Nuclear Physics Question 2
Question 2 - 24 January - Shift 2
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 thongo $\times 10^{25}$ $MeV$.
(Given $N_A=6 \times 10^{23}$ )
Show Answer
Answer: (6)
Solution:
Formula: Binding energy
No. of mole $=\frac{120}{240}=\frac{1}{2}$
No. of molecules $=\frac{1}{2} \times N_A^{al}$
Energy released $=\frac{1}{2} \times 6 \times 10^{23} \times 200$
$=6 \times 10^{25} MeV$
