Thank you! In a photoelectric interaction, a gamma-ray photon transfers all of i

Question

Thank you!

In a photoelectric interaction, a gamma-ray photon transfers all of its energy to a bound electron in one of the atoms of the crystal. If this electron is a K-shell electron, this will leave its atom with a K-shell vacancy. When this vacancy is filled by an electron from a higher shell, a characteristic X-ray is produced. Sometimes this X-ray escapes the NaI crystal without interacting. If this happens often, it will produce an"iodine escape peak" in the measured spectrum, at an energy that is less than the photopeak energy by an amount equal to an iodine K_alpha X-ray (28.6 keV). At what energy would you expect to find an iodine escape peak in a gamma energy spectrum of^57Co, which emits a 122-keV gamma ray? There are just as many sodium atoms as iodine atoms in a NaI crystal. Why would you not expect to see a"sodium escape peak" in a spectrum generated with a^57Co source?

Explanation / Answer

(a) We would expect the escape peak to be identically separated i.e., 122 - 28.6 = 93.4 keV

(b) Although there are as many sodium atoms as iodine atoms in a NaI crystal, but the occurrence of a "sodium escape peak" is far more negligible (7 times less) than the iodine escape peak. Also, the energy supplied to the K-shell electron that exits the shell is synonymous with the iodine electron potential.

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