Question 50: Atoms & Nuclei

Question

In a nuclear reaction 2 deuteron nuclei combine to form a helium nucleus. The energy released in $$\mathrm{MeV}$$ will be: (Given mass of deuteron $$=2.01355 \mathrm{~amu}$$. and mass of helium nucleus $$=4.0028 \mathrm{~amu}$$.

Accepted Answer

To find the energy released in this nuclear reaction, we use the concept of mass defect and Einstein's mass-energy equivalence principle. The mass defect is the difference between the mass of the reactants and the mass of the products in a nuclear reaction. According to Einstein's relation, $$E=mc^2$$, where $$E$$ is the energy released, $$m$$ is the mass defect, and $$c$$ is the speed of light in vacuum.
Given data:
Mass of deuteron = $$2.01355\, \mathrm{amu}$$
Mass of helium nucleus = $$4.0028\, \mathrm{amu}$$
The reaction involves two deuteron nuclei combining into one helium nucleus. First, let's calculate the total mass of the reactants (2 deuterons):
$$	ext{Total mass of reactants} = 2 	imes 	ext{mass of deuteron} = 2 	imes 2.01355\, \mathrm{amu}$$
$$	ext{Total mass of reactants} = 4.0271\, \mathrm{amu}$$
Now, let's find the mass defect ($$\Delta m$$), which is the difference between the mass of the reactants and the mass of the product (helium nucleus):
$$\Delta m = 	ext{Total mass of reactants} - 	ext{mass of product}$$
$$\Delta m = 4.0271\, \mathrm{amu} - 4.0028\, \mathrm{amu}$$
$$\Delta m = 0.0243\, \mathrm{amu}$$
To convert the mass defect from atomic mass units (amu) to kilograms (kg), we use the conversion factor $$1\, \mathrm{amu} = 1.660539040 	imes 10^{-27}\, \mathrm{kg}$$:
$$\Delta m \,(\mathrm{kg}) = 0.0243\, \mathrm{amu} 	imes 1.660539040 	imes 10^{-27} \, \mathrm{kg/amu}$$
$$\Delta m \, (\mathrm{kg}) = 4.03311 	imes 10^{-29}\, \mathrm{kg}$$
The energy released can now be calculated using the formula $$E=mc^2$$, where $$c = 2.99792458 	imes 10^8 \, \mathrm{m/s}$$ (speed of light):
$$E = (4.03311 	imes 10^{-29}\, \mathrm{kg}) 	imes (2.99792458 	imes 10^8 \, \mathrm{m/s})^2$$
$$E = 3.62894 	imes 10^{-12}\, \mathrm{J}$$
To convert energy from Joules to Mega-electronvolts ($$\mathrm{MeV}$$), we use $$1\, \mathrm{J} = 6.242 	imes 10^{12}\, \mathrm{MeV}$$:
$$E = 3.62894 	imes 10^{-12}\, \mathrm{J} 	imes 6.242 	imes 10^{12}\, \mathrm{MeV/J}$$
$$E = 22.65\, \mathrm{MeV}$$
The energy released is approximately $$22.65\, \mathrm{MeV}$$, which is closest to Option C, $$22.62\, \mathrm{MeV}$$. Therefore,
Option C
is the correct answer.

Answer

24.3 MeV

Answer

2.262 MeV

Answer

22.62 MeV

Answer

0.0243 MeV

Atoms & Nuclei

In a nuclear reaction 2 deuteron nuclei combine to form a helium nucleus. The energy released in $\mathrm{MeV} will be: (Given mass of deuteron =2.01355 \mathrm{~amu}. and mass of helium nucleus =4.0028 \mathrm{~amu}$.

Choose the correct answer: