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Jacey Yang 1F
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Joined: Fri Aug 09, 2019 12:17 am


Postby Jacey Yang 1F » Mon Oct 14, 2019 7:39 pm

A lamp rated at 32 W (1W = 1 J.s-1) emits violet light of wavelength 420 nm. How many photons of violet light can the lamp generate in 2.0 s? How many moles of photons are emitted in that time interval?

I'm not sure how to go about solving this problem. Can someone please explain?

Michelle Le 1J
Posts: 50
Joined: Fri Aug 09, 2019 12:16 am

Re: 1B.9

Postby Michelle Le 1J » Mon Oct 14, 2019 9:37 pm

Since they gave you the watts of the lamp, you would convert that to J/s and multiply that by 2 because of the 2 seconds in order to get the Joules. You would just need to find the energy of the photon emitted using E=hc/wavelength and then use that answer (which is in Joules/photon) and multiply that to the 64 J to find the number of photons generated. In order to find the moles, you would then proceed with Avogadro's number.

Ziyan Wang 3J
Posts: 51
Joined: Wed Sep 18, 2019 12:22 am

Re: 1B.9

Postby Ziyan Wang 3J » Mon Oct 14, 2019 11:03 pm

Energy per photon=h*c/lambda=6.63*10^-34*3.00*10^8/(420*10^-9)=4.74*10^-19
Total energy= 32*2=64
# of photons= Total energy/energy per photon=64/(4.74*10^-19)=1.35*10^20

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Joined: Fri Aug 02, 2019 12:16 am

Re: 1B.9

Postby Mariah » Tue Oct 15, 2019 1:23 pm

Can anyone explain how the units should turn out? Or how they work throughout the problem because I am having trouble with that.

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Joined: Sat Sep 07, 2019 12:16 am

Re: 1B.9

Postby ramiro_romero » Fri Oct 25, 2019 1:09 am

To find photons, you first multiply Watts by 2 seconds and get 64 J/2sec as the Total Energy. Then, you find the energy per photon by using the formula E=(hc)/lambda, (energy per photon should be 4.717 x 10^-19). Divide energy per photon by total energy, and you get the number of photons [1.35 x 10^20 photons].

To find moles of photons, you divide the number of photons(1.35 x 10^20) by avogrados number (6.022 x 10^23), and get [2.25 x 10^-4 mol photons]

nshahwan 1L
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Joined: Fri Aug 30, 2019 12:18 am

Re: 1B.9

Postby nshahwan 1L » Wed Nov 06, 2019 11:52 am

I understand how to do this, but could someone explain why you divide the energy of the photon from the given watts?

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