The energy released per fission of nucleus of $$^{240}$$X is 200 MeV. The energy released if all the atoms in 120 g of pure $$^{240}$$X undergo fission is _____ $$\times 10^{25}$$ MeV.
(Given $$N_A = 6 \times 10^{23}$$)

$$n = \frac{\text{Given Mass}}{\text{Molar Mass}} = \frac{120 \text{ g}}{240 \text{ g/mol}} = 0.5 \text{ mol}$$

$$N = n \times N_A = 0.5 \times 6 \times 10^{23} = 3 \times 10^{23} \text{ atoms}$$

$$E_{\text{total}} = N \times (\text{Energy per fission})$$

$$E_{\text{total}} = 6 \times 10^{25} \text{ MeV}$$