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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To determine the identity of the target that would be needed to generate x-rays of wavelength as low as 10^-11 meters, we need to consider the principles of x-ray generation.
X-rays are produced through a process called bremsstrahlung radiation, which occurs when high-energy electrons are decelerated or deflected by the target material. As a result of this interaction, x-rays with a range of energies and wavelengths are emitted.
The energy of an x-ray is related to its wavelength through the equation: E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength.
In this case, we know that the wavelength (λ) is 10^-11 meters. Let's substitute this value into the equation to determine the energy (E) of the x-ray.
E = (6.626 x 10^-34 J·s) * (3 x 10^8 m/s) / (10^-11 m)
E ≈ 1.99 x 10^-7 J
Now, to generate x-rays of such a high energy, we need a target material that can provide a strong interaction with the incident high-energy electrons. The ability of a material to interact strongly with electrons is characterized by its atomic number (Z).
Generally, materials with higher atomic numbers (Z) provide stronger interactions due to the increased nuclear charge. This enhanced interaction leads to the emission of higher-energy x-rays.
Thus, to generate x-rays with wavelengths as low as 10^-11 meters, we would require a target material with a relatively high atomic number (Z) to ensure a strong interaction and the emission of x-rays at that energy level.
In summary, to determine the identity of the target material needed to generate x-rays with a wavelength of 10^-11 meters, we would choose a material with a relatively high atomic number. The specific material would depend on the energy level required for the x-rays, which corresponds to the target's atomic number.