Question

Is the ka characteristic radiation emitted by lanthanum (La: Z=57) more effectively absorbed by a 1-mm layer of gadolinium (Gd: Z=64) or by a 1-mm layer of lanthanum? Please explain numerically. (The k-edge of Gd is at 50 keV, while the k-edge of La is at 39 keV).

Answer 1

The ka or k-alpha characteristic radiation is nothing but the X ray radiation. From Moseley's law, we know that

$\sqrt{}$f = k₁ . (Z - k₂)

where f is the frequency of k alpha radiation and Z is the atomic number.

This means that frequency is directly proportional to the atomic number. Greater the atomic number, larger the frequency.

But energy is directly proportional to frequency as E = h${\nu }$

This means that higher the atomic number, greater is the energy of k alpha radiation emitted.

Given that Gd has greater atomic number (64) than La (57), the ka emitted by La will be smaller than that of Gd and hence easily absorbed by Gd because electrons of Gd will not get excited at 39 keV but at 50 keV.