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CUET · PHYSICS · PYQ PAPER 2025

"The nucleus \( ^{23}_{10}\text{Ne} \) decays by \(\beta\)- emission. The maximum kinetic energy of the emitted electrons is :
Given that : \( m(^{23}_{10}\text{Ne}) = 22.994466 \text{ u} \), \( m(^{23}_{11}\text{Na}) = 22.989770 \text{ u} \) and \( 1 \text{ u} = 931 \text{ MeV/c}^2 \)

  1. A 3.472 MeV
  2. B 4.732 MeV
  3. C 2.375 MeV
  4. D 4.372 MeV
Verified Solution

Answer & Solution

Correct Answer

(D) 4.372 MeV

Step-by-step Solution

Detailed explanation

\( \Delta m = m(^{23}_{10}\text{Ne}) - m(^{23}_{11}\text{Na}) = 22.994466 \text{ u} - 22.989770 \text{ u} = 0.004696 \text{ u} \)…