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:

$V_1= 3.4*10^7m/s$

Explanation:

From the question we are told that

Nucleus diameter $d=5.50-fm$

a 12C nucleus

Required kinetic energy $K=2.30 MeV$

Generally initial speed of proton must be determined,applying the law of conservation of energy we have

            $K_2 +U_2=K_1+U_1$

where

$K_1$ =initial kinetic energy

$K_2$ =final kinetic energy

$U_1$ =initial electric potential

$U_2$ =final electric potential

mathematically

   $U_2 = \frac{Kq_pq_c}{r_2}$

where

$r_f$=distance b/w charges

$q_c$=nucleus charge $=6(1.6*10^-^1^9C)$

$K$=constant

$q_p$=proton charge

Generally kinetic energy is know as

         $K=\frac{1}{2} mv^2$

Therefore

         $U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2} mv_1^2 +U_1$

Generally equation for radius is $d/2$

Mathematically solving for radius of nucleus

         $R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})$

         $R=2.75*10^-^1^5m$

Generally we can easily solving mathematically substitute into v_1

   $q_p$$=6(1.6*10^-^1^9C)$

   $K_1=9.0*10^9 N-m^2/C^2$

   $U_1= 0$

   $R=2.75*10^-^1^5m$

   $K=2.30 MeV$

   $m= 1.67*10^-^2^7kg$

   $V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }$

    $V_1= 3.4*10^7m/s$

Therefore the proton must be fired out with a speed of $V_1= 3.4*10^7m/s$