A single 215Rn nucleus, initially at rest, emits an alpha particle as it decays

Question

A single 215Rn nucleus, initially at rest, emits an alpha particle as it decays into a 211Po nucleus. There are no other decay products, and the kinetic energy of the alpha particle is measured to be 1.390 × 10-12 J. What is the total amount of nuclear potential energy, in joules, that was released in this process?

delta PE (nuclear) = _________________

What is the difference between the original mass of the 215Rn nucleus and the total mass of the fragments (211Po and alpha particle) after the decay?

delta m=_________________

Explanation / Answer

KE -= mv^2 /2

1.390 x 10^-12 = (4 x 1.66 x 10^-27 ) v^2 / 2

v = 2.046 x 10^7 m/s

decay process is due to internal forces hence momentum of system will be conserved.

using momentum conservation,

0 = m1v1 - m2v2

0 = (4u x 2.046 x 10^7 ) - (211u v2)

v2 = 3.88 x 10^5 m/s

initial PE = final total KE

= [ 1.390 x 10^-12 ] + [ (211 x 1.66 x 10^-27 ) ( 3.88 x 10^5)^2 /2 ]

= 1.415 x 10^-12 J .......Ans

b) deltaM c^2 = 1.415 x 10^-12 J

deltaM = 1.57 x 10^-29 kg Or 0.00947 amu

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