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 × 10⁴kJ C) 1.32 × 10⁶ kJ D) 1.32 × 10⁹kJ E) none of the above

Question

Quizplus · Accepted Answer

First, we need to calculate how many particles are in 0.0025 mol of calcium-39:

0.0025 mol x 6.022 x 10^23 particles/mol = 1.505 x 10^21 particles

Since each calcium-39 nucleus decays by emitting one positron, we have the same amount of positrons emitted:

1.505 x 10^21 positrons

The total energy emitted is:

(1.505 x 10^21 positrons) x (5.49 MeV/positron) = 8.25945 x 10^12 MeV

We need to convert MeV to kJ:

8.25945 x 10^12 MeV x (1.602 x 10^-19 J/MeV) x (1 kJ/1000 J) = 1.32 x 10^16 kJ

Therefore, the answer is C, 1.32 x 10^16 kJ.

Calcium-39 Undergoes Positron Decay