{ }_{92}^{238} {\A} \to{ }_{90}^{234} {\B}+{ }_2^4 {\D}+ {\Q} In the given nuclear reaction, the approximate amount of energy released will be: [Given, mass of { }_{92}^{238} {\A}=238.05079 \times 931.5 {\MeV} \/ {\c}^2, \text{ mass of }{ }_{90}^{234} {\B}=234.04363 \times 931.5 {\MeV} \/ {\c}^2 \text{, } \text{ mass of } { }_2^4 {\D}=4.00260 \times 931.5 {\MeV} \/ {\c}^2] [13-Apr-2023 shift 1]

JEE Mains 13-Apr-2023 Shift 1 Solved Paper

Section: Physics

Question : 41 of 90

Marks: +1, -0

‌92238A→‌90234B+‌24D+Q
In the given nuclear reaction, the approximate amount of energy released will be: [Given, mass of ‌92238A=238.05079×931.5‌MeV∕c2,
‌‌ mass of ‌‌90234B=234.04363×931.5‌MeV∕c2‌, ‌
‌‌ mass of ‌‌24D=4.00260×931.5‌MeV∕c2]

[13-Apr-2023 shift 1]

4.25‌MeV
5.9‌MeV
3.82‌MeV
2.12‌MeV

Validate

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