The radioactive decay of ^238 U produces alpha particles with a kinetic energy of 4.17 MeV. (a)

The radioactive decay of ^238 U produces alpha particles...

The radioactive decay of ${ }^{238} \mathrm{U}$ produces alpha particles with a kinetic energy of $4.17 \mathrm{MeV}$. (a) At what speed do these alpha particles move? (b) Put yourself in the place of Rutherford and Geiger. You know that alpha particles are positively charged (from the way they are deflected in a magnetic field). You want to measure the speed of the alpha particles using a velocity selector. If your magnet produces a magnetic field of $0.30 \mathrm{~T}$, what strength electric field would allow the alpha particles to pass through undeflected? (c) Now that you know the speed of the alpha particles, you measure the radius of their trajectory in the same magnetic field (without the electric field) to determine their charge-to-mass ratio. Using the charge and mass of the alpha particle, what would the radius be in a $0.30$ -T field? (d) Why can you determine only the ratio $q / m$ by this experiment, but not the individual values of $q$ and $m$ ?

The radioactive decay of ${ }^{238} \mathrm{U}$ produces alpha particles with a kinetic energy of $4.17 \mathrm{MeV}$. (a) At what speed do these alpha particles move? (b) Put yourself in the place of Rutherford and Geiger. You know that alpha particles are positively charged (from the way they are deflected in a magnetic field). You want to measure the
speed of the alpha particles using a velocity selector. If your magnet produces a magnetic field of $0.30 \mathrm{~T}$, what strength electric field would allow the alpha particles to pass through undeflected? (c) Now that you know the speed of the alpha particles, you measure the radius of their trajectory in the same magnetic field (without the electric field) to determine their charge-to-mass ratio. Using the charge and mass of the alpha particle, what would the radius be in a $0.30$ -T field? (d) Why can you determine only the ratio $q / m$ by this experiment, but not the individual values of $q$ and $m$ ?

Key Concepts

Kinetic Energy and Velocity Relation

This concept involves using the kinetic energy equation (KE = ½ mv²) to calculate the speed of a particle from its kinetic energy. It also requires converting energy units (for instance, from electron volts or MeV to joules) and using the known mass of the particle to determine its velocity.

Velocity Selector

A velocity selector employs perpendicular electric and magnetic fields such that particles moving at a specific speed experience equal and opposite forces, resulting in no net deflection. The condition E = vB is used, ensuring that only particles with velocity v pass through undeflected.

Charged Particle Motion in a Magnetic Field

This concept explains the motion of charged particles subject to a magnetic field, where they experience a Lorentz force perpendicular to their velocity. This force causes them to move in a circular path, and the radius of curvature can be calculated using the formula r = (mv)/(qB), linking mass, charge, velocity, and magnetic field strength.

Determination of Charge-to-Mass Ratio

Many experiments only allow the measurement of the ratio of charge to mass (q/m) because both parameters appear as a product in the motion equations of charged particles in electromagnetic fields. Without independent measurements of either the charge or the mass, it is impossible to separate and determine their individual values.

Transcript

00:01 In this question, we are looking at the alpha particle that was emitted by a radioactive decay of uranium 238.

00:10 So this is the nucleus and it emits an alpha particle.

00:18 Alpha particle is emitted with a kinetic energy of 4 .17m .v.

00:29 Now we want to convert it into speed.

00:30 We assume that it is classical, not relativistic, not as fast.

00:40 So this is actually, we know that it's related to the speed by half and v square.

00:50 So v, we can calculate by taking 2 times of the k -e over m, take the square root.

01:00 Right and mass of the alpha particle mass of alpha is four atomic units approximately converting it into si units which is in kg right stand apartment as 27 kg and then substituting that back into this equation here get the velocity to be about 1 .4 to times 10 power of 7 meters per second right and remember that the k -e you have to convert it into si unit as well which is in juice and to do that you multiply the k e by 10 to power 6 times 1 .6 times 10 to power minus 19 right is to convert to juice now for the next part we know that they are there's a magnetic field being applied all right so the force of the magnetic field, fb, is related to the speed, the charge qv, and finally the strength of the magnetic field.

02:30 Right, we know the charge of the alpha particles, which is two times of the charge of the electron, and the velocity is over here, we already found it.

02:47 B is given to be 0 .3.

02:49 3 -0 -t.

02:52 Right? and we want to balance it out with the electric field.

02:57 For you to pass undeflected, it should be equal to your force of the electric field.

03:04 Electric field is actually q times e is the force, where e is the electric field.

03:13 So if you were to equate these two together, actually you find that q is cancelled out.

03:19 And what we're left with is that e must be equals to v times b.

03:30 Substitute the values in, you should get 4 .3 percent of power of 6 volts per meter, right? this is in si units.

03:46 Now the last part, we are looking at it, traversing the same magnetic field, but without the electric field.

03:56 So what will happen in a as the charges pass through undeflected and comes into the place where there is no more electric field, it will undergo circular motion.

04:23 Circle motion with a center over here...

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