When a lithium-7 nucleus is bombarded by a proton, two alpha particles are produ

Question

When a lithium-7 nucleus is bombarded by a proton, two alpha particles are produced. Initially the proton must overcome the Coulomb repulsion of the nucleus. (Hint: The following atomic masses may be useful: 73Li has 7.016000 u, 42He has 4.002603 u and 1H has 1.007825 u.)

(a) If the lithium nucleus and proton can be treated as spherically symmet- ric charge distributions separated by a distance 3.5 × 1015m, what is the minimum energy required for the proton in MeV ?

(b) What is the energy released in the final state in MeV ?

(c) Could this reaction happen naturally when a sample of solid lithium is placed in a flask of hydrogen gas ?

Explanation / Answer

lithium nucleus has a positive charge of 3*e.

where e=1.6*10^(-19) C

coloumb's force=k*q1*q2/distance^2

where k=coloumb's constant=9*10^9

so couloumb's force=9*10^9*3*1.6*10^(-19)*1.6*10^(-19)/(3.5*10^(-15))^2)=56.4244 N

energy =k*q1*q2/distance

=9*10^9*3*1.6*10^(-19)*1.6*10^(-19)/(3.5*10^(-15))

=1.9749*10^(-13) J

=1.2343 MeV

part b:

mass difference=mass of proton+mass of lithium-sum of two alpha particles

=1.007825+7.016-2*4.002603

=0.018619 u

total energy released=mass difference*speed of light^2

=0.018619*1.67*10^(-27)*(3*10^8)^2

=17.49 MeV

part c:

as the final energy released is greater than energy required to overcome the couloumb force, the reaction can happen naturally.

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