A beam of alpha paricles is incident on a target of lead. `A` particular alpha paticles comes in 'head- on' to a particular lead nucleus and stops `6.50 xx 10^(-14)` m away from the center of the nucleus. (This point is well outside the nucleus.) Assume that the lead nucleus, which has `82` protons, remains at rest. The mass of alpha particle is `6.64 xx 10^(-27)kg`
Calculate the electrostatic potential energy at the instant when the alpha particle stops?

Options

A. `36.3 MeV`
B. `45.0 MeV`
C. `3.63 MeV`
D. `40.0 MeV`

Answer

C

Explanation

If the particles are treated as point charges,
`U=(1)/(4 pi epsilon_(0))(q_(1)q_(2))/(r)` where `Q_(1)=2e, q_(2)=82 e`, and `r=6.5xx10^(-14)m`
` :. U= 9xx10^(-9)x((2xx82)(1.602xx10^(-19))^(2))/(6.50xx10^(-14)m)=5.82xx10^(-13)J`
`=3.63MeV`