Tungsten x-ray target replaced by molybdenum.

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The discussion centers on the impact of replacing a tungsten target with a molybdenum target in an X-ray tube operating at 50kV. The minimum wavelength of X-rays emitted can be calculated using the formula λ min = hc/eV, yielding a value of 2.481×10^(-11) m for tungsten. When substituting molybdenum, the minimum wavelength increases due to its lower atomic number (Z=42) compared to tungsten (Z=74), which affects the energy conversion during electron collisions. The justification for this change lies in the differences in atomic structure and energy transfer during inelastic collisions.

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In an Xray tube containing tungsten target , 50kV voltage is applied. Calculate the minimum wavelength of xray emitted. If the target is replaced by molybdenum will the min. wavelength change. Justify your answer?
I used λ min= hc/eV= 2.481×10^ (-11) m for the 1st part.
For the 2nd part I am clueless. I infer from observing graphs given in textbooks that molybdenum should have higher λmin than tungsten. Z=74 for tungsten and Z=42 for molybdenum. But can't come up with a rigid justification.
Should i use
ΤΜο)=( ΖΜο-1/ΖΤ-1)^2??
 
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I think you must have missed something from the question statement. What is the 50kV applied to?

Presumably it is applied to an electron that comes along and smacks into the target. So, where does the maximum amount of energy (and so the minimum wavelength) for the photon produced come from?

Think about an inelastic collision between an electron and the nucleus of some target atom that starts at rest in the lab frame. What will the maximum possible amount of energy be that could be converted to a photon? Think about the electron and the nucleus colliding and at that moment giving up the maximum possible amount of kinetic energy to the photon. How much is that maximum? On what does that maximum depend?

Hint: It's not Z for the nucleus. Z is the number of protons. There are other things in a nucleus.
 

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