A nucleus whose mass is 3.726656 × 10 25 kg undergoes spontaneous \"alpha\" deca

Question

A nucleus whose mass is 3.726656 × 1025 kg undergoes spontaneous "alpha" decay. The original nucleus disappears and there appear two new particles: a He-4 nucleus of mass 6.640678×1027 kg (an "alpha particle" consisting of two protons and two neutrons) and a new nucleus of mass 3.660098 × 1025 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 = ?

Explanation / Answer

Mass difference btwn original nucleus and resulting partiles after decay

=3.726656 × 1025  - [6.640678 x 10-27 + 3.660098 x 10-25]

= 1.5122 x 10-29 kg

1 amu = 931.46 MeV

converting MeV to J/Kg = (931.46MeV)(1.6 x 10-13 J/MeV)[1/(1.66 x 10-27kg)] = 8.98662 x 1016 J/Kg

so,

K = [1.5122 x 10-29 kg][ 8.98662 x 1016 J/Kg] = 1.358956576 e-12 J = 8482607.574 ev (1 j = 6.242e+18)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.