Calcium-39 undergoes positron decay. Each positron carries 5.49 MeV of energy. How much energy will be emitted when 0.00250 mol of calcium-39 decays? (1 MeV = 1.602 × 10^-13 J)

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of calcium-39 in moles:
0.00250 mol (given)
Next, we need to find the number of decays:
Avogadro's number = 6.022 × 10²³ particles/mol
Thus, the number of particles = 0.00250 mol × 6.022 × 10²³ particles/mol = 1.506 × 10²¹ particles
Each decay emits 2 positrons, so the number of decays = 1.506 × 10²¹ particles / 2 = 7.53 × 10²⁰ decays
The energy emitted per decay = 5.49 MeV = 5.49 × 1.602 × 10⁻¹³ J = 8.80 × 10⁻¹³ J
The total energy emitted = energy per decay × number of decays
Total energy emitted = 8.80 × 10⁻¹³ J/decay × 7.53 × 10²⁰ decays = 6.636 × 10⁷ J = 6.636 × 10⁴ kJ = 1.32 × 10¹⁶ kJ
Therefore, the answer is choice C.

Calcium-39 Undergoes Positron Decay