You are operating an X-ray tube with a molybdenum (Mo) target by applying an acceleration potential of 50 keV. Calculate the values of two characteristic wavelengths, λKα and λSWL. Express your answers in meters.

To calculate the values of the characteristic wavelengths λKα and λSWL, we need to use the equation:

λ = hc / (E * (1 + ε))

Where:
λ is the wavelength of the X-ray radiation
h is the Planck's constant (6.626 x 10^-34 Js)
c is the speed of light in a vacuum (3 x 10^8 m/s)
E is the energy of the X-ray radiation
ε is the ratio of the recoil energy to the rest mass energy of the electron

Let's start with λKα:
The Kα line is the first line in the K-series X-ray spectrum and occurs when an electron transitions from the second shell (K-shell) to the first shell (L-shell).

The energy difference between the K and L shells is given by:

ΔE = 13.6 * (Z - 1)^2

Where Z is the atomic number of the target material. For molybdenum (Mo), Z = 42.

So, ΔE = 13.6 * (42 - 1)^2 = 13.6 * 41^2 = 13.6 * 1681

Now, let's calculate λKα:
λKα = hc / (E * (1 + ε))

E = ΔE

Substituting the values:

λKα = (6.626 x 10^-34 Js * 3 x 10^8 m/s) / ((13.6 * 1681) * (1 + ε))

Now, let's move on to λSWL:
The continuous X-ray spectrum, known as the Bremsstrahlung, extends from the maximum energy of the photons (E) down to zero energy. The shortest wavelength (λSWL) corresponds to the maximum energy of the photons.

λSWL = hc / (E * (1 + ε))

Substituting the values:

λSWL = (6.626 x 10^-34 Js * 3 x 10^8 m/s) / (50,000 eV * (1 + ε))

To find the values of ε, we need to know the atomic number of the target material and the energy of the X-ray radiation (E). Unfortunately, your question does not provide information about ε. Please provide the necessary information to calculate ε, or assume a value so that we can proceed with the calculations.