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In an X-ray tube, electrons strike a metal target and produce X-rays, which then have to pass out of the evacuated tube through a thin “window.” The window material must satisfy two crucial conditions: it must be mechanically strong enough to withstand the pressure difference, and it must absorb the least possible fraction of the emerging X-ray beam so that intensity losses are minimal.
To analyse the absorption of X-rays by a material, we recall the exponential attenuation law, stated as
$$I \;=\; I_0 \, e^{-\mu x},$$
where $$I_0$$ is the incident intensity, $$I$$ is the transmitted intensity after passing through a thickness $$x,$$ and $$\mu$$ is the linear absorption coefficient of the material. A smaller $$\mu$$ means smaller attenuation for the same thickness, hence a better window material.
Now, the absorption coefficient $$\mu$$ depends strongly on the atomic number $$Z$$ of the material; in general, $$\mu$$ increases rapidly with increasing $$Z$$ because heavier nuclei interact more strongly with X-rays. Therefore, to minimise absorption, we want a metal with the lowest possible atomic number among those that are mechanically suitable.
Let us list the atomic numbers:
$$\text{Mg: } Z = 12,\quad \text{Na: } Z = 11,\quad \text{Be: } Z = 4,\quad \text{Ca: } Z = 20.$$
Clearly, beryllium ($$Z = 4$$) has by far the lowest atomic number of the four. Owing to this low $$Z,$$ its absorption coefficient $$\mu$$ is correspondingly small, and the factor $$e^{-\mu x}$$ remains close to unity even for practical window thicknesses. Thus, beryllium allows most of the generated X-ray photons to exit the tube with minimal loss while still providing adequate mechanical strength because of its high stiffness-to-weight ratio.
All other listed metals (magnesium, sodium, calcium) have higher atomic numbers, resulting in significantly greater absorption of X-rays and therefore are not preferred for the window.
So, the only metal that meets the requirement of very low X-ray absorption while retaining mechanical integrity is beryllium.
Hence, the correct answer is Option C.
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