A star has 10^{40} deuterons. It produces ene | vedclass

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

Question

(C) The net reaction is obtained by adding the two processes:$3(_1H^2) 	o _2He^4 + p + n$Calculate the mass defect $\Delta m$:$\Delta m = 3 	imes m(

Vedclass · Accepted Answer

$10^{12} \ s$

Vedclass · Answer

(C) The net reaction is obtained by adding the two processes:$3(_1H^2) 	o _2He^4 + p + n$Calculate the mass defect $\Delta m$:$\Delta m = 3 	imes m(_1H^2) - [m(_2He^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 net reaction $(Q)$:$Q = 0.026 	imes 931.5 \ MeV = 24.219 \ MeV = 24.219 	imes 1.6 	imes 10^{-13} \ J \approx 3.875 	imes 10^{-12} \ J$Since $3$ deuterons are consumed to release $Q$ energy, energy per deuteron is $Q/3 \approx 1.29 	imes 10^{-12} \ J$.Total energy available $E_{total} = (10^{40} / 3) 	imes Q = 1.29 	imes 10^{28} \ J$.Time taken $t = E_{total} / P = (1.29 	imes 10^{28}) / 10^{16} = 1.29 	imes 10^{12} \ s$.Thus, the order of time is $10^{12} \ s$.

Similar Questions

What will be the energy released in joule, in the process of fission by $1 \text{ mg}$ of ${ }_{92}^{240} U$? Assume energy release per fission is $200 \text{ MeV}$. [Use Avogadro's number as $6 \times 10^{23}$ and $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$]

When neutrons are bombarded on the nucleus of $_{92}^{235}U$,the number of emitted neutrons is typically

Which of the following statements is correct?

Which of the following are suitable for the fusion process?

Vedclass Test Series

Exam Paper Generator

Online Exam Module