Show that a conventional x-ray generator is incapable of generating x-rays of wavelength as low as 10-11m. Do this by determining the identity of the target (in atomic number, Z) which would be needed to generate such x-rays.
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CHEAT!
Do NOT help this person. They are trying to cheat in a midterm exam.
tttt you should be ashamed of yourself too for giving someone else an answer.
To determine the identity of the target (in atomic number, Z) required to generate x-rays with a wavelength as low as 10^(-11) meters, we need to consider the relationship between photon energy and wavelength.
The energy of a photon (E) is given by the equation E = hc/λ, where h is Planck's constant (approximately 6.626 x 10^(-34) J·s), c is the speed of light (approximately 3 x 10^(8) m/s), and λ is the wavelength of the x-rays.
In order to produce x-rays with a wavelength of 10^(-11) meters, we can rearrange this equation to solve for energy:
E = hc/λ
E = (6.626 x 10^(-34) J·s) × (3 x 10^(8) m/s) / (10^(-11) m)
E ≈ 1.988 x 10^(-15) J
Now, let's consider the concept of characteristic x-rays. When an electron in an atom undergoes a transition from a higher energy level to a lower energy level, it emits an x-ray photon with energy equal to the energy difference between the two levels. These x-rays are called characteristic x-rays, and their energy can be calculated using the equation:
E = 13.6 eV × (Z^2 / n_f^2 - Z^2 / n_i^2)
In this equation, E is the energy of the x-ray photon, Z is the atomic number of the target material, n_i is the principal quantum number of the initial energy level, and n_f is the principal quantum number of the final energy level.
For our case, we need to find the Z value that would give us an energy of approximately 1.988 x 10^(-15) J. Let's assume the initial and final energy levels are the K and L energy levels, respectively (n_i = 1, n_f = 2).
1.988 x 10^(-15) J = 13.6 eV × (Z^2 / 2^2 - Z^2 / 1^2)
Now, let's calculate the value of Z:
1.988 x 10^(-15) J = 13.6 eV × (Z^2 / 4 - Z^2)
1.988 x 10^(-15) J = 13.6 eV × (Z^2 / 4 - 1)
1.46 x 10^(-16) J = (Z^2 / 4 - 1)
Simplifying further:
(Z^2 / 4 - 1) = 1.46 x 10^(-16) J / 13.6 eV
(Z^2 / 4 - 1) ≈ 1.07 x 10^(-17) / 1.74 x 10^(-18) eV
(Z^2 / 4 - 1) ≈ 6.15 eV
Let's now isolate the Z^2 term:
Z^2 / 4 ≈ 6.15 eV + 1
Z^2 ≈ 28.6 eV
Taking the square root of both sides:
Z ≈ √(28.6 eV)
Z ≈ 5.35
Therefore, in order to generate x-rays with a wavelength as low as 10^(-11) meters, we would need a target material with an atomic number (Z) of approximately 5.35. Since atomic numbers are always whole numbers, we can conclude that a conventional x-ray generator, which typically uses target materials with higher atomic numbers, is incapable of generating x-rays with such short wavelengths.