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

Answer:

The value is  $u = 3.23 *10^{7} \ m/s$

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  $Q_2 = \frac{12}{2} * 1.60 *10^{-19} =9.6*10^{-19} \ C$

    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

         $r = \frac{d}{2}$

=>      $r = \frac{5.50 *10^{-15}}{2}$

=>      $r = 2.75 *10^{-15} \ m$

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

       $KE_I = \frac{1}{2} * m * u ^2$

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

         $PE_f = \frac{k * Q_1 * Q_2}{r}$

Here $Q_1$ is the charge on the proton with a value of $Q_1 = 1.60 *10^{-19} \ C$

So

        $PE_f = \frac{9*10^{9} *(1.60 *10^{-19} ) * ( 9.6 *10^{-19})}{ 2.75 *10^{-15}}$

=>     $PE_f = 5.027 *10^{-13 } \ J$

So  

         $KE_i + PE_i = KE_f + PE_f$

=>       $\frac{1}{2} * m * u ^2 +0 = 3.68*10^{-13} + 5.027 *10^{-13 }$

Here m is the mass of the moving proton with value $m = 1.67*10^{-27} \ kg$

So

       $\frac{1}{2} * 1.67*10^{-27} * u ^2 +0 = 3.68*10^{-13} + 5.027 *10^{-13 }$

=>      $u = \sqrt{\frac{3.68*10^{-13} + 5.027 *10^{-13 }}{0.5 * 1.67*10^{-27}} }$

=>       $u = 3.23 *10^{7} \ m/s$