Question

yield of some photon energies is greater than 100%. does anyone know how it is possible?

Answer 1

where did u study it?

Answer 2

for a 92Nb radionuclide, i have seen that yield is around 134% for a 561 keV energy.

Answer 3

The quantum efficiency (QE), or incident photon to converted electron (IPCE) ratio,[1] of a photosensitive device or a charge-coupled device (CCD) is the percentage of photons hitting the device's photoreactive surface that produce charge carriers. It is measured in electrons per photon or amps per watt.[2] QE is a measurement of a device's electrical sensitivity to light. Since the energy of a photon is inversely proportional to its wavelength, QE is often measured over a range of different wavelengths to characterize a device's efficiency at each photon energy level. The QE for photons with energy below the band gap is zero. Photographic film typically has a QE of much less than 10%, while CCDs can have a QE of well over 90% at some wavelengths.


A solar cell's quantum efficiency value indicates the amount of current that the cell will produce when irradiated by photons of a particular wavelength. If the cell's quantum efficiency is integrated over the whole solar electromagnetic spectrum, one can evaluate the amount of current that the cell will produce when exposed to sunlight. The ratio between this energy-production value and the highest possible energy-production value for the cell (i.e., if the QE were 100% over the whole spectrum) gives the cell's overall energy conversion efficiency value. Note that in the event of multiple exciton generation (MEG), quantum efficiencies of greater than may be achieved since the incident photons have more than twice the band gap energy and can create two or more electron-hole pairs per incident photon.

Answer 4

http://en.wikipedia.org/wiki/File:Solarcellige-en.svg

Answer 5

It is mainly in the case of quantum efficiency of solar energy.

Answer 6

By the way welcome to open study :D

Answer 7

i m asking regarding the basic physics behind it and not related to any experimental setup or due to the detector used.
thanks

Answer 8

ur question is too high for me to define
btw
http://physics.aps.org/synopsis-for/10.1103/PhysRevLett.108.097403
The last help:)