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 for an X-ray tube with a molybdenum (Mo) target and an acceleration potential of 50 keV, we can make use of two key formulas.
1. For the Kα line wavelength:
The Kα line wavelength (λKα) corresponds to the transition between the K-shell and the L-shell. This can be calculated using Moseley's law:
λKα = λK / (Z - eff)^2
where λK is the K-shell constant equal to 0.71 pm (picometers), Z is the atomic number of the target material, and eff is the effective nuclear charge.
For molybdenum, Z = 42. The effective nuclear charge (eff) is usually estimated by the formula:
eff = Z - σ
where σ is the screening constant. For molybdenum, σ is approximately 1.
Using these values, we can calculate λKα:
λKα = (0.71 pm) / (42 - 1)^2
2. For the shortest wavelength (λSWL):
The shortest wavelength (λSWL) corresponds to the highest energy X-rays emitted by the X-ray tube and can be calculated using the formula:
λSWL = hc / eV
where h is Planck's constant (6.626 x 10^-34 J s), c is the speed of light (3 x 10^8 m/s), and eV is the energy in electron volts.
Converting the acceleration potential of 50 keV to joules:
50 keV = 50 * 1000 eV = 50,000 eV = 50,000 * 1.602 x 10^-19 J
Now, we can calculate λSWL:
λSWL = (6.626 x 10^-34 J s * 3 x 10^8 m/s) / (50,000 * 1.602 x 10^-19 J)
Calculating these values will give you the accurate wavelengths λKα and λSWL in meters.