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

(4) ~ 10²⁰

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

Mass defect Δm [2 x 2.04178 - 4.00388] = 0.02568 amuu

Energy released = 0.02568 x 931 MeV = 23.91 MeV

Since efficiency of the process is 30% actual output is 23.91 x 30/100 = 7.173 MeV

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

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

No. of deutron atomsrequired to produce this output