(b) Deuterium undergoes fusion as per the reaction: 12H+12H⟶23He+01n+3⋅27MeVFind the duration for which an electric bulb of 500 W can be kept glowing by the fusion of 100 g of deuterium.

Number of atoms in 100g deuterium, 1002×6.023×10−23Energy Released- 3.272 x1.6 x 10-13 JTime required t = Total energy(E) / Power of lamp (P)= 500 x 60 x 60 x 24 x 365t = 1.5768 x 1010 years.

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

(b) Deuterium undergoes fusion as per the reaction:
$_{1}^{2} H +_{1}^{2} H ⟶_{2}^{3} He +_{0}^{1} n + 3 \cdot 27 MeV$
Find the duration for which an electric bulb of 500 W can be kept glowing by the fusion of 100 g of deuterium.

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

answer is 1.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Number of atoms in 100g deuterium, $\frac{100}{2} \times 6 .023 \times 10^{- 23}$
Energy Released- $\frac{3.27}{2} x 1.6 x 10^{- 13} J$
Time required t = Total energy(E) / Power of lamp (P)
= 500 x 60 x 60 x 24 x 365
t = 1.5768 x 10¹⁰ years.