A photon of energy \(240.0 \mathrm{keV}\) is scattered by a free electron. If the recoil electron has a kinetic energy of \(190.0 \mathrm{keV},\) what is the wavelength of the scattered photon?

In this exercise, we are given: - Initial photon energy, E₁ = 240 keV - Kinetic energy of the recoil electron, KE = 190 keV We need to find the wavelength of the scattered photon. To do this, we will use the Compton scattering formula, which is given by: $$λ₂ = λ₁ + \frac{h}{m_ec}(1 - \cos{θ})$$ Here, λ₁ is the initial wavelength of the photon, λ₂ is the final wavelength of the photon, h is the Planck's constant (6.626 x 10^(-34) Js), me is the mass of the electron (9.109 x 10^(-31) kg), c is the speed of light (3 x 10^8 m/s), and θ is the scattering angle. E₁ (initial photon energy) and KE (recoil electron energy) are related by E₁ = KE + E₂, where E₂ is the energy of the scattered photon. We can convert the photon energies into wavelengths using the formula, E = (hc)/λ, and then solve for the final wavelength λ₂.

Using the energy to wavelength conversion formula, we can find the initial wavelength, λ₁, by rearranging the formula: $$λ₁ = \frac{hc}{E₁}$$ Substituting the values, we have: $$λ₁ = \frac{(6.626 \times 10^{-34} Js) (3 \times 10^{8} m/s)}{240 \times 10^3 eV \times 1.6 \times 10^{-19} J/eV}$$ $$λ₁ ≈ 8.14 × 10^{-12} m$$

We know that the initial photon energy (E₁) is the sum of the scattered photon energy (E₂) and the kinetic energy of the recoil electron (KE). So, we can then find E₂: $$E₂ = E₁ - KE$$ $$E₂ = (240 - 190) keV$$ $$E₂ = 50 keV$$

Now, we can use the conversion formula E = (hc)/λ to find the wavelength of the scattered photon, λ₂: $$λ₂ = \frac{hc}{E₂}$$ $$λ₂= \frac{(6.626 \times 10^{-34} Js) (3 \times 10^{8} m/s)}{50 \times 10^3 eV \times 1.6 \times 10^{-19} J/eV}$$ $$λ₂ ≈ 4.03 × 10^{-12} m$$ The wavelength of the scattered photon (λ₂) is approximately 4.03 × 10⁻¹² m.