What is the de Broglie wavelength of an electron with kinetic energy $7.0 \mathrm{TeV} ?$

First, we need to convert the kinetic energy from tera-electron volts (TeV) to electron volts (eV). Since 1 TeV is equal to \(10^{12}\) eV, we can multiply the given energy value by \(10^{12}\) to get the energy value in eV: Energy (in eV) = \(7.0\,\text{TeV} \times 10^{12}\,\frac{\text{eV}}{\text{TeV}} = 7.0 \times 10^{12}\,\text{eV}\)

Now that we have the kinetic energy in eV, we can use the equation for the relativistic energy of a particle to find its momentum. The equation is: E = \(\sqrt{(p c)^2 + (m_e c^2)^2}\) Where E is the energy of the electron, p is its momentum, c is the speed of light, and \(m_e\) is the electron's rest mass. Rearranging the equation to find the momentum p, we get: p = \(\frac{1}{c} \sqrt{E^2 - (m_e c^2)^2}\) Plug in the known values for E, \(m_e\) and c: - E = \(7.0 \times 10^{12}\,\text{eV}\) - \(m_e = 0.511\,\text{MeV}\) (Note that we need to convert this value to eV by multiplying by \(10^6\)) - c = \(3.0 \times 10^8\, \text{m/s}\) We first need to convert the energy values to Joules. The conversion factor from eV to Joules is \(1.602 \times 10^{-19}\, \text{J/eV}\): - E (in J) = \((7.0 \times 10^{12}\,\text{eV}) \times (1.602 \times 10^{-19}\, \text{J/eV})\) - \(m_e\) (in J) = \((0.511\times10^6\,\text{eV}) \times (1.602\times10^{-19}\, \text{J/eV})\) Now, calculate the momentum p: p = \(\frac{1}{c} \sqrt{E^2 - (m_e c^2)^2}\)

Finally, we can find the electron's de Broglie wavelength using the formula: \(\lambda = \frac{h}{p}\) Where \(\lambda\) is the de Broglie wavelength, h is the Planck's constant (\(6.626 \times 10^{-34}\, \text{Js}\)), and p is the momentum we calculated in the previous step. Plug in the values for h and p, then solve for \(\lambda\): \(\lambda = \frac{h}{p}\) This will give us the de Broglie wavelength of the electron with a kinetic energy of \(7.0\,\text{TeV}\).