Length and velocity of a pulse of particles We are given a pulse of protons of duration $10^{-7}s$ and energy $2KeV$. I know I am supposed to use the uncertainty principle to solve this. I need to get the length and the indetermination of the velocity for this pulse.
I suppose the pulse has $N$ particles, so the kinetic energy would be $T=\frac{1}{2}Nm_p v^2=2KeV$ (I can assume the pulse is non relativistic). I also say that $x=vt$ for the length of the pulse. Then I know that $\Delta x \Delta v = \hbar /2 $ for the minimum unceratainty in velocity.
I think I should also use $\Delta E \Delta t \geq \hbar/2$ to get 2 equations to get the number of particles... but I'm really not sure about how to use it because we are not given any uncertainty in $E$ or $t$.
I have no idea how to continue from this point on... Maybe if I say $\Delta x= \Delta v t$, then I could say $\Delta v = \sqrt{\hbar / 2t}$ or something, but then I could not get $x$ from anywhere... I'm confused with this problem
 A: The length of the pulse
Classical kinetic energy gives you the velocity, just like you calculated. Time is $10^{-7}s$ and $x=vt$. Multiply the velocity $v=\sqrt{\frac{4\mathrm{keV}}{Nm_p}}$ you calculated by time and you've got $x$. That's your length of the pulse.
Edit: Alternative interpretation. If 2 keV refers to the energy of the pulse and not each individual proton then "length" might refer to the de Brogie wavelength of the pulse, calculated $\lambda=\frac E{hc}$.
The indetermination of the velocity of the pulse
The length of the pulse is calculated from two values $10^{-7}\mathrm{s}$ and $2\mathrm{keV}$. Each of these is given with one significant figure. In other words, the uncertainty of time is no more than $\frac{10^{-7}\mathrm{s}}{10}=10^{-8}\mathrm{s}$ and the uncertainty of the energy is no more than $\frac{2\mathrm{keV}}{10}=0.2\mathrm{keV}$. If we handwave some error propagation then we can calculate $x$ to one significant figure. This gives us an upper bound $\Delta x\leq \frac x{10}$. Once you have an upper bound on $\Delta x$ then you can plug it into Heisenberg's Uncertainty Principle.
$$\Delta x\Delta p\geq\frac\hbar2$$
$$\Delta p\geq\frac\hbar{2\Delta x}\geq\frac\hbar{2\frac x{10}}=\frac{5\hbar}{x}$$
$$p=mv$$
$$\Delta p=m\Delta v$$
$$m\Delta v\geq\frac{5\hbar}x$$
$$\Delta v\geq\frac{5\hbar}{mx}$$
