Question

A pure Si detector operating in photoconductive mode and at room temperature absorbs a 1.04MeV gamma-ray photon produced by the decay of a 136Cs nucleus. The detector has an active area of 1 cm2 and is 0.5 cm thick. (a) Compute the number of electrons generated by the absorption of this photon. (b) Compare this number with that expected to be in the CB in the absence of any illumination (this would be the background signal). Will the signal from the gamma ray be detectable? (c) How would you enhance the signal over background?

Answer 1

We know that the rest mass of an electron is approximately 0.52Mev.

(a) For the given gamma-ray photon of energy 1.04 MeV,

the maximum electrons can be produced in Si detector = 1.04/0.52 = 2 (following energy conservation law).

(b) For a pure Si material, there will be no electron in the conduction band (CB) at absolute temperature. But at room temperature, some electrons of valence band will jump to the conduction band due to the thermal energy. Experimentally it is observed that there are approximately 1010 electrons per cubic centimetre in the CB of the pure Si material at room temperature.

The active volume of the given Si detector = active area X thickness =1X0.5 = 0.5 cm3

Therefore the available electrons in the CB of the given Si detector = 0.5 X 1010 = 5X109

This is the background signal due to the thermal energy at room temperature which is much larger than the electrons produced by the absorption of photons (2). Therefore the signal due to the gamma-ray will not be detectable.

(c) To enhance the signal over background, we'll enhance the signal to noise ratio (SNR). Sometimes it is done by using a lock-in amplifier or sometimes by a low noise amplifier. It can also be enhanced by averaging the measurements. For details, one can search for the signal to noise ratio (SNR).