Alpha particle close to the nucleus

[Juli]

Homework Statement

What is the impact parameter of an alpha particle with kinetic energy 4 MeV that is deflected by the angle $\theta = 15°$ when scattered by a gold nucleus (Z =79)?

Relevant Equations

$$p = \frac{k}{mv_0^2}cot\frac{\theta}{2} $$ with $$k = \frac{2Ze^2}{4\pi\epsilon_0}$$

Hello everyone, while studying I found this task in my textbook.
Solving this problem with the help of the formula seems quite straightforward. But I get a different result than the solution the textbook offers.
I get: Around $5∗10^{−15}m$ (which is a typical solution for a radius of a nucleus)
Textbook says: $2.16∗10^{−13}m$The point where I think I probably could be mistaken, is the velocity $v_0$. I calculated it with $E= \frac{1}{2}m∗v^2$ with $E=4MeV$.
Is that wrong? I get $v_0=1.44∗10^7\frac{m}{s}$ (which I think is already relativistic, so I think there is my mistake?)
Can anyone verify the solution of the textbook?
I would be very grateful for any help, since I'm quite confused.

 

[TSny]

A kinetic energy of 4 Mev is nonrelativistic for an alpha particle, which has a rest mass energy of about 3.7 Gev. For the speed of the alpha particle, I get 1.38 x 10⁷ m/s, which is a little less than your value. This is about 5% the speed of light.

Your formula looks correct. I get the textbook's answer for the impact parameter. The only way we can identify your mistake is for you to show your calculation explicitly with all the numerical values and units for the various quantities.

 

[Steve4Physics]

Can anyone verify the solution of the textbook?

Like @TSny, I get the same answer as the text book.

Your value for $v_0 (1.44\times 10^7m/s)$ is about 4% bigger than the correct value (a small but signficant difference). You might want to sort out why.

In fact there is no need to work out $v_0$. In the formula $p = \frac{k}{mv_0^2}cot\frac{\theta}{2}$ note that $mv_0^2$ is simply twice the kinetic energy, i.e. $mv_0^2 = 8MeV$.

 

[Juli]

Thank you both so much.

Like @TSny, I get the same answer as the text book.

Your value for $v_0 (1.44\times 10^7m/s)$ is about 4% bigger than the correct value (a small but signficant difference). You might want to sort out why.

In fact there is no need to work out $v_0$. In the formula $p = \frac{k}{mv_0^2}cot\frac{\theta}{2}$ note that $mv_0^2$ is simply twice the kinetic energy, i.e. $mv_0^2 = 8MeV$.

Especially the point about $mv^2$ was very helpful. There was my mistake, I get the right solution now.