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Question: A ray of energy 14.2 Me V is emitted from a \(^{60}Co\) nucleus. The recoil energy of the Co-nucleus...

A ray of energy 14.2 Me V is emitted from a 60Co^{60}Co nucleus. The recoil energy of the Co-nucleus is nearly:

A) 3×10−16J B) 3×10−15J C) 3×10−14J D) 3×10−13JA)\,3 \times {10^{ - 16}}J\\\ B)\,3 \times {10^{ - 15}}J\\\ C)\,3 \times {10^{ - 14}}J\\\ D)\,3 \times {10^{ - 13}}J
Explanation

Solution

In the above question we have to apply the concepts of two topics, emissions and energy-momentum relation. The ray of energy is emitted from the 60Co^{60}Conucleus, therefore we know the cases for momentum of the both particles. Using that relation, we will calculate the recoil energy of the nucleus.

Complete step by step answer:
As we know if a ray of energy is emitted by a nucleus, the momentum of the ray emitted and the momentum of the nucleus after the emission will be equal.
Therefore, using the above assertion, we can form the equation as:
Pr=PCo′{P_r} = {P_{Co'}}, where Pr{P_r}is the momentum of the ray emitted and PCo′{P_{Co'}} is the momentum of the Co nucleus after the emission.
Now we have the above relation now we have to apply the energy momentum relation:
We know that energy momentum relation states:
E2=(pc)2+(mc)2{E^2} = {(pc)^2} + {(mc)^2}, we know that for a photon m=0, because its rest mass is zero but p is not equal to zero.
Hence the relation becomes:
E=pcE = pc, where c is the speed of light in free space.
And hence:
p=Ecp = \dfrac{E}{c}
Now if we use this value of momentum in the momentum conservation equation calculated earlier, we get:
Pr=Ec{P_r} = \dfrac{E}{c}
Now we know the momentum of the radiation emitted, now we will use the relation between momentum and kinetic energy to calculate the kinetic recoil energy of the nucleus
The relation between kinetic energy and momentum is:
K.E=P22mK.E = \dfrac{{{P^2}}}{{2m}}, where P is momentum and m is the mass.
Now we will put the calculated value of P in the above equation:
K.E = 12E2c2mCoK.E\, = \,\dfrac{1}{2}\dfrac{{{E^2}}}{{{c^2}{m_{Co}}}}.
Now we have the final equation we needed to solve the problem and we have the values that we have to put in this question.
Now we have to only convert the values of the parameters to suitable units before putting them in the equation.
E=14.2MeV ⇒E=14.2×106×1.6×10−19 ⇒E=14.2×1.6×10−13J E = 14.2MeV\\\ \Rightarrow E = 14.2 \times {10^6} \times 1.6 \times {10^{ - 19}}\\\ \Rightarrow E = 14.2 \times 1.6 \times {10^{ - 13}}J
Mass of Co nucleus: 60×1.67×10−27kg60 \times 1.67 \times {10^{ - 27}}kg
Now we put the values in the equation to get the answer to the problem:

⇒recoil energy = (14.2×1.6×10−13)22×(3×108)2×60×1.67×10−27 ⇒recoil energy = 5.1619×10−241.8036×10−8 ⇒recoil energy =2.86×10−16J ∴≈3×10−16J\Rightarrow recoil\,energy\, = \,\dfrac{{{{(14.2 \times 1.6 \times {{10}^{ - 13}})}^2}}}{{2 \times {{(3 \times {{10}^8})}^2} \times 60 \times 1.67 \times {{10}^{ - 27}}}}\\\ \Rightarrow recoil\,energy\, = \,\dfrac{{5.1619 \times {{10}^{ - 24}}}}{{1.8036 \times {{10}^{ - 8}}}}\\\ \Rightarrow recoil\,energy\,= 2.86 \times {10^{ - 16}}J\\\ \therefore\approx 3 \times {10^{ - 16}}J

Therefore, the recoil energy of the nucleus is 3×10−16J3 \times {10^{ - 16}}Jhence the correct option is (A).

Note: While solving the question the concept of energy momentum relation for a photon is very important. While calculating the rounding off of the numbers is to be done only in the final answer.The above problem can also be solved only by considering the power of 10 as the number before that is the same in all the options.