If three helium nuclei combine to form a carbon nucleus then the energy released in this reaction is .......... × 10–2 MeV.
(Given 1 u = 931 MeV/c2 , atomic mass of helium = 4.002603 u)
(Given 1 u = 931 MeV/c2 , atomic mass of helium = 4.002603 u)
Correct Answer: 727
Solution and Explanation
Given reaction:
\[ ^3_2\text{He} \longrightarrow ^{12}_6\text{C} + \gamma \text{ rays} \]
Mass defect:
\[ \Delta m = (3m_{\text{He}} - m_{\text{C}}) \]
Calculating:
\[ \Delta m = (3 \times 4.002603 - 12) = 0.007809 \, \text{u} \]
Energy released:
\[ \text{Energy} = 931 \Delta m \, \text{MeV} \] \[ = 7.27 \, \text{MeV} = 727 \times 10^{-2} \, \text{MeV} \]
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