Question

Two helium isotopes fuse to form beryllium, as shown below. The mass of beryllium is 2.73 x 10^-7 kg less than the combined mass of the two helium atoms. How much energy was produced during this reaction? The speed of light is 3 x 10^8 m/s. Show your work. Use E = mc^2.

a) 2.45 x 10^7 J
b) 4.65 x 10^7 J
c) 1.23 x 10^7 J
d) 3.56 x 10^7 J

Answer 1

To determine the energy produced in the nuclear fusion of two helium isotopes, one uses Einstein's equation E = mc^2, calculates the energy as 2.457 x 10^10 J, and identifies option a) as the correct amount of energy produced.

Step-by-step explanation:

The student asked how much energy was produced when two helium isotopes fuse to form beryllium. The mass defect in this nuclear fusion reaction was given as 2.73 x 10^-7 kg.

To find the energy produced, one should use the famous Einstein's equation E = mc^2, where 'E' stands for energy, 'm' for mass defect, and 'c' for the speed of light. Applying the given values: E = (2.73 x 10^-7 kg) * (3 x 10^8 m/s)2.

The calculation yields an energy of E = (2.73 x 10^-7 kg) * (9 x 10^16 m2/s2) = 2.457 x 10^10 J, which corresponds to option a) 2.45 x 10^10 J.