Question 11 of 13intermediate🔧 ApplyNumerical3 marks

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

12H+12H→23He+n+3.27MeV^{2}_{1}\text{H} + ^{2}_{1}\text{H} \rightarrow ^{3}_{2}\text{He} + n + 3.27\text{MeV}
Correct Answer

The electric lamp can be kept glowing for approximately 50,000 years.

Exercise: EXERCISES | Q: 13.8 | (Page 17)
For More Understanding

Explanation

This numerical problem requires calculating the total energy released from fusion of 2.0 kg of deuterium and then finding how long a 100W lamp can run on that energy. The key steps involve finding the number of deuterium atoms, determining the number of fusion reactions (since 2 atoms fuse per reaction), calculating total energy released, and finally finding the time using Power = Energy/Time.

Solution Steps

  1. Step 1: Calculate number of deuterium atoms in 2.0 kg Number of moles = 2.0 kg / (2.014102×102.014102 \times 10⁻³ kg/mol) = 992.99 moles Number of atoms = 992.99×6.022992.99 \times 6.022 × 10²³ = 5.976×105.976 \times 10²⁶ atoms

  2. Step 2: Calculate number of fusion reactions In the reaction ²₁H + ²₁H → ³₂He + n + 3.27 MeV, two deuterium atoms fuse per reaction. Number of reactions = 5.976×105.976 \times 10²⁶ / 2 = 2.988×102.988 \times 10²⁶ reactions

  3. Step 3: Calculate total energy released Total energy = 2.988×102.988 \times 10²⁶ × 3.27 MeV = 9.77×109.77 \times 10²⁶ MeV Converting to Joules: Energy = 9.77×109.77 \times 10²⁶ × 1.602×101.602 \times 10⁻¹³ J = 1.566×101.566 \times 10¹⁴ J

  4. Step 4: Calculate time Time = Energy / Power = 1.566×101.566 \times 10¹⁴ J / 100 W = 1.566×101.566 \times 10¹² s Time in years = 1.566×101.566 \times 10¹² / (365.25×24365.25 \times 24 × 3600) ≈\approx 4.96×104.96 \times 10⁴ years ≈\approx 50,000 years

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