Get Answers to all your Questions

header-bg qa

Q. 13.19 How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as

                   _{1}^{2}\textrm{H}+_{1}^{2}\textrm{H}\rightarrow _{2}^{3}\textrm{He}+n+3.27\; MeV

                 

Answers (1)

best_answer

The energy liberated on the fusion of two atoms of deuterium= 3.27 MeV

Number of fusion reactions in 2 kg of deuterium = NA\times500

The energy liberated by fusion of 2.0 kg of deuterium atoms E

\\=3.27\times 10^{6}\times 1.6\times 10^{-19}\times 6.023\times 10^{23}\times 500\\=1.576\times 10^{14}\ J 

Power of lamp (P)= 100 W

Time the lamp would glow using E amount of energy is T=

\\=\frac{E}{P}\\ =\frac{1.576\times 10^{14}}{100\times 3600\times 24\times 365}

=4.99\times104 years

Posted by

Sayak

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads