As part of his discovery of the neutron in 1932 , James...

As part of his discovery of the neutron in 1932 , James Chadwick determined the mass of the newly identified particle by firing a beam of fast neutrons, all having the same speed, at two different targets and measuring the maximum recoil speeds of the target nuclei. The maximum speeds arise when an elastic head-on collision occurs between a neutron and a stationary target nucleus. (a) Represent the masses and final speeds of the two target nuclei as $m_{1}, v_{1}, m_{2},$ and $v_{2}$ and assume Newtonian mechanics applies. Show that the neutron mass can be calculated from the equation $$m_{n}=\frac{m_{1} v_{1}-m_{2} v_{2}}{v_{2}-v_{1}}$$ (b) Chadwick directed a beam of neutrons (produced from a nuclear reaction) on paraffin, which contains hydrogen. The maximum speed of the protons ejected was found to be $3.30 \times 10^{7} \mathrm{m} / \mathrm{s} .$ Because the velocity of the neutrons could not be determined directly, a second experiment was performed using neutrons from the same source and nitrogen nuclei as the target. The maximum recoil speed of the nitrogen nuclei was found to be $4.70 \times 10^{6} \mathrm{m} / \mathrm{s}$. The masses of a proton and a nitrogen nucleus were taken as $1.00 \mathrm{u}$ and $14.0 \mathrm{u}$, respectively. What was Chadwick's value for the neutron mass?

Key Concepts

Elastic Collision

An elastic collision is one in which both momentum and kinetic energy are conserved. This concept is fundamental in analyzing interactions between particles, as it allows the derivation of relationships between the initial and final velocities and masses of the colliding bodies.

Conservation of Momentum

This principle asserts that the total momentum of a closed system remains constant if no external forces act on it. In collision problems, this law enables the calculation of unknown particle masses or velocities by equating the momentum before and after the impact.

Conservation of Kinetic Energy

Applicable to elastic collisions, this law states that the total kinetic energy of a system remains unchanged through the collision process. It provides a second key equation that, together with momentum conservation, allows for the complete characterization of a collision event in terms of mass and velocity.

Experimental Techniques in Particle Physics

This concept involves deducing the properties of subatomic particles by analyzing the results of controlled collisions. By measuring the recoil speeds of target nuclei when struck by particles, physicists can infer unknown quantities such as particle masses using theoretical models based on conservation principles.

Transcript

00:02 Question 42, excuse me, is based on an experiment done by james chadwick as he discovered the neutron.

00:11 So essentially, he fired a beam of fast neutrons at the same speed at two different targets and measuring the maximum recoil speed of the target nuclei.

00:20 This maximum speed occurred during an elastic head -on collision.

00:25 Excuse me.

00:26 So in the scenario, he fired neutrons at two different sources, two different masses, sorry, i should say one in three.

00:33 Where each scenario was described as follows where the neutron hit the object and they both traveled in the same direction as it was an unelastic and elastic head -out collision.

00:49 So based here, if we represent the final masses and final speeds of the two -targeted nuclei m1 v1 and m2, and we assume the typical newtonian mechanics will apply and there's no other quantum or atomic forces to deal with, we want to show the that the mass of the neutron can be found by this relationship.

01:07 So essentially we want to prove this based on these two different experiments.

01:13 So if we just look at one experiment, you know that because it's an elastic head -on collision, that the initial momentum must equal to the final momentum, and of course the initial energy must equal the final energy.

01:29 So looking at the momentum, so say this is for experiment one.

01:33 Initially, nucleus one is at rest.

01:35 So the initial momentum is just the mass of the neutron times its velocity, which equals, again, the mass of the neutron multiplied by its new velocity.

01:45 Maybe i'll call v prime plus the mass 1 and v1.

01:50 So again, m1 v1 are the final velocities.

01:53 Sorry, m1 is the math, which doesn't change.

01:57 And v1 is the final velocity.

02:00 So i can represent this the mass of the neutron as its initial velocity minus final velocity, including to m1.

02:09 M1v1.

02:12 Okay, so i'm a little stuck here.

02:14 So now i'm going to use the energy, a conservation energy.

02:21 So i have 1 -half mnv squared is initial, because again, the stationary objects are stationary, so there's no kinetic energy.

02:30 That's one -half m -n v -prime squared, plus one -half m -1v -1 squared.

02:41 So all these half terms canceled.

02:43 And again i can represent the mass of the neutron as v squared minus v squared minus v prime squared that's too big of the two v prime squared and that equals m1 v1 squared so now this expression this v squared minus v prime squared i can rewrite this as the product of two equations of v plus v prime multiplied by v minus v prime.

03:31 Right, and just verify v squared minus vvv prime plus vv prime minus v prime squared.

03:39 Good, so that makes sense.

03:50 And if we note, based here, mn, v minus v prime, is just this term and this term together.

04:00 So i'm going to replace this, i will insert this m1v1 into this equation here.

04:05 So therefore i have m1 v1 times v plus v prime equals m1 v1 squared.

04:19 So the question will have a lot of algebra, so i hope i can kind of guide you on a simple enough way.

04:26 But yeah, so now i can cancel out one of my m1 terms, so my m1 term and one of my v1 terms.

04:37 So i can represent that the final velocity of target 1 is simply the initial velocity of the of the proton plus its final velocity.

04:49 Or maybe more appropriately, i'm going to replace, say, v prime, this final velocity of the proton is v1, sorry, neutron is v1 minus v.

05:01 And now, i'll color code this, i'm going to insert this term back into this line, such that mn equals v minus v1 minus v, equals again m1v1.

05:26 So that's mn 2.

05:31 2v minus v1 equals m1 v1.

05:42 So i have this expression, this is just for reaction 1, i can do a similar thing for reaction 2.

05:51 The only difference is, of course, i have 2v minus v2 equals m2 v2...

Please give Ace some feedback