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

Answer:

$\Delta V = 1.8 \times 10^7 V$

Explanation:

GIVEN

diameter = 15 fm  =$15 \times 10^{-15}$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

$K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}$

on solving these equations we get kinetic energy initial

$KE_{i} = 5.65\times 10 ^{-12} \times \frac {1 eV}{1.6 \times 10^{-19}}$

$KE_{i} = 35.33$ 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

$\Delta V = \frac{35.33}{2} MV\\\Delta V = 1.8 \times 10^7 V$