A star initially has 10^{40} deuterons. It pr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The net reaction is: $3(_1H^2) 	o _2He^4 + p + n$.Mass defect $\Delta m = 3 	imes M(H^2) - [M(He^4) + M(p) + M(n)]$$\Delta m = 3(2.014) - [4.001

Vedclass · Accepted Answer

$10^{12} \, sec$

Vedclass · Answer

(C) The net reaction is: $3(_1H^2) 	o _2He^4 + p + n$.Mass defect $\Delta m = 3 	imes M(H^2) - [M(He^4) + M(p) + M(n)]$$\Delta m = 3(2.014) - [4.001 + 1.007 + 1.008] = 6.042 - 6.016 = 0.026 \, amu$.Energy released per reaction $Q = 0.026 	imes 931.5 \, MeV \approx 24.22 \, MeV$.$Q = 24.22 	imes 10^6 	imes 1.6 	imes 10^{-19} \, J \approx 3.875 	imes 10^{-12} \, J$.Since $3$ deuterons are used to produce $Q$ energy, energy per deuteron is $E_d = Q/3 = 1.29 	imes 10^{-12} \, J$.Total energy available $E_{total} = N 	imes E_d = 10^{40} 	imes 1.29 	imes 10^{-12} = 1.29 	imes 10^{28} \, J$.Time taken $t = E_{total} / P = (1.29 	imes 10^{28}) / 10^{16} = 1.29 	imes 10^{12} \, s$.Thus, the time is of the order of $10^{12} \, s$.

Similar Questions

The binding energy per nucleon for a deuteron and helium are $1.1 \, MeV$ and $7 \, MeV$ respectively. If two deuteron nuclei fuse to form one helium nucleus,the energy released is $...... \, MeV$.

In the process of a nuclear explosion,in which form is the maximum energy released?

In any fission process,the ratio $\frac{\text{mass of fission products}}{\text{mass of parent nucleus}}$ is

The scientific principle that forms the basis of the Tokamak technology is

In a nuclear fission reaction of an isotope of mass $M$,three similar daughter nuclei of the same mass are formed. The speed of a daughter nucleus in terms of mass defect $\Delta M$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module