A star initially has 1040 deuterons. It produces energy via the processes 12H+12H→13H+P and 12H+13H→ 24He+n. If the average power radiated by the star is 1016W, the deuteron supply of the star is exhausted in a time of the order of [M(2H) = 2.104 amu, M(n) = 1.008 amu, M(p) = 1.008 amu and M(4He) = 1.008 amu]

3 12H→24He+n+p∆m=3×2.104-4.001-1.008-1.007=0.026 amuEnergy released = 0.026×931 MeV = 0.026×931×1.6×10-13 J=3.87×10-12 JTotal Energy =10403×3.87×10-12 J=1.29×1028 Javg power=1016 J/st=1.29×1028 J1016 J/s=1.29×1012 order of 1012 s

A star initially has 1040 deuterons. It produces energy via the processes 12H+12H→13H+P and 12H+13H→ 24He+n. If the average power radiated by the star is 1016W, the deuteron supply of the star is exhausted in a time of the order of [M(2H) = 2.104 amu, M(n) = 1.008 amu, M(p) = 1.008 amu and M(4He) = 1.008 amu]

3 12H→24He+n+p∆m=3×2.104-4.001-1.008-1.007=0.026 amuEnergy released = 0.026×931 MeV = 0.026×931×1.6×10-13 J=3.87×10-12 JTotal Energy =10403×3.87×10-12 J=1.29×1028 Javg power=1016 J/st=1.29×1028 J1016 J/s=1.29×1012 order of 1012 s

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A star initially has 10⁴⁰ deuterons. It produces energy via the processes $_{1}^{2} H +_{1}^{2} H \to_{1}^{3} H + P$ and $_{1}^{2} H +_{1}^{3} H {\overset{}{\to}}_{2}^{4} He + n$ . If the average power radiated by the star is 10¹⁶W, the deuteron supply of the star is exhausted in a time of the order of
[M(²H) = 2.104 amu, M(n) = 1.008 amu, M(p) = 1.008 amu and M(⁴He) = 1.008 amu]

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

10⁶ s

10⁸ s

10¹² s

10¹⁶ s

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{rcl} 3 (_{1}^{2} H) \to_{2}^{4} He + n + p \\ ∆ m & = & 3 \times 2.104 - [4.001 - 1.008 - 1.007] = 0.026 amu \\ Energy released & = & 0.026 \times 931 MeV = 0.026 \times 931 \times 1.6 \times 10^{- 13} J = 3.87 \times 10^{- 12} J \\ T o t a l E n e r g y & = & \frac{10^{40}}{3} \times 3.87 \times 10^{- 12} J = 1.29 \times 10^{28} J \\ avg power & = & 10^{16} J / s \\ t & = & \frac{1.29 \times 10^{28} J}{10^{16} J / s} = 1.29 \times 10^{12} \\ order of 10^{12} s \end{array}$