JEE Mains · Physics · STD 12 - 13. Nuclei
The energy released when \(\dfrac{7}{17.13}\) kg of \(^{7}_{3}\text{Li}\) is converted into \(^{4}_{2}\text{He}\) by proton bombardment is \(\alpha \times 10^{32}\) eV. The value of \(\alpha\) is _______. (Nearest integer) (Mass of \(^{7}_{3}\text{Li} = 7.0183\) u, mass of \(^{4}_{2}\text{He} = 4.004\) u, mass of proton \(= 1.008\) u and \(1\) u \(= 931\) MeV/c\(^2\) and Avogadro number \(= 6.0 \times 10^{23}\))
- A 6
- B 12
- C 18
- D 24
Answer & Solution
Correct Answer
(A) 6
Step-by-step Solution
Detailed explanation
The nuclear reaction for the proton bombardment of Lithium is: \(^{7}_{3}\text{Li} + ^{1}_{1}\text{H} \rightarrow 2 ^{4}_{2}\text{He}\) The mass defect (\(\Delta m\)) for the reaction is: \(\Delta m = m(^{7}_{3}\text{Li}) + m(^{1}_{1}\text{H}) - 2m(^{4}_{2}\text{He})\)…
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