A nucleus whose mass is 3.806096 × 10-25 kg undergoes spontaneous \"alpha\" deca

Question

A nucleus whose mass is 3.806096 × 10-25 kg undergoes spontaneous "alpha" decay. The original nucleus disappears and there appear two new particles: a He-4 nucleus of mass 6.640678×10-27 kg (an "alpha particle" consisting of two protons and two neutrons) and a new nucleus of mass 3.739609 × 10-25 kg (note that the new nucleus has less mass than the original nucleus, and it has two fewer protons and two fewer neutrons) In these calculations, it is very important to keep at least 7 figures in your intermediate calculations. (a) When the alpha particle has moved far away from the new nucleus (so the electric interactions are negligible), what is the combined kinetic energy of the alpha particle and new nucleus? Kalpha + Knucleus (b) How many electron volts is this? Kalpha + Knucleus = In contrast to this nuclear reaction, chemical reactions typically involve only a few ev ev

Explanation / Answer

(a)
Calculate the mass difference between the original nucleus and the
resulting particles after decay:
3.806096e-25 - [3.739609e-25 kg + 6.640678e-27 kg] = 8.022e-30 kg

Thus,
K = mc¬¬¬2 = (8.022e-30)(3e+8)2 = 7.2198e-13 J

(b)
In eV = 7.2198e-13 * 6.242e+18 = 4.507e+6 eV

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