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

Answer

c

Explanation

If the particles are treated as point charges,
`U=(1)/(4)pi epsilon_(0) (q_(1) q_(2))/(r)`
`Q_(1) =2e ("alpha particle"), q_(2)=82e ("gold nucleus")`,
`r=6.5 xx 10^(14)m`
`:. U=(8.987 xx 10^(8)N m^(2) C^(2))`
`xx(92xx82)(1.602xx10^(19C))/(6.50xx10^(-14)m)=5.82 xx10^(-13) J`
or `U=5.82xx10^(-28) J`
`xx((1 eV)/(1.602xx10^(-19)Jh))=3.63xx10^(6)eV=3.63 MeV`.