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 \( 1 \text{ u} = 931 \text{ MeV/c}^2 \)
- A 3.472 MeV
- B 4.732 MeV
- C 2.375 MeV
- D 4.372 MeV
Answer & Solution
Correct Answer
(D) 4.372 MeV
Step-by-step Solution
Detailed explanation
\( \Delta m = m(^{23}_{10}\text{Ne}) - m(^{23}_{11}\text{Na}) = 22.994466 \text{ u} - 22.989770 \text{ u} = 0.004696 \text{ u} \)…
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