Answer:
The value is [tex]u = 3.23 *10^{7} \ m/s[/tex]
Explanation:
From the question we are told that
The diameter of the nucleus is [tex]d = 5.50 \ fm = 5.50 *10^{-15} \ c[/tex]
The charge of the proton that makes up the nucleus is [tex]Q_2 = \frac{12}{2} * 1.60 *10^{-19} =9.6*10^{-19} \ C[/tex]
The energy to be impacted is [tex]KE_f = 2.30 \ MeV = 2.30 *10^{6} \ eV = 2.30 *10^{6} * 1.60 *10^{-19} = 3.68*10^{-13} \ J[/tex]
Generally the radius of the nucleus is mathematically represented as
[tex]r = \frac{d}{2}[/tex]
=> [tex]r = \frac{5.50 *10^{-15}}{2}[/tex]
=> [tex]r = 2.75 *10^{-15} \ m[/tex]
Generally from the law energy conservation we have that
[tex]Initial \ total \ energy \ of the \ proton = final \ total \ energy \ of the \ proton[/tex]
i.e
[tex]T_i = T_f[/tex]
Here
[tex]T_i = KE_i + PE_i[/tex]
Here [tex]KE_i[/tex] is the initial kinetic energy which is mathematically represented as
[tex]KE_I = \frac{1}{2} * m * u ^2[/tex]
Here [tex]PE_i[/tex] is the initial potential energy of the proton and the value is 0 J given that the proton is moving
Also [tex]T_f[/tex] is mathematically represented as
[tex]T_f = KE_f + PE_f[/tex]
Here
[tex]PE_f[/tex] is the final potential energy which is mathematically represented as
[tex]PE_f = \frac{k * Q_1 * Q_2}{r}[/tex]
Here [tex]Q_1[/tex] is the charge on the proton with a value of [tex]Q_1 = 1.60 *10^{-19} \ C[/tex]
So
[tex]PE_f = \frac{9*10^{9} *(1.60 *10^{-19} ) * ( 9.6 *10^{-19})}{ 2.75 *10^{-15}}[/tex]
=> [tex]PE_f = 5.027 *10^{-13 } \ J[/tex]
So
[tex]KE_i + PE_i = KE_f + PE_f[/tex]
=> [tex]\frac{1}{2} * m * u ^2 +0 = 3.68*10^{-13} + 5.027 *10^{-13 }[/tex]
Here m is the mass of the moving proton with value [tex]m = 1.67*10^{-27} \ kg[/tex]
So
[tex]\frac{1}{2} * 1.67*10^{-27} * u ^2 +0 = 3.68*10^{-13} + 5.027 *10^{-13 }[/tex]
=> [tex]u = \sqrt{\frac{3.68*10^{-13} + 5.027 *10^{-13 }}{0.5 * 1.67*10^{-27}} }[/tex]
=> [tex]u = 3.23 *10^{7} \ m/s[/tex]
