Answer
a) See the answer below.
b) See the answer below.
c) $47.9\;\rm MeV$
Work Step by Step
a)
We know that the magnetic field forces the particles to move on a circular path which means that the net force exerted on the particles inside the hollow metal dees is given by
$$\sum F=F_B=ma_r$$
Thus,
$$qvB =\dfrac{mv^2}{r}$$
$$q B =\dfrac{mv }{r}$$
Solving for $r$;
$$r =\dfrac{mv }{q B}\tag 1\\
$$
We can assume that the motion of the particles and the voltage are matching.
The period of the circular path is given by
$$T=\dfrac{2\pi r}{v}$$
Recall that $f=1/T$, hence,
$$\dfrac{1}{f}=\dfrac{2\pi r}{\cdot v}$$
So,
$$ {f}=\dfrac{v}{2\pi r}$$
Plugging from (1);
$$ {f}=\dfrac{v}{2\pi }\cdot \dfrac{q B}{mv }$$
Therefore,
$$\boxed{\boxed{ {f}= \dfrac{q B}{ 2\pi m }}}$$
b) Since the gap is small, then we can assume that the electric field is constant and uniform.
We assumed that the motion of the particles and the voltage are matching which means that the maximum voltage occurs in the gap. Thus, the particles receive energy in the gap to increase their speed.
The energy received is given by
$$\Delta E=KE=qV$$
And since the particle go inside the gap two times during one complete circular revolution, so the net energy gained is given by
$$\Delta E_{tot}=+2KE$$
$$\boxed{\boxed{\Delta E_{tot}=+2qV_0}}$$
c)
We know that the maximum kinetic energy is given by
$$KE_{max}=\frac{1}{2}mv_{max}^2\tag 2$$
We know the mass of the proton, So we need now to find its velocity.
Solving (1) for $v$;
$$\dfrac{rq B}{m } =v$$
Plugging into (2);
$$KE_{max}=\frac{1}{2}m\left[\dfrac{rq B}{m } \right]^2$$
$$KE_{max}=\frac{1}{2m} \left[ rq B \right]^2$$
Plugging the known;
$$KE_{max}=\frac{1}{2\cdot1.67\times10^{-27} } \left[2 \cdot1.6 \times10^{-19}\cdot 0.5 \right]^2=\bf 7.66\times10^{-12}\;\rm J$$
Thus,
$$KE_{max}= \dfrac{7.66\times10^{-12}}{1.6\times10^{-19}}\;\rm eV=\color{red}{\bf 47.9}\;\rm MeV$$
