What is the de Broglie wavelength of a proton whose kinetic energy is \(2.0 \mathrm{MeV} ? 10.0 \mathrm{MeV} ?\)

The kinetic energy is given in MeV (mega-electron volts). We need to convert it to Joules (J) before any calculation. We know that 1 eV = \(1.602 \times 10^{-19}\) J, so we can convert MeV to Joules as follows: 1 MeV = \(10^6\) eV = \(10^6 \times 1.602 \times 10^{-19}\) J = \(1.602 \times 10^{-13}\) J. Now, we can do the conversion for the given kinetic energies: For 2.0 MeV: \(2.0 \, \mathrm{MeV} \times \frac{1.602 \times 10^{-13} \mathrm{J}}{\mathrm{MeV}} = 3.204 \times 10^{-13} \mathrm{J}\) For 10.0 MeV: \(10.0 \, \mathrm{MeV} \times \frac{1.602 \times 10^{-13} \mathrm{J}}{\mathrm{MeV}} = 1.602 \times 10^{-12} \mathrm{J}\)

We will now use the kinetic energy formula, which relates the kinetic energy (E) and momentum (p) of a particle: \(E = \frac{p^2}{2m}\) Here, m is the mass of the proton, which is approximately \(1.67 \times 10^{-27}\) kg. We can rearrange the formula to find the momentum: \(p = \sqrt{2Em }\) For 2.0 MeV: \(p_{2.0\, \mathrm{MeV}} = \sqrt{2 \times 3.204 \times 10^{-13}\, \mathrm{J} \times 1.67 \times 10^{-27}\, \mathrm{kg}} = 4.429 \times 10^{-20} \mathrm{kg\:m/s}\) For 10.0 MeV: \(p_{10.0\, \mathrm{MeV}} = \sqrt{2 \times 1.602 \times 10^{-12}\, \mathrm{J} \times 1.67 \times 10^{-27}\, \mathrm{kg}} = 9.893 \times 10^{-20} \mathrm{kg\:m/s}\)

Now that we have the momentum of the proton, we can find the de Broglie wavelength (λ) using the formula: \(λ = \frac{h}{p}\) In this equation, h is the Planck's constant, which is approximately \(6.626 \times 10^{-34}\) Js. For 2.0 MeV: \( λ_{2.0\, \mathrm{MeV}} = \frac{6.626 \times 10^{-34}\, \mathrm{J\:s}}{4.429 \times 10^{-20} \mathrm{kg\:m/s}} = 1.497 \times 10^{-14}\, \mathrm{m}\) For 10.0 MeV: \( λ_{10.0\, \mathrm{MeV}} = \frac{6.626 \times 10^{-34}\, \mathrm{J\:s}}{9.893 \times 10^{-20} \mathrm{kg\:m/s}} = 6.700 \times 10^{-15}\, \mathrm{m}\) So, the de Broglie wavelength of a proton with a kinetic energy of 2.0 MeV is approximately \(1.497 \times 10^{-14}\) m and for a kinetic energy of 10.0 MeV, it's approximately \(6.700 \times 10^{-15}\) m.