temperature of the star to be 0 K, the average energy per electron in the star is \( \epsilon_0 = \frac{3h^2}{10m} (3n)^{2/3 \), where \( m \) is the mass of the electron. The electronic pressure in the star is:}

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The pressure in a degenerate electron gas (such as in a white dwarf star) is given by a formula proportional to the energy density, where the proportionality constant depends on the dimension and the type of particle.
Updated On: Nov 18, 2025
  • \( n \epsilon_0 \)
  • \( 2n \epsilon_0 \)
  • \( \frac{1}{3} n \epsilon_0 \)
  • \( \frac{2}{3} n \epsilon_0 \)
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the electronic pressure.
In a white dwarf star, the electronic pressure is related to the energy density of the electrons. The total energy per unit volume is given by the electron's average energy per electron, \( \epsilon_0 \), and the number density \( n \). Using the standard equation for pressure in terms of energy density, we find that the electronic pressure is \( P = \frac{1}{3} n \epsilon_0 \).
Step 2: Conclusion.
Thus, the correct answer is option (C).
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