Question 49

A radionuclide of atomic mass $$300\text{ amu}$$ on fission liberates $$200\text{ MeV}$$ of energy. The energy liberated when $$1\text{ g}$$ of this radionuclide undergoes fission is (given Avogadro number $$= 6 \times 10^{23}$$)

Solution

$$1\text{ g}$$ contains $$\frac{1}{300} \times 6 \times 10^{23} = 2 \times 10^{21}$$ nuclei. Energy per fission $$= 200\text{ MeV} = 200 \times 1.6 \times 10^{-13} = 3.2 \times 10^{-11}\text{ J}$$. Total energy $$= 2 \times 10^{21} \times 3.2 \times 10^{-11} = 6.4 \times 10^{10}\text{ J}$$.

Was this solution helpful?

Get AI Help

50,000+ JEE Students Trusted Our Score Calculator

Predict your JEE Main percentile, rank & performance in seconds

Ask AI