how much Kinetic Energy How much kinetic energy in MeV must

how much Kinetic Energy

How much kinetic energy, in MeV, must an particle have to just \"touch\" the surface of a U nucleus? Assume that the 2U nucleus doesn\'t move. Express your answer using two significant figures. KE. MeV Submit My Answers Give Up

Solution

THe charge on the nucleus of the uranium atom is 92e where e = 1.6x10^-19 coul. The charge on teh alpha is 2e. So you require that the energy of the alpha be such that it is equal to potential of the U nucleus:

V = k92e/a0 where V is the electrical potential (J/coul) and k = 9x10^9 N-m^2/C^2 and a0 = radius of U238 nucleus.

The energy E is just E = 2e*V = 184ke^2/a0

a₀ r₀A here r₀ is 1.2*10^-15 m

r=1.2*10^-15 (4)^1/3 =1.90*10^{15 (for} alpha particle)

r=1.2*10^-15 (232)^1/3 = 7.3*10^-15 (for uranium)

a₀=sum of these two radius

a₀=9.2*10^-15 m

so E=184ke^2/(9.2*10^-15 )

=184*9*10⁹ *1.6*10^-19*1.6*10^-19/(9.2*10^-15) =460.8*10^-14=4.60*10^-12 joule=

we know 1eV=1.6*10^-19 joule

and 1 MeV=1.6*10^-13 joule

E= 4.6*10^-12/(1.6*10^-13)=28.75 meV