Question

To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. The proton must impact the nucleus with a kinetic energy of 2.30 MeV. Assume the nucleus remains at rest. With what speed must the proton be fired toward the target?

Answer 1

The value is
`u = 3.23 *10^(7) \ m/s`

Step-by-step explanation:

From the question we are told that

The diameter of the nucleus is
`d = 5.50 \ fm = 5.50 *10^(-15) \ c`

The charge of the proton that makes up the nucleus is
`Q_2 = (12)/(2) * 1.60 *10^(-19) =9.6*10^(-19) \ C`

The energy to be impacted is
`KE_f = 2.30 \ MeV = 2.30 *10^(6) \ eV = 2.30 *10^(6) * 1.60 *10^(-19) = 3.68*10^(-13) \ J`

Generally the radius of the nucleus is mathematically represented as

`r = (d)/(2)`

=>
`r = (5.50 *10^(-15))/(2)`

=>
`r = 2.75 *10^(-15) \ m`

Generally from the law energy conservation we have that

`Initial \ total \ energy \ of the \ proton = final \ total \ energy \ of the \ proton`

i.e

`T_i = T_f`

Here

`T_i = KE_i + PE_i`

Here
`KE_i` is the initial kinetic energy which is mathematically represented as

`KE_I = (1)/(2) * m * u ^2`

Here
`PE_i` is the initial potential energy of the proton and the value is 0 J given that the proton is moving

Also
`T_f` is mathematically represented as

`T_f = KE_f + PE_f`

Here

`PE_f` is the final potential energy which is mathematically represented as

`PE_f = (k * Q_1 * Q_2)/(r)`

Here
`Q_1` is the charge on the proton with a value of
`Q_1 = 1.60 *10^(-19) \ C`

So

`PE_f = (9*10^(9) *(1.60 *10^(-19) ) * ( 9.6 *10^(-19)))/( 2.75 *10^(-15))`

=>
`PE_f = 5.027 *10^(-13 ) \ J`

So

`KE_i + PE_i = KE_f + PE_f`

=>
`(1)/(2) * m * u ^2 +0 = 3.68*10^(-13) + 5.027 *10^(-13 )`

Here m is the mass of the moving proton with value
`m = 1.67*10^(-27) \ kg`

So

`(1)/(2) * 1.67*10^(-27) * u ^2 +0 = 3.68*10^(-13) + 5.027 *10^(-13 )`

=>
`u = \sqrt{(3.68*10^(-13) + 5.027 *10^(-13 ))/(0.5 * 1.67*10^(-27)) }`

=>
`u = 3.23 *10^(7) \ m/s`