The value of Q is
Q=(mA−mB−mD)×931.5MeV⇒Q=(238.05079−234.04363−4.00260)×931.5MeV⇒Q=4.25MeV
JEE Main 2023 — Physics Modern Physics
92238A→90234B+D24+Q
In the given nuclear reaction, the approximate amount of energy released will be:
[Given, mass of 92238A=238.05079×931.5MeVc−2, mass of B90234=234.04363×931.5MeVc−2, mass of D24=4.00260×931.5MeVc−2 ]
Held on 13 Apr 2023 · Verified 6 Jul 2026.
3.82MeV
5.9MeV
2.12MeV
4.25MeV
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