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.
lambda min = 1239.8 pm/kilovolts
You can read more about it here.
http://en.wikipedia.org/wiki/Bremsstrahlung
DrBob222 but its asking the energy not the wavelength,can you explain better pls?
7.47*10^3
thanks
To calculate the minimum energy required for a ballistic electron to produce Ni Kα radiation, we need to determine the energy difference between two electronic transitions.
First, we need to find the energy of the Ni Kα radiation. The wavelength of Ni Kα radiation is well-known and is equal to 1.5418 Å (angstroms). We can convert this wavelength to energy using the equation:
E = hc/λ
Where:
E = energy
h = Planck's constant (6.626 x 10^-34 J·s)
c = speed of light (2.998 x 10^8 m/s)
λ = wavelength (in meters)
Converting the wavelength to meters:
λ = 1.5418 x 10^-10 m
Substituting the values into the equation:
E = (6.626 x 10^-34 J·s * 2.998 x 10^8 m/s) / (1.5418 x 10^-10 m)
E ≈ 12.735 keV (kilo-electron volts)
Now we can calculate the minimum energy required for a ballistic electron to produce Ni Kα radiation. The energy of an electron is equal to its charge (e) multiplied by the voltage (V) it is accelerated through:
Ee = eV
Where:
Ee = energy of the electron
e = elementary charge (1.602 x 10^-19 C)
V = voltage (in volts)
Substituting the values into the equation:
Ee = (1.602 x 10^-19 C) * (7000 V)
Ee ≈ 1.1214 keV (kilo-electron volts)
Therefore, the minimum energy required for a ballistic electron to produce Ni Kα radiation is approximately 1.1214 keV.