A neutron of kinetic energy is E = 22.4 MeV,the height of the potential energy barrier is U₁ = 36.2 MeV and the width of the barrier is b - a = 8.4 fm.The kinetic energy of the neutron is increased by 20.0%.The neutron's probability of tunneling through the barrier increases by a factor of

Question

Quizplus · Accepted Answer

The transmission probability $T$ through the barrier is given by: 
$$T = \exp\left(-\frac{2\pi}{\hbar}\sqrt{2m(E-U_0)}\frac{(b-a)}{2}ight)$$
Where $m$ is the mass of the neutron, $E$ is the neutron's kinetic energy and $U_0$ is the height of the potential energy barrier.

Increasing the neutron's kinetic energy by 20% means its new energy is $E_f = 1.2E_i = 26.88$ MeV. The transmission probability with the new energy is: 
$$T_f = \exp\left(-\frac{2\pi}{\hbar}\sqrt{2m(E_f-U_0)}\frac{(b-a)}{2}ight)$$
The ratio of the new transmission probability to the initial one is: 
$$\frac{T_f}{T_i} = \exp\left(-\frac{2\pi}{\hbar}\sqrt{2m\left(\frac{6}{5}E_i-U_0ight)}\frac{(b-a)}{2}ight)$$
Plugging in the given values, we get: 
$$\frac{T_f}{T_i} = \exp\left(-\frac{2\pi}{\hbar}\sqrt{2(1.674	imes 10^{-27})(3	imes 10^8)^2\left(\frac{6}{5}(22.4	imes 10^6)- 36.2	imes 10^6ight)}\frac{(8.4	imes 10^{-15})}{2}ight) \approx 11.5$$
Therefore, the answer is B.

A Neutron of Kinetic Energy Is E = 22