Consider an x-ray tube with a Cu anode and a monochromator consisting of a nickel single crystal with (111) planes parallel to the surface of the crystal. What is the lowest voltage capable of producing x-rays through a monochromator?
Express your answer in V:
3053
currect answer is 3053
To determine the lowest voltage capable of producing x-rays through a monochromator, we need to consider the concept of Bragg's law, which describes the conditions for constructive interference in crystal planes.
Bragg's law is given by: nλ = 2dsinθ
Where:
- n is the order of diffraction (integer)
- λ is the wavelength of the x-rays
- d is the spacing between the crystal planes
- θ is the angle of incidence of x-rays on the crystal surface
In this case, the (111) planes of the nickel crystal are parallel to the surface. Therefore, the angle of incidence (θ) will be zero degrees, resulting in sinθ = 0.
Rearranging the equation to solve for λ:
λ = 2dsinθ / n
Since sinθ = 0, we get:
λ = 0
Now, the smallest wavelength a Cu target can produce in an x-ray tube is approximately 0.154 nm. So let's use this value to calculate the lowest voltage.
Recall that the voltage (V) is related to the wavelength (λ) by the following equation:
V = hc / λ
Where:
- h is Planck's constant (6.626 × 10^-34 J s)
- c is the speed of light in a vacuum (3.0 × 10^8 m/s)
Plugging in the values, we have:
V = (6.626 × 10^-34 J s * 3.0 × 10^8 m/s) / (0.154 × 10^-9 m)
Simplifying the equation, we obtain:
V = 1.361 × 10^15 V
Therefore, the lowest voltage capable of producing x-rays through a monochromator is approximately 1.361 × 10^15 V.