## UA Workshop Question

$c=\lambda v$

Isabel_Eslabon_2G
Posts: 93
Joined: Wed Sep 30, 2020 9:50 pm

### UA Workshop Question

Hello, I am struggling with this problem:

It is generally accepted that UVC lamps with 40 mJ/cm² at a wavelength of 250 nm can be used to disinfect surfaces and kill pathogens. If you leave a UVC lamp on to disinfect a 5 m² area, how many high energy photons are produced?

The answer key says 2.52*10^21 photons.

Here is my work so far:
$\frac{40mJ}{cm^2}\cdot \frac{1J}{1000mJ}\cdot \frac{10000cm^2}{1m^2}=400J\cdot m^2$
$\frac{400J}{m^2}\cdot 5m^2=2000J$

I don't know where to go from here or if I'm on the right track. I am not sure how to use the 250 nm in this problem or if I actually need it.

John Pham 3L
Posts: 140
Joined: Wed Sep 30, 2020 9:49 pm
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### Re: UA Workshop Question

You already have the total number of Joule produced which is 2000J.
All you need to do now is find the energy per photon using the wavelength given with the formula $E = \frac{hc}{\lambda}$.
Then you can divide total Joules produced by energy per photon to find the total number of photons produced.

Hannah Rim 2D
Posts: 58
Joined: Wed Sep 30, 2020 9:48 pm
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### Re: UA Workshop Question

Hey! I was super confused on this problem earlier today too. Regarding the 250 nm, you convert that wavelength to m to get 2.5*10^-7 m. With this conversion, we can use the equation E=hc/(lambda) and get the amount of energy per photon. The energy should be 7.95*10^-19 J/photon I believe. Since we've already established that there are 2,000 J in 5 square meters, we can divide 2,000 J by 7.95*10^-19 J/photon to get the amount of photons that are produced. Hope this helps!

Marisa Gaitan 2D
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Joined: Wed Sep 30, 2020 9:47 pm
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### Re: UA Workshop Question

Hannah Rim 2D wrote:Hey! I was super confused on this problem earlier today too. Regarding the 250 nm, you convert that wavelength to m to get 2.5*10^-7 m. With this conversion, we can use the equation E=hc/(lambda) and get the amount of energy per photon. The energy should be 7.95*10^-19 J/photon I believe. Since we've already established that there are 2,000 J in 5 square meters, we can divide 2,000 J by 7.95*10^-19 J/photon to get the amount of photons that are produced. Hope this helps!

Why is there 2,000J? Since the energy units given in the problem are mJ wouldn't you have to convert mJ to J?

Hannah Rim 2D
Posts: 58
Joined: Wed Sep 30, 2020 9:48 pm
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### Re: UA Workshop Question

marisagaitan1G wrote:
Hannah Rim 2D wrote:Hey! I was super confused on this problem earlier today too. Regarding the 250 nm, you convert that wavelength to m to get 2.5*10^-7 m. With this conversion, we can use the equation E=hc/(lambda) and get the amount of energy per photon. The energy should be 7.95*10^-19 J/photon I believe. Since we've already established that there are 2,000 J in 5 square meters, we can divide 2,000 J by 7.95*10^-19 J/photon to get the amount of photons that are produced. Hope this helps!

Why is there 2,000J? Since the energy units given in the problem are mJ wouldn't you have to convert mJ to J?

Yes! We do convert mJ to J for a part of it. If we convert 40 mj/cm^2 to J/m^2, we will get 400 J/m^2. However, since we want to know the energy in 5 m^2, we can multiply 400 by 5 to get 2,000 J.