The energy released per fission of nucleus of $^{240}$X is 200 MeV. The energy released if all the atoms in 120g of pure $^{240}$X undergo fission is ____________ $\times$ 10$^{25}$ MeV.
(Given $\mathrm{N_A=6\times10^{23}}$)
Answer (integer)
6
Solution
$120 \mathrm{~g}$ of $^{240}X$ will have $\frac{1}{2}$ mole of $X$
<br/><br/>
Number of atom of $X=\frac{1}{2} \times N_{A}=3 \times 10^{23}$ atom
<br/><br/>
Energy released $=3 \times 10^{23} \times 200 ~ \mathrm{MeV}$
<br/><br/>
$=6 \times 10^{25} ~\mathrm{MeV}$
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Nuclear Fission and Fusion
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