Question

In a nuclear fusion reaction two 2H atoms are combined to produce one 4He.

(a) Calculate the decrease in rest mass in unified mass units.

(b) How much energy is released in this reaction?

(c) How many such reactions must take place per second to produce 59 W of power?

Answer 1

a)= 0.025602u

b) = 23.848MeV

c) N = 1.546 × 10¹³

Step-by-step explanation:

The reaction is

²₁H + ²₁H ⇄ ⁴₂H + Q

a) The mass difference is

Δm = 2m(²₁H) - m (⁴₂H)

= 2(2.014102u) - 4.002602u

= 0.025602u

b) Use the Einstein mass energy relation ship

The enegy release is the mass difference times 931.5MeV/U

E = (0.025602) (931.5)

= 23.848MeV

c)

the number of reaction need per seconds is

N = Q/E

= 59W/ 23.848MeV

`= (59)/((23.848 * 10^6 )(1.6 * 10^1^9) ) \\\\= 1.546 * 10^1^3`

N = 1.546 × 10¹³