JEE MAIN - Physics (2023 - 24th January Evening Shift - No. 22)
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
6
Explanation
$120 \mathrm{~g}$ of $^{240}X$ will have $\frac{1}{2}$ mole of $X$
Number of atom of $X=\frac{1}{2} \times N_{A}=3 \times 10^{23}$ atom
Energy released $=3 \times 10^{23} \times 200 ~ \mathrm{MeV}$
$$ =6 \times 10^{25} ~\mathrm{MeV} $$
Number of atom of $X=\frac{1}{2} \times N_{A}=3 \times 10^{23}$ atom
Energy released $=3 \times 10^{23} \times 200 ~ \mathrm{MeV}$
$$ =6 \times 10^{25} ~\mathrm{MeV} $$
Comments (0)
