Question

The fusion reaction 2₁H²→ ₂He⁴ + energy is proposed to use for the production of industrial power. Assuming the efficiency of the process to be 30%, the number of deutrons atoms per second required to produce an output of 50000 kW is of the order of [Given mass of ₁H² = 2.01478 amu and mass of ₁He² = 4.00388 amu]. 

(1) ~ 10¹⁷ 

(2) ~ 10¹⁸ 

(3) ~ 10¹⁹ 

(4) ~ 10²⁰

Answer 1

Correct option : (4) ~ 10²⁰

Explanation: 

The reaction is represented by 2₁H² → ₂He⁴ + energy 

Mass defect Δm [2×2.04178 - 4.00388] = 0.02568 amu 

Energy released = 0.02568 × 931 MeV = 23.91 MeV

Since efficiency of the process is 30% actual output is 23.91 x 30/100 = 7.17 MeV

Actual output per deutron atom is 7.173/2 = 3.587 MeV = 3.587×1.6×10^–12 J

Output required = 50000 kW = 5×10⁷ J = 5×10⁷ J 

No. of deutron atoms required to produce this output