Question

What amount of energy, in kJ, is released when a mole of Co-57 undergoes electron capture? Recall that 1 J = 1 kgm2/s2.

mass of nucleus Co-57: 56.93629 amu

mass of nucleus Fe-57: 56.93539 amu

mass of electron: 0.000549 amu

Answer 1

The reaction involved is

Co-57 + e^- ---------> Fe-57

Energy = (mass defect)*velocity of light

Mass defect = Mass of Fe-57 - (Mass of Co-57 + mass of electron)

= 56.93539 amu - (56.93629 amu - 0.000549 amu)

= -0.001449 amu

In units of kg, 1 amu = 1.66054 x 10^-27 kg

Mass defect = -0.001449 amu * 1.66054 x 10^-27 kg/amu

= - 2.4061 x 10^-30 kg

velocity of light = 2.9979 x 10⁸ m/s

Energy = mass defect * (velocity)²

= (- 2.4061 x 10^-30 kg)*( 2.9979 x 10⁸ m/s)²

= -21.624 x 10^-14 kgm²/s²

= -21.624 x 10^-14 J/atom

So, amount of energy released by electron capture of 1 atom of Co-57 = 2.16 x 10^-10 kJ

For 1 mole of Co-57, the energy released will be 2.16 x 10^-10 kJ * 6.022 x 10²³

= 1.30 x 10¹⁴ kJ (Answer)