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user7951
user7951

Thanks all for the responses! (Sorry for the delayed upvotes and answer selection, I thought I did those already).

It was recently pointed out to me that there's a way to do this in a principled manner. Let's assume the lamp is 4W$4\ \mathrm W$ strong and distributes its power over 10 plates (1000 mm$^2$$1000\ \mathrm{mm^2}$). The energy of a 254nm$254\ \mathrm{nm}$ photon is $7.8 \times 10^{-19}$ Joules$7.8 \times 10^{-19}\ \mathrm J$. If the lamp is at 4W$4\ \mathrm W$, it's putting out 4 J/s$4\ \mathrm{J/s}$ of energy. This corresponds to $$ \frac{4 J}{s} * \frac{\mathrm{photon}}{7.8 \times 10^{-19} J} = 5.11 \times 10^{18}$$$$ \frac{4\ \mathrm J}{\mathrm s}\times\frac{\text{photon}}{7.8 \times 10^{-19}\ \mathrm J} = 5.11 \times 10^{18}$$ photons per second.

Now, if the absorbance of the initiator is 0.0001, we have that the absorbed fraction of the photons is: $$ \% Abs = 1 - \%T = 1 - \frac{I_1}{I_0} = 1 - 10^{-A} = 2 \times 10^{-4} $$

Since we have $5.11 \times 10^{18}$ photons per second are incoming, roughly $1 \times 10^{15}$ of them will be absorbed, which should be more than enough to initiate polymerization.

Of course, this is a very, very rough calculation, but I thought it was cool that you could, in theory, calculate that this experiment should work (which it did!)

Thanks all for the responses! (Sorry for the delayed upvotes and answer selection, I thought I did those already).

It was recently pointed out to me that there's a way to do this in a principled manner. Let's assume the lamp is 4W strong and distributes its power over 10 plates (1000 mm$^2$). The energy of a 254nm photon is $7.8 \times 10^{-19}$ Joules. If the lamp is at 4W, it's putting out 4 J/s of energy. This corresponds to $$ \frac{4 J}{s} * \frac{\mathrm{photon}}{7.8 \times 10^{-19} J} = 5.11 \times 10^{18}$$ photons per second.

Now, if the absorbance of the initiator is 0.0001, we have that the absorbed fraction of the photons is: $$ \% Abs = 1 - \%T = 1 - \frac{I_1}{I_0} = 1 - 10^{-A} = 2 \times 10^{-4} $$

Since we have $5.11 \times 10^{18}$ photons per second are incoming, roughly $1 \times 10^{15}$ of them will be absorbed, which should be more than enough to initiate polymerization.

Of course, this is a very, very rough calculation, but I thought it was cool that you could, in theory, calculate that this experiment should work (which it did!)

It was recently pointed out to me that there's a way to do this in a principled manner. Let's assume the lamp is $4\ \mathrm W$ strong and distributes its power over 10 plates ($1000\ \mathrm{mm^2}$). The energy of a $254\ \mathrm{nm}$ photon is $7.8 \times 10^{-19}\ \mathrm J$. If the lamp is at $4\ \mathrm W$, it's putting out $4\ \mathrm{J/s}$ of energy. This corresponds to $$ \frac{4\ \mathrm J}{\mathrm s}\times\frac{\text{photon}}{7.8 \times 10^{-19}\ \mathrm J} = 5.11 \times 10^{18}$$ photons per second.

Now, if the absorbance of the initiator is 0.0001, we have that the absorbed fraction of the photons is: $$ \% Abs = 1 - \%T = 1 - \frac{I_1}{I_0} = 1 - 10^{-A} = 2 \times 10^{-4} $$

Since we have $5.11 \times 10^{18}$ photons per second are incoming, roughly $1 \times 10^{15}$ of them will be absorbed, which should be more than enough to initiate polymerization.

Of course, this is a very, very rough calculation, but I thought it was cool that you could, in theory, calculate that this experiment should work (which it did!)

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chipbuster
chipbuster
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Thanks all for the responses! (Sorry for the delayed upvotes and answer selection, I thought I did those already).

It was recently pointed out to me that there's a way to do this in a principled manner. Let's assume the lamp is 4W strong and distributes its power over 10 plates (1000 mm$^2$). The energy of a 254nm photon is $7.8 \times 10^{-19}$ Joules. If the lamp is at 4W, it's putting out 4 J/s of energy. This corresponds to $$ \frac{4 J}{s} * \frac{\mathrm{photon}}{7.8 \times 10^{-19} J} = 5.11 \times 10^{18}$$ photons per second.

Now, if the absorbance of the initiator is 0.0001, we have that the absorbed fraction of the photons is: $$ \% Abs = 1 - \%T = 1 - \frac{I_1}{I_0} = 1 - 10^{-A} = 2 \times 10^{-4} $$

Since we have $5.11 \times 10^{18}$ photons per second are incoming, roughly $1 \times 10^{15}$ of them will be absorbed, which should be more than enough to initiate polymerization.

Of course, this is a very, very rough calculation, but I thought it was cool that you could, in theory, calculate that this experiment should work (which it did!)