Question

Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach the surface of a 15-fmfm-diameter 238U238U nucleus

Answer 1

`\Delta V = 1.8 * 10^7 V`

Explanation:

GIVEN

diameter = 15 fm =
`15 * 10^(-15)`m

we use here energy conservation

`K_(i)+U_(i) =K_(f)+U_(f)`

there will be some initial kinetic energy but after collision kinetic energy will zero

`K_(i) + 0 = 0 + (1)/(4 \pi \epsilon _(0)) ((2e)(92e))/(7.5 * 10^(-15))`

on solving these equations we get kinetic energy initial

`KE_(i) = 5.65* 10 ^(-12) * \frac {1 eV}{1.6 * 10^(-19)}`

`KE_(i) = 35.33` J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e

and gains kinetic energy K =e∆V ..........(ii)

by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the
`{238}_U` nucleus after being accelerated through a potential difference ∆V

equating (i) and second equation we get

e∆V = 35.33 Me V

`\Delta V = (35.33)/(2) MV\\\Delta V = 1.8 * 10^7 V`