Calcium 39 Undergoes Positron Decay
Calcium 39 is subject to the dissolution of positron. Each positron has an energy of 5.49 MeV. How much energy will there be ...? 3
Calcium 39 is subject to the dissolution of positron. Each positron is 5.49 MeV. C has energy.
How much energy is released when 0.0025 mole of calcium 39 is destroyed?
(1 MeV = 1602 x 10 - 13 J)
A) 13.2 KJol
B) 1.32 x 10 4 kJ
C) 1.32 x 10 6 kJ
D) 1.32 x 10 9 kJ
E) None of these cycles are correct.
The correct answer is c.
Can anyone tell in detail when they got this answer?
Thank you very much
(0.0025 mol) x (6.022 x 10 23 atoms / mol) x (1 positron / 1 atom) x (5.49 MeV / positron) x
(1.602 x 10 13 J / MeV) = 1.32 x 10 9 J = 1.32 x 10 6 kJ
So answer C)
