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A nucleus of mass m+Δm is at rest and decays into two daughter nuclei of equal mass each
M2. The speed of light is c. The speed of daughter nuclei is
a. cΔmM+Δm
b. c2ΔmM
C. cΔmM
d. cΔmM+Δm

Answer
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Hint: In this question, first use the law of conservation of momentum and then find the velocity of the daughter nuclei is the same. Then find the total kinetic energy of the two daughter nuclei and mass defect and then find the velocity of the two daughter nuclei.

Complete step by step answer:
A nucleus decays into two parts and the mass of each nucleus is M2. The mass of the parent nucleus is M+Δm .
Let us assume that the speed of daughter nuclei is V1 and V2 respectively.
Hence conservation of momentum, M2V1=M2V2V1=V2
Now the mass defect is M+Δm(M2+M2)=Δm
As the product of mass defect and the square of the speed of light is the total kinetic energy.
So, the kinetic energy of the two daughter nuclei is E1=12.M2.V12 and E2=12.M2.V22
Hence the total kinetic energy of the two daughter nuclei is E1+E2=12.M2.V12+12.M2.V22
As the velocity of the two daughter nuclei is same V1=V2
Thus, the total kinetic energy of the two daughter nuclei is M2V12 .
Δmc2=M2V12
V1=c2ΔmM
The speed of the two daughter nuclei is c2ΔmM.
Hence option (b) is the correct answer.

Note: As we know that the law of conservation of momentum states that in an isolated system, when the two objects collide with each other the total momentum of two objects before the collision is equal to the total momentum after the collision. Momentum is neither destroyed nor created; it transforms into one form to another.