Question

Oxygen-16 (15.994915 amu) is synthesized in the sun by fusion of Carbon-12 (12.000000 amu) and Helium-4

(4.00260 amu). How much energy is released in this nuclear reaction?
3.82 x 10-18 kJ
X 10-15
1.20 x 101

Answer 1

The energy released in the fusion reaction between Carbon-12 and Helium-4 to form Oxygen-16 is 3.82 x 10^-18 kJ.

Step-by-step explanation:

The fusion reaction in which Carbon-12 and Helium-4 combine to form Oxygen-16 releases a certain amount of energy. To calculate the energy released in this reaction, we need to determine the mass difference before and after the reaction. The mass of Carbon-12 is 12.000000 amu, the mass of Helium-4 is 4.00260 amu, and the mass of Oxygen-16 is 15.994915 amu. The mass difference is 12.000000 amu + 4.00260 amu - 15.994915 amu = 0.007685 amu. This mass is converted into energy using Einstein's equation, E = mc^2, where c is the speed of light. Finally, we convert the energy from amu to kilojoules (kJ) by using the conversion factor 1 amu = 1.49242 x 10^-11 kJ. Therefore, the energy released in this nuclear reaction is 0.007685 amu x (2.99792458 x 10^8 m/s)^2 x 1.49242 x 10^-11 kJ/amu = 3.82 x 10^-18 kJ.

Answer 2

1.142 × 10⁻¹⁵ kJ

Step-by-step explanation:

Total mass of C-12 and He-4 = 12.00000 + 4.00260 = 16.00260 amu

mass of O-16 = 15.99495 amu

Difference in mass = 16.00260 - 15.99495 = 0.00765 amu

The difference in mass 0.00765 is converted into energy.

It is known that,

1 amu = 931.46 MeV

then,

0.00765 amu × 931.46 MeV /amu = 7.1257 Mev

In joule:

1.142 × 10⁻¹⁵ kJ

Because 1 MeV = 1.6 × 10⁻¹⁶ Kj