Question

a) What is γ for a proton having an energy of 1.00 Trev, produced by the Fermi lab accelerator? (b) Find its momentum. (c) What is the proton’s wavelength?

Short answer

(a) The proton produced by Fermi lab is \[\gamma = 1066\]

(b) The required momentum is \(3.41 \times {10^{ - 13}}\;{\rm{kg}} \cdot {\rm{m}} \cdot {{\rm{s}}^{ - 1}}\)

(c) The proton’s wavelength is \[\lambda = 1.94 \times {10^{ - 21}}\;{\rm{m}}\]

Step-by-step solution

Definition the formulas:

The relativistic total energy is given by

\[E = \gamma m{c^2}\]

Here, γ is the relativistic constant, mc² is the rest mass energy of the particle.

Here the rest mass energy of the electron is\[m{c^2} = 0.512\;{\rm{MeV}}\].

The relativistic momentum is given by\[p = \gamma mu\]

Here, mis the rest mass of the object, and uis its velocity relative to an observer.

The relativistic factor is given by

\[\gamma = \frac{1}{{\sqrt {1 - {{\left( {\frac{u}{c}} \right)}^2}} }}\]

Here, u is the speed of the object.

Now from the above equation, we get

\[{\bf{u = c}}\sqrt {{\bf{1 - }}{{\left( {\frac{{\bf{1}}}{{\bf{\gamma }}}} \right)}^{\bf{2}}}} \]

Now from above equations and solve further as:

\[\begin{array}{c}{\bf{p = \gamma mu}}\\{\bf{ = \gamma mc}}\sqrt {{\bf{1 - }}{{\left( {\frac{{\bf{1}}}{{\bf{\gamma }}}} \right)}^{\bf{2}}}} \end{array}\]

Consider the formula for the photon momentum is as follows:

\[p = \frac{h}{\lambda }\]

Solve for the wavelength as:

\[{\bf{\lambda = }}\frac{{\bf{h}}}{{\bf{p}}}\]

Finding electron emerging from the Stanford Linear Accelerator with a total energy of 50.0 GeV

(a)

Determine the accelerating energy of the proton is as follows:

\[\begin{array}{c}E = 1.00{\rm{TeV}}\\ = 1.00 \times {10^{12}}{\rm{eV}}\\ = 1.00 \times {10^{12}} \times 1.602 \times {10^{ - 19}}{\rm{CV}}\\ = 1.602 \times {10^{ - 7}}\;{\rm{J}}\end{array}\]

Determine the proton rest mass energy as follows:

\[\begin{array}{c}m{c^2} = 1.67 \times {10^{ - 27}}\;{\rm{kg}} \times {\left( {3.00 \times {{10}^8}\;{\rm{m}}\;{{\rm{s}}^{ - 2}}} \right)^2} \times \frac{1}{{1.602 \times {{10}^{ - 19}}{\rm{C}}}}\\ = 938.2 \times {10^6}{\rm{eV}}\\ = 938.2{\rm{MeV}}\end{array}\]

Determine the value of therelativistic constant.

\[\begin{array}{c}\gamma = \frac{E}{{m{c^2}}}\\ = \frac{{1.00 \times {{10}^{12}}{\rm{eV}}}}{{938.2{\rm{MeV}}}}\\ = \frac{{1.602 \times {{10}^7}{\rm{eV}}}}{{938.2 \times {{10}^6}{\rm{eV}}}}\\ = 1066\end{array}\]

Therefore, Fermi labis γ = 1066.

Finding the momentum 

(b)

Determine the value of the momentum:

\(\begin{array}{c}p = 1066\left( {1.67 \times {{10}^{ - 27}}} \right)\left( {3 \times {{10}^8}\;{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}} \right)\sqrt {1 - {{\left( {\frac{1}{{1066}}} \right)}^2}} \\ = 3.41 \times {10^{ - 13}}\;{\rm{kg}} \cdot {\rm{m}} \cdot {{\rm{s}}^{ - 1}}\end{array}\)

Therefore, Momentum is \(3.41 \times {10^{ - 13}}\;{\rm{kg}} \cdot {\rm{m}} \cdot {{\rm{s}}^{ - 1}}\).

Calculating electron’s wavelength

c)

Determine the wavelength of the proton is:

\[\begin{array}{c}\lambda = \frac{h}{p}\\ = \frac{{6.626 \times {{10}^{ - 34}}{\rm{Js}}}}{{3.41 \times {{10}^{ - 13}}\;{\rm{kg}}\;{\rm{m}}\;{\rm{s}}}}\\ = 1.94 \times {10^{ - 21}}\;{\rm{m}}\end{array}\]

Therefore, Electron’s wavelength is\[\lambda = 1.94 \times {10^{ - 21}}\;{\rm{m}}\].