# You have been able to get a pa

You have been able to get a part time job with a medical physicsgroup investigating ways to treat inoperable brain cancer. One formof cancer therapy being studied uses slow neutrons to knock aparticle (either a neutron or a proton) out of the nucleus of theatoms which make up cancer cells. The neutron knocks out theparticle it collides with in an inelastic collision. The rest ofthe nucleus does not move. After a single proton or neutron isknocked out of the nucleus, the nucleus decays, killing the cancercell. To test this idea, your research group decides to measure thechange in internal energy of a nitrogen nucleus after a neutroncollides with one of the neutrons in its nucleus and knocks it out.The change in internal energy should be equal to the difference inkinetic energy before and after the collision. In the experiment,one neutron goes into the nucleus along the horizontal with a speedof 2.0 x 107 m/s. You detect two neutrons coming out, one at anangle of 30° above and the other 15° below the horizontal. You cannow calculate the change in internal energy of the nucleus.

Im asking for a very detailed solution, including a physicsdiagram, the equations needed for the solution and the solutionlaid out. Please help, thank you.

Answer:

Let the mass of a neutron be **m**

Before Collision:-

Let initial speed(given) of neutron be **v** = 2 *10^7 m/s

**Momentum = mv in right direction**

**and Momentum = 0 in upward direction**

After Collision :-

Let speed of one neutron at angle of 30 deg be **v1m/s** and other at angle of 15 deg be **v2m/s**

**Momentum = mv1 Cos 30 + mv2 Cos 15 in rightdirection**

**Momentum = mv1 Sin 30 – mv2 Sin 15 in upwarddirection**

NOW BY **Conservation of momentum principal in rightdirection, before and after collision momentum should beequal,**

hence,

**mv** = **mv1 Cos 30 + mv2 Cos 15 ……[Ist equation]**

**Also in upward direction**

**0 = mv1 Sin 30 – mv2 Sin 15 …… [2ndequation]**

ON SOLVING EUATION 1 AND EQUATION 2 WE GET

v1 = 2 v Sin 15 / (2 Sin15 Cos 30 + Cos 15) = 2 (v) 0.366 = 2 (2 * 10^7) 0.366 = 1.46 * 10^7 m/s

So **v1** = **1.46 * 10^7 m/s**

v2 = v / (2sin15Cos30 + Cos15) = v / 1.414 = 2* 10^7 /1.414 =1.414 * 10^7 m/s

So **v2 = 1.414 * 10^7 m/s**

KE = Kinetic Energy

Now, Change in internal energy = KE ( before collision )- KE (after collision )

KE(before) = 1/2 m v^2

m= mass of neutron = 1.6 * 10^-27 kg

v = 2 * 10^7 m/s so v^2 = 4 * 10^14 m/s

So, KE (Before) = 1/2 * (1.6*10^-27 )* (4*10^14) = 3.2 * 10^-13Joules

KE(After collision) = 1/2 m(v1)^2 + 1/2 m (v2)^2 = 1/2 m [(v1)^2 + (v2)^2 ]

= 1/2 (1.6 * 10^-27 kg) [ (**1.46 * 10^7**)^2 +(**1.414 * 10^7)**^2 ]

= 3.3 * 10^-13 Joules

**Change in internal energy = KE ( before collision )- KE( after collision )**

**= 3.2 * 10^-13 Joules – 3.3 * 10^-13 Joules = – 0.1 *10^-13 Joules**