It is known that ZnWO4, (Eg approx 3.6-3.8 eV), owing to direct band gap should show negligible photocatalytic activity the in visible range and only in UV range. On the other hand, beta ZnMoO4 (Eg approx 3.2-3.4), with indirect band gap shows photocatalytic activity in visible range. Upon substitution of Molybdenum in ZnWO4, the mixed solid solution should ideally show better efficiency in Visible range. But how do we conclude that the mixed solid solution of ZnW(1-x)Mo(x)O4 will show better photocatalysis in which system (UV or visible)? What should be the ideal experimental apparatus?
Hi Avi,
In my opinion, based on your band gap energy (close to TiO2) and its activity data....
the ratio of W to Zn is a key due to its activity in visible or UV range. In addition, the photocatalytic activity is also dependent on type of compound you want to break-down its bond (dyes or chemical conversion) and the UV or Visible light energy used. in general, as lower the band gap energy the more active in the visible region.
I hope this information will be useful,
Regard,
Rudy
I think UV-Vis irradiation is better than UV and Visible for that material.
@Schindra kumar Ray If I carry out the reaction under UV-Visible irradiation, how may we conclude that which of the part of the Em spectrum is actually responsible for creating the hydroxyl radical and oxygen anion via electron transport and hole generation.
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