Re: Conversion of K2ptcl4 to K2ptcl6
Jeffrey D. Mathias ( KOUKLIS_KERIK@aphub.aerojetpd.com)
Thu, 16 January 1997 3:58 PM
Philippe Monnoyer wrote:
>
>
> Why do you want to get K2PtCl6 instead of using K2PtCl4 ? I would think you
> loose some sensitivity with K2PtCl6 , am I right ? The ferrous oxalate you
> form while exposing as a given number of electrons to reduce the platinum
> ions. That number is proportional to the amount of light that you gave to
> the layer. The difference between K2PtCl4 and K2PtCl6 is that they contain
> respectively Pt2+ and Pt4+. To reduce Pt2+ to metallic black platinum, you
> need 2 electrons from the ferrous oxalate. To reduce Pt4+ you need 4
> electrons. So, for a given exposure, you will get twice the black reduced
> platinum with K2PtCl4 vs K2PtCl6.
> Does it match to your experiences ?
>
Philippe,
Perhaps you misunderstood. I do not wish to use K2PtCl6 for it is not
hardly soluble in water. However, your comment on the electrons is
interesting, and indeed one photon will only kick out one electron.
Jeff
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