In the given nuclear reaction, the approximate amount of energy released will be: [2023]

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Solution

Q.
${}_{92}^{238}A \to {}_{90}^{234}B + {}_{2}^{4}D + Q$

In the given nuclear reaction, the approximate amount of energy released will be: [2023]

[Given, mass of ${}_{92}^{238}A = 238.05079 \times 931.5 MeV / c^{2}$

mass of ${}_{90}^{234}B = 234.04363 \times 931.5 MeV / c^{2}$

mass of ${}_{2}^{4}D = 4.00260 \times 931.5 MeV / c^{2}$ ]

1 3.82 MeV

2 5.9 MeV

3 2.12 MeV

4 4.25 MeV

Ans.
(4)

$Q = (m_{A} - m_{B} - m_{D}) \times 931.5 MeV$

$= (238.05079 - 234.04363 - 4.00260) \times 931.5$

$\Rightarrow 4.25 MeV$

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