Calculate the amount of energy released per gram of nuclei when 2 protium nuclei react to form deuterium nucleus (protium = 1.00782 amu, deuterium = 2.01410 amu).
The answer in the book is -2.0 x 10^10 J, but I don't have any idea on how to do this. Please help, and thank you so much.
I get close to that answer if we neglect the fact that an extra electron produced is ignored, AND I think you transposed numbers for the mass of the proton. I think
(2.0141 - 2*1.00728) = ? amu
Change to kg.
?amu x (1/6.022E23*1000) = kg
E = kg*c^2 and that gives me 6.86E-14 per atom.
Convert to mols = 6.86E-14 x 6.022E23 = -4.13E10J/2.0141 grams or 2.05E10 J/g.
To calculate the amount of energy released per gram of nuclei during a nuclear reaction, we can use Einstein's famous equation, E = mc², where E is the energy, m is the mass, and c is the speed of light.
First, let's find the mass difference between the reactants and the products.
The mass difference (Δm) is given by:
Δm = (mass of reactants) - (mass of products)
In this case, the reactants are two protium nuclei (1H) with a mass of 1.00782 amu each, and the product is a deuterium nucleus (2H) with a mass of 2.01410 amu.
The total mass of the reactants is:
Mass of reactants = 2 * (mass of protium nucleus)
Mass of reactants = 2 * 1.00782 amu
The mass of the product is:
Mass of product = mass of deuterium nucleus
Mass of product = 2.01410 amu
Now, we can calculate the mass difference:
Δm = (2 * 1.00782 amu) - 2.01410 amu
Next, we need to convert the mass difference from atomic mass units (amu) to kilograms (kg), as units of energy are specified in joules (J).
To convert 1 amu to kg, we will use the conversion factor:
1 amu = 1.66054 × 10^-27 kg
Thus, the mass difference in kilograms is:
Δm (in kg) = Δm (in amu) * (1.66054 × 10^-27 kg/1 amu)
Now we can calculate the energy released using Einstein's equation, E = mc², where c = 2.998 × 10^8 m/s is the speed of light:
E = (Δm in kg) * (c²)
E = (Δm (in amu) * (1.66054 × 10^-27 kg/1 amu)) * (2.998 × 10^8 m/s)²
Substituting the values into the equation, we get:
E = [(2 * 1.00782 amu - 2.01410 amu) * (1.66054 × 10^-27 kg/1 amu)] * (2.998 × 10^8 m/s)²
Now, we can calculate this energy using a calculator. The result should be approximately -2.0 x 10^10 J, as given in the book. The negative sign indicates that energy is released during the reaction.
Please note that the actual calculated value may vary slightly due to rounding errors or different values for atomic masses.