A 5.30 MeV ? particle (charge = +2e) happens, by chance, to be headed directly t

Question

A 5.30 MeV ? particle (charge = +2e) happens, by chance, to be headed directly toward the nucleus of an atom of gold (charge = +79e). How close does it get before it comes momentarily to rest and revers its course? Neglect the recoil of (relatively massive) gold nucleus.

26. A 5.30 MeV particle (charge +2e) happens, by chance, to be headed directly toward the nucleus of an atom of gold (charge = +79e). How close does it get before it comes momentarily to rest and revers its course? Neglect the recoil of (relatively massive) gold nucleus. Lord iO 9 2

Explanation / Answer

I am providing the explanation of above solution.

both particles (alpha and necleus) are positively charged. So both repel each other.
but alpha particle have speed towards necleus so it will move towards neclues as
it have initially have some kinetic energy (mv^2 /2). But we will take it as reference pooint
for potential energy. But as as it goes toward necleus , it loses its K.E. equal to potential
energy difference it gets ( kq1q2/d). as it goes close to necleus , its potential energy gets
on incerease and K.E. gets decreasing .
at the point when its all K.E. gets converted in P.E. it stops (zero K.E> means no speed)

thus from energy conservation we can write ,
initial total energy = final total energy
initial K.E. + P.E. = final K.E. + P.E.
mu^2 /2 + 0 = 0 + kq1q2 /d
from here you can find distance d as solved in above solution.

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