Calculate the wavelength of a) a \(2.00 \mathrm{eV}\) photon, and b) an electron with kinetic energy \(2.00 \mathrm{eV}\).

The energy given is in electron volts (eV). We need to convert it to joules (J) to use it in the appropriate formulas. Use the conversion factor: \(1 \mathrm{eV} = 1.60218 \times 10^{-19} \mathrm{J}\). \(2.00 \mathrm{eV} \times \frac{1.60218 \times 10^{-19} \mathrm{J}}{1 \mathrm{eV}} = 3.20436 \times 10^{-19} \mathrm{J}\)

Now that we have the energy in joules, we can use the Planck-Einstein relation to calculate the wavelength of the photon: \(E = h\frac{c}{\lambda}\), where \(E\) is the energy of the photon, \(h\) is the Planck constant (\(6.62607 \times 10^{-34} \mathrm{Js}\)), \(c\) is the speed of light (\(2.998\times 10^8 \mathrm{m/s}\)), and λ is the wavelength. Rearrange the formula to solve for λ: \(\lambda = \frac{hc}{E}\) Now, substitute the values: \(\lambda = \frac{(6.62607 \times 10^{-34} \mathrm{Js})(2.998\times 10^8 \mathrm{m/s})}{3.20436 \times 10^{-19} \mathrm{J}} = 6.21541 \times 10^{-7} \mathrm{m}\) The wavelength of the \(2.00 \mathrm{eV}\) photon is approximately \(6.215\times 10^{-7} \mathrm{m}\).

We need to use the de Broglie wavelength formula to calculate the wavelength of the electron: \(\lambda = \frac{h}{p}\), where \(\lambda\) is the wavelength, \(h\) is the Planck constant, and \(p\) is the momentum of the electron. However, we don't have the momentum; we have the kinetic energy. To find the momentum, we can use the following relation between kinetic energy and momentum: \(E_K = \frac{p^2}{2m}\), where \(E_K\) is the kinetic energy, \(p\) is the momentum, and \(m\) is the mass of the electron (\(9.109\times 10^{-31} \mathrm{kg}\)). Rearrange the formula to solve for \(p\): \(p = \sqrt{2mE_K}\) Now, substitute the values: \(p = \sqrt{2(9.109\times 10^{-31} \mathrm{kg})(3.20436 \times 10^{-19} \mathrm{J})} = 1.75834 \times 10^{-24} \mathrm{kg\cdot m/s}\) Now that we have the momentum, we can use the de Broglie formula to calculate the wavelength: \(\lambda = \frac{h}{p} = \frac{6.62607 \times 10^{-34} \mathrm{Js}}{1.75834 \times 10^{-24} \mathrm{kg\cdot m/s}} = 3.76919 \times 10^{-10} \mathrm{m}\) The wavelength of an electron with kinetic energy \(2.00 \mathrm{eV}\) is approximately \(3.77 \times 10^{-10} \mathrm{m}\).