Solution

26-55. I am solving this one here since we were asked not to cover it in class.

Comment: 1eV is the energy gained by a particle with charge e, going through a potential difference of 1V, or

$$1\textrm{eV} = (\textrm{e})(1\textrm{V}) = (1.602 \times 10^{-19} \textrm{C})(1\textrm{V}) = 1.602 \times 10^{-19} \textrm{J}$$

The first problem takes a little bit of work. We need to relate the bulk properties of current and such, to the microscopic ones - the fact that we have individual particles flying about. The book has a nice derivation of the current density of a stream of particles $J=nqv$, $n$ the number density, $q$ the charge,and $v$ speed. This simply says that current is moving charge. The faster and more numerous the charge, the more current density you get. This is related to the current by $i=JA$. Thus we have

$$i= nqAv = \frac{N}{\textrm{Vol}}qAv$$

Where $N$ is the total number in the volume Vol of the beam. The reason for making this substitution is that we want the total number ($N$), and we want to eliminate the beam dimensions (as these are unknown). Note that volume is area times length - so if we can find a length somewhere, the dimensions can be eliminated. Thus we note that the speed of the particles is related to the length $v=L/t$. Thus finally

$$i=\frac{Nq}{Vol}\frac{AL}{v} = \frac{Nq}{t} \to N=\frac{it}{q}$$

Don't forget that q=2e, twice the electric charge.

In the previous part we found the number of alphas for a given amount of time. Now we want the number in a given length. We can do this simply if we can convert between length and time. This is done conveniently by the velocity, thus we have

$$N=\frac{it}{q}=\frac{iL}{qv}$$

But we also need the speed, which can be found from the kinetic energy

$$v=\sqrt{\frac{2KE}{m}}$$

Where $m$ is the alpha mass, which is 4 times the proton mass (since an alpha is 2 protons and 2 neutrons). Just don't forget to convert eV to Jules. Note - the velocity can by calculated quicker if the mass is given in units of MeV/$c^2$, as is common in particle physics (it always helps to use consistent units throughout your calculations!)

The energy the alphas gain from the potential difference becomes kinetic energy (energy conservation). Thus

$$KE=qV \to V=\frac{KE}{q}$$

Notice that the electric charge e can cancel the e in the eV in the numerator, simplifying this calculation.