Question:

When the speed of light becomes \( \frac{2}{3} \) of its present value, then the energy released in a given atomic explosion would

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When \( c \to \frac{2}{3}c \), energy \( E \to \frac{4}{9}E \), so decrease = \( \frac{5}{9} \).
Updated On: Apr 23, 2025
  • decrease by a factor \( \frac{2}{3} \)
  • decrease by a factor \( \frac{4}{9} \)
  • decrease by a factor \( \frac{5}{9} \)
  • decrease by a factor \( \frac{\sqrt{5}}{9} \)
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The Correct Option is C

Solution and Explanation

Energy released in atomic explosion is given by Einstein’s equation: \[ E = mc^2 \] If speed of light becomes \( \frac{2}{3}c \), then: \[ E' = m \left( \frac{2}{3}c \right)^2 = m \cdot \frac{4}{9} c^2 = \frac{4}{9} E \] So the energy decreases by a factor: \[ 1 - \frac{4}{9} = \frac{5}{9} \]
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