What is the minumum potential that must be applied across an x-ray tube in order to observe each of the following? What is the min of the continuous spectrum in each case?
(a) K line of potassium
____kV
____nm
(b) K line of tin
____kV
____nm
(c) L line of cesium
____kV
____nm
To determine the minimum potential required to observe specific X-ray lines and the minimum of the continuous spectrum, we need to refer to the energy level diagram of the elements in question.
(a) K line of potassium:
The K-line of an element corresponds to the transition of an electron in the K-shell of that element. The energy of the K-line can be calculated using the formula:
E_K = (13.6 eV) * (Z^2) / n^2
where:
- E_K is the energy of the K-line in electron volts (eV)
- Z is the atomic number of the element (in this case, potassium has an atomic number of 19)
- n is the shell number (1 for the K-shell)
To convert the energy to wavelength (λ) in nanometers (nm), we can use the equation:
λ = (1240 eV·nm) / E
Now, let's calculate the minimum potential and the wavelength for potassium:
E_K = (13.6 eV) * (19^2) / 1^2 = 5240.4 eV
λ = (1240 eV·nm) / 5240.4 eV = 0.236 nm
Therefore, the minimum potential required to observe the K-line of potassium is approximately 5.24 kV (5240.4 eV), and the minimum wavelength in the continuous spectrum is approximately 0.236 nm.
(b) K line of tin:
We can follow the same steps as above to calculate the minimum potential and wavelength for the K-line of tin.
For tin, the atomic number is 50, and we are again considering the K-shell transition (n = 1):
E_K = (13.6 eV) * (50^2) / 1^2 = 34,000 eV
λ = (1240 eV·nm) / 34,000 eV = 0.0365 nm
Thus, the minimum potential required to observe the K-line of tin is approximately 34 kV (34,000 eV), and the minimum wavelength in the continuous spectrum is roughly 0.0365 nm.
(c) L line of cesium:
To determine the minimum potential and wavelength for the L-line of cesium, we need to consider the transition from the L-shell.
Cesium has an atomic number of 55, and we'll take the L-shell transition (n = 2):
E_L = (13.6 eV) * (55^2) / 2^2 = 805 eV
λ = (1240 eV·nm) / 805 eV = 1.54 nm
Hence, the minimum potential required to observe the L-line of cesium is around 0.805 kV (805 eV), and the minimum wavelength in the continuous spectrum is approximately 1.54 nm.