An x-ray generator operated at a plate voltage of 7000 V is incapable of producing Ni Kα radiation. Calculate the minimum energy required for a ballistic electron to produce Ni Kα radiation. Express your answer in units of eV.
To two digit accuracy Moseley's law is
10.2 ev x (atomic number -1)^2 =
10.2 x *(27)^2 = ? eV
7435.8
To calculate the minimum energy required for a ballistic electron to produce Ni Kα radiation, we need to use the concept of energy conservation and the equations related to X-rays.
The energy of an X-ray photon can be expressed in terms of its wavelength using the formula:
E = hc/λ
Where:
E is the energy of the X-ray photon,
h is Planck's constant (6.626 x 10^-34 J·s),
c is the speed of light (3.0 x 10^8 m/s),
λ is the wavelength of the X-ray photon.
Now, let's find the wavelength of Ni Kα radiation. The Kα X-ray radiation from Nickel (Ni) occurs when an electron transitions from the L shell to the K shell. The wavelength of this radiation is commonly known to be 1.54 Å (angstroms), which is equivalent to 1.54 x 10^-10 m.
Next, we can substitute the value of λ into the energy equation to find the minimum energy required for the ballistic electron:
E = hc/λ
E = (6.626 x 10^-34 J·s) * (3.0 x 10^8 m/s) / (1.54 x 10^-10 m)
By simplifying this expression, we can calculate the value of E in joules. However, the question asks us to express the answer in electron-volts (eV). We can convert the answer from joules to electron-volts using the conversion factor: 1 eV = 1.6 x 10^-19 J.
Therefore, the final step is to convert the energy from joules to electron-volts:
E(eV) = E(J) / (1.6 x 10^-19 J)
Calculating the value will provide the minimum energy required for a ballistic electron to produce Ni Kα radiation in eV.