Stationary Target X-Ray Tubes - Energy

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An x-ray tube with a stationary target operates at 100kV and 50mA for 0.2 seconds, resulting in a power output of 5000 W and a total energy of 1000 J. The discussion highlights confusion regarding the distribution of this energy between heat and x-rays. It is noted that typically only about 1% of the energy is converted into x-rays, suggesting approximately 10 J is emitted as x-rays while the remaining 990 J is dissipated as heat. The conversation emphasizes that the exact calculation of x-ray energy output requires more specific data, such as the atomic number of the target material. Overall, the estimation of energy distribution is primarily based on operational figures rather than precise calculations.
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Homework Statement


An x-ray tube with a stationary target is operated with a constant voltage and current of 100kV and 50mA for 0.2s . Approximately how much energy will be deposited as heat in the target and how much emitted as x-rays?

Homework Equations


##P=IV## ?

The Attempt at a Solution


This is not coursework but preparation for an exam, it is is a past paper question.

I'm just a bit confused really as I can't find any reference to how to calculate the amount as heat/x-rays etc.

But the Energy in 1/5s would just be
<br /> P=IV=(50 \times 10^{-3})(1 \times 10^5)=5000 W
So therefore in 1/5s of a second the energy would be 1000J.

But how do I work out how much of that is heat or x-rays? Is that even correct as it? The only reference to anything at all relating to this in the lecture notes/slides says that usually only around 1% is x-rays, so I don't know if the question just wants me to say that and therefore say ~10J as x-rays and 990J as heat?
 
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ChrisJ said:
The only reference to anything at all relating to this in the lecture notes/slides says that usually only around 1% is x-rays

i think the quoted estimate of 1% of energy being emitted as X-rays is an operational figure and the details of the estimate can not be done with the available data.
There are formulae for estimating the continuous X-ray intensity and it is proportional to Z the atomic number, tube current and square of the Potential difference applied.about one percent of the KE of electrons are used up in producing X-rays.
 
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