92238 A → 90234 B + D 24 + Q In the given nuclear reaction, the...

$92238 A \to 90234 B + D 24 + Q$

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

[Given, mass of $92238 A = 238 . 05079 \times 931 . 5MeV{c}^{- 2}$ , mass of $B 90234 = 234 . 04363 \times 931 . 5MeV{c}^{- 2}$ , mass of $D 24 = 4 . 00260 \times 931 . 5MeV{c}^{- 2}$ ]

Options

A $3 . 82MeV$
B $5 . 9MeV$
C $2 . 12MeV$
D $4 . 25MeV$

Correct answer

D. $4 . 25MeV$

Step-by-step solution

The value of $Q$ is
$Q = ( {m}_{A} - {m}_{B} - {m}_{D} ) \times 931 . 5MeV ⇒ Q = ( 238 . 05079 - 234 . 04363 - 4 . 00260 ) \times 931 . 5MeV ⇒ Q = 4 . 25MeV$

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