Calcium-39 undergoes positron decay. Each positron carries 5.49 MeV of energy. How much energy will be emitted when 0.0025 mol of calcium-39 decays? A)13.2 kJ B)1.32 × 10 ⁴kJ C)1.32 × 10 ⁶ kJ D)1.32 × 10 ⁹kJ E)None of these choices are correct.

Question

Quizplus · Accepted Answer

First, we need to use the Avogadro's number to convert 0.0025 mol of calcium-39 to the number of atoms:

0.0025 mol x (6.022 x 10^23 atoms/mol) = 1.506 x 10^21 atoms

Each decay of calcium-39 produces one positron, so the total number of positrons produced is also 1.506 x 10^21.

The total energy emitted can be calculated by multiplying the number of positrons by the energy carried by each positron:

Total energy = number of positrons x energy per positron
Total energy = (1.506 x 10^21) x (5.49 MeV)
Total energy = 8.27 x 10^10 MeV

Finally, we need to convert the answer to joules (J):

8.27 x 10^10 MeV x (1.602 x 10^-13 J/MeV) = 1.32 x 10^16 J = 1.32 x 10^16 kJ = 1.32 x 10^13 MJ

We are given answer choices in kJ, so we need to divide our answer by 1000 to get:

Total energy = 1.32 x 10^13 kJ or 1.32 x 10^16 J

Therefore, the best choice is C.

Calcium-39 Undergoes Positron Decay