Ncert Exercises & Solutions

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

\({ }_{1}^{2} \mathrm{H}+{ }_{2}^{2} \mathrm{H} \rightarrow{ }_{2}^{3} \mathrm{He}+\mathrm{n}+3.27 \mathrm{MeV}\)

Hint: 2 atoms of deuterium produces 3.27 MeV energy.
Step 1: Find the total number of atoms in 2 kg of deuterium.
The given fusion reaction is :

\({ }_{1}^{2} \mathrm{H}+{ }_{2}^{2} \mathrm{H} \rightarrow{ }_{2}^{3} \mathrm{He}+\mathrm{n}+3.27 \mathrm{MeV}\)

Amount of deuterium, m=2 kg

1 mole, i.e., 2 g of deuterium contains

6.023 \times 10^{23}

atoms.

∴

2.0 kg of deuterium contains

\frac{6.023 \times 10^{23}}{2} \times 2000 = 6.023 \times 10^{26}

atoms
Step 2: Find the energy generated by 2 kg of deuterium.

It can be inferred from the given reaction that when two atoms of deuterium fuse, 3.27 MeV energy is released.

∴

Total energy released in the fusion reaction:

E = \frac{3.27}{2} \times 6.023 \times 10^{26} M e V

= \frac{3.27}{2} \times 6.023 \times 10^{26} \times 1.6 \times 10^{- 19} \times 10^{6}

= 1.576 \times 10^{14} J

Step 3: Find the time for which the lamp glows.
Power of the eletric lamp, P=100 W=100 J/s

Hene, the energy consumed by the lamp per second=100 J

The total time for which the electric lamp will glow =

\frac{1.576 \times 10^{14}}{100} \sec = \frac{1.576 \times 10^{14}}{100 \times 60 \times 60 \times 24 \times 365} \approx 4.9 \times 10^{4} y e a r s

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions