protons at Fermilab

[logistic_guy]

What is the wavelength, and maximum resolving power attainable, using $\displaystyle 900$-$\displaystyle \text{GeV}$ protons at Fermilab?

 

[khansaheb]

What is the wavelength, and maximum resolving power attainable, using $\displaystyle 900$-$\displaystyle \text{GeV}$ protons at Fermilab?

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Please share your work/thoughts about this problem

 

[logistic_guy]

We first need to calculate the total energy.

$\displaystyle E = K + mc^2$

where $\displaystyle mc^2$ is the rest energy of the proton.

Since the rest energy is very small compared to the kinetic energy, we can safely ignore it. (You can calculate it if you want. It will not affect the answer!)

Then, we have:

$\displaystyle E = K = 900 \ \text{GeV}$

The momentum formula is $\displaystyle p = \frac{E}{c}$ and the wavelength formula is $\displaystyle \lambda = \frac{h}{p}$, then

$\displaystyle \lambda = \frac{hc}{E}$

where $\displaystyle h$ is Planck's Constant: $\displaystyle h = 6.626 \times 10^{-34} \ \text{J} \cdot \text{s}$

And $\displaystyle hc = 1.24 \times 10^{-12} \ \text{MeV} \cdot \text{m}$ (I will derive this result in future posts.)

Then, the wavelength is:

$\displaystyle \lambda = \frac{1.24 \times 10^{-12} \ \text{MeV} \cdot \text{m}}{900 \ \text{GeV}} \times \frac{\text{GeV}}{10^3 \ \text{MeV}} = \textcolor{blue}{1.38 \times 10^{-18} \ \text{m}}$

The power is given by:

$\displaystyle P = \frac{cE}{\lambda} = \frac{(299792458 \ \text{m/s})(900 \times 10^{9} \ \text{eV})}{1.38 \times 10^{-18} \ \text{m}} \times \frac{1.6 \times 10^{-19} \ \text{J}}{\text{eV}} = \textcolor{blue}{3.13 \times 10^{19} \ \text{W}}$