Atoms & Nuclei

JEE Main level, single correct, medium

198 of 239

Question 198 of 239

92238A→90234B+D24+Q92238A\rightarrow 90234B+D24+Q
In the given nuclear reaction, the approximate amount of energy released will be:
[Given, mass of 92238A=238.05079×931.5MeVc−292238A=238.05079\times 931.5MeV{c}^{-2}, mass of B90234=234.04363×931.5MeVc−2B90234=234.04363\times 931.5MeV{c}^{-2}, mass of D24=4.00260×931.5MeVc−2D24=4.00260\times 931.5MeV{c}^{-2} ]

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Answer: (D) 4.25MeV4.25MeV

The value of QQ is
Q=(mA−mB−mD)×931.5MeV⇒Q=(238.05079−234.04363−4.00260)×931.5MeV⇒Q=4.25MeVQ=({m}_{A}-{m}_{B}-{m}_{D})\times 931.5MeV \Rightarrow Q=(238.05079-234.04363-4.00260)\times 931.5MeV \Rightarrow Q=4.25MeV

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