JEE Main 2023PhysicsNuclear PhysicsMediumMCQ

JEE Main 2023 — Nuclear Physics Question with Solution

JEE Main 2023 (13 Apr Shift 1)

Question

 92238A→ 90234B+D24+Q

In the given nuclear reaction, the approximate amount of energy released will be:

[Given, mass of  92238A=238.05079×931.5 MeV c-2, mass of  B90234=234.04363×931.5 MeV c-2, mass of D24=4.00260×931.5MeV c-2 ]

Choose an option

Show full solutionCorrect option: D
Correct answer
D4.25MeV

Step-by-step explanation

The value of Q is 

Q=(mA-mB-mD)×931.5 MeV⇒Q=(238.05079-234.04363-4.00260)×931.5 MeV⇒Q=4.25 MeV

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About this question

This is a previous-year question from JEE Main 2023, covering the Nuclear Physics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.