How would I approach this problem?
A liquid is exposed to infrared radiation with a wavelength of 9.74×10−4 cm. Assume that all the radiation is absorbed and converted to heat. How many photons are required for the liquid to absorb 32.74 J of heat?
Do I need to use both E=hv and C=lambda(V)?
HW problem 5
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 Posts: 63
 Joined: Wed Sep 30, 2020 9:52 pm
Re: HW problem 5
Yes, that is exactly correct.
Conceptually, the equations we are solving are PER photon. So at the end we would just need to figure out how many of the energy per photon we would need to get the 32.74J.
First you would want to find the frequency so do c=lambda v
this would give you 3.08 x 10^11 hz
Then you would use the E=hv equation.
This should give you 2.04 x 10^22 J
then lastly divide 32.74 joules by the energy per photon, which is 2.04 x 10^22 J to get 1.61 x10^23 photons!
Conceptually, the equations we are solving are PER photon. So at the end we would just need to figure out how many of the energy per photon we would need to get the 32.74J.
First you would want to find the frequency so do c=lambda v
this would give you 3.08 x 10^11 hz
Then you would use the E=hv equation.
This should give you 2.04 x 10^22 J
then lastly divide 32.74 joules by the energy per photon, which is 2.04 x 10^22 J to get 1.61 x10^23 photons!

 Posts: 60
 Joined: Wed Sep 30, 2020 9:51 pm
Re: HW problem 5
Karina Rodriguez 2H wrote:How would I approach this problem?
A liquid is exposed to infrared radiation with a wavelength of 9.74×10−4 cm. Assume that all the radiation is absorbed and converted to heat. How many photons are required for the liquid to absorb 32.74 J of heat?
Do I need to use both E=hv and C=lambda(V)?
Yes. I saw another reply walking through the math, so I wanted to explain more conceptually why you would use those two equations.
The question has a body (the liquid) and an energy source (the infrared radiation), and it is asking how many discrete (or individual) units of energy (number of photons) are needed for the liquid to capture some amount of energy (as heat).
So, to solve this question, you would use E=hv because that describes the amount of energy you have per discrete unit of energy.
However, to use this equation, you need to know v, and you are not given v, you are given the wavelength (lambda). This requires you to use c=lambda*v, which allows you to find to find v, because c is constant.
Once you have found v, you can calculate the number of energy per photon. The final step is to find the number of photons using the total energy absorbed.
By dividing the energy you have by the energy per photon (energy/(energy/photon)) leaves you with just photons, which is the desired answer.
I hope this helped explain conceptually what this question is asking for.

 Posts: 63
 Joined: Wed Sep 30, 2020 10:00 pm
Re: HW problem 5
Karina Rodriguez 2H wrote:How would I approach this problem?
A liquid is exposed to infrared radiation with a wavelength of 9.74×10−4 cm. Assume that all the radiation is absorbed and converted to heat. How many photons are required for the liquid to absorb 32.74 J of heat?
Do I need to use both E=hv and C=lambda(V)?
Another way of doing it would be using just the one equation, E= hc/lambda, which gives you the energy per photon. From there you can continue by dividing the given heat energy absorbed by the energy per photon that was just found. Also, something to keep in mind is converting the wavelength in cm to m because that messed me up at first.

 Posts: 76
 Joined: Wed Sep 30, 2020 9:33 pm
Re: HW problem 5
Rohit Srinivas 1C wrote:Yes, that is exactly correct.
Conceptually, the equations we are solving are PER photon. So at the end we would just need to figure out how many of the energy per photon we would need to get the 32.74J.
First you would want to find the frequency so do c=lambda v
this would give you 3.08 x 10^11 hz
Then you would use the E=hv equation.
This should give you 2.04 x 10^22 J
then lastly divide 32.74 joules by the energy per photon, which is 2.04 x 10^22 J to get 1.61 x10^23 photons!
Before solving the problem, do we need to convert the wavelength in cm to m? or is it okay to leave it in cm?

 Posts: 69
 Joined: Wed Sep 30, 2020 10:03 pm
Re: HW problem 5
Hi!
The first thing you need to do is calculate the energy with the equation E=hc/lambda. Then, because it is asking for the photons required to absorb a specific energy, you will do the given energy divided by what you just solved.
The first thing you need to do is calculate the energy with the equation E=hc/lambda. Then, because it is asking for the photons required to absorb a specific energy, you will do the given energy divided by what you just solved.

 Posts: 63
 Joined: Wed Sep 30, 2020 10:00 pm
Re: HW problem 5
Minjoo_Park_3F wrote:Rohit Srinivas 1C wrote:Yes, that is exactly correct.
Conceptually, the equations we are solving are PER photon. So at the end we would just need to figure out how many of the energy per photon we would need to get the 32.74J.
First you would want to find the frequency so do c=lambda v
this would give you 3.08 x 10^11 hz
Then you would use the E=hv equation.
This should give you 2.04 x 10^22 J
then lastly divide 32.74 joules by the energy per photon, which is 2.04 x 10^22 J to get 1.61 x10^23 photons!
Before solving the problem, do we need to convert the wavelength in cm to m? or is it okay to leave it in cm?
Yes, you need to convert the wavelength from cm to m or else you will get the wrong answer. Just divide the wavelength by 100 before you plug it into the equation.
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