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Finding Wavelength using Energy

  1. Jul 26, 2011 #1
    1. The problem statement, all variables and given/known data
    The energy required to dislodge electron from sodium and metal via the photoelectric effect is 275 kJ/mol. What wavelength in nm of light has sufficient energy per photon to dislodge an electron from the surface of sodium?


    2. Relevant equations
    How do I convert 272 kJ/mol to J so that it cancels out Plank's Constant and the light leaving me with m which I can convert to nm?


    3. The attempt at a solution
    I kept trying and failing. The book has the answer as 435 nm and unfortunately, it doesn't show the work. I'd like to know how it got the 435 nm.
     
  2. jcsd
  3. Jul 26, 2011 #2

    Pengwuino

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    Gold Member

    Well, of course the kJ = 1000 J, the mole needs to be canceled out by how many atoms are in a mole. Use Avagadro's number for this.
     
  4. Jul 27, 2011 #3

    Yeah I get how to convert the KJ -> J. But the mole is the problem. And I've tried avagdro's but the answer still doesn't match what's in the book. Where I am currently in the chapter, the book hasn't even introduced Avagadro's number yet so I wasn't sure if that's really what was needed but even after trying it out, I'm not getting the correct answer. Look:

    275,000 x 6.022E10^23 / mole = 1.656E10^29 J/atom

    (3.00x10E8 m/s)(6.626E10-34 j*s)
    ------------------------------------------ =/=435 nm (even after you convert m -> nm)
    (1.656E1029 j)

    ------------
    I just don't get it
     
  5. Jul 27, 2011 #4

    Pengwuino

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    Whoa, think about what you're doing for a second. What are the units for Avagadro's number? Does 1.656x10^29 Joules of energy PER atom sound even vaguely reasonable? You're multiplying when you should be dividing.
     
  6. Jul 27, 2011 #5
    But aren't you trying to get the moles to cancel out so therefore you set up the equation like this?:
    ............................. 6.022E10^23
    275E10^3 J/moles ------------------------
    ....................................1 mole

    moles cancels with moles and you multiple 275,000 to 6.022E10^23. That's what I've been doing this whole time.


    In the back of my textbook Avagadro's number = 6.022E10^23/mol
     
  7. Jul 27, 2011 #6
    This is actually very simple if you know what Avogadro's number is. It's the number of atoms (or molecules or ions or whatevers) in a mole. Now, if I told you that is costs $4 for a dozen cans of coke, how would you figure out the price per can? You would divide the total price by the number of cans in a dozen, right? So how do you get the energy per atom, if you know the energy per mole?
     
  8. Jul 27, 2011 #7
    Last edited by a moderator: May 5, 2017
  9. Jul 27, 2011 #8
    You're supposed to be calculating energy per photon. That means you take the amount of energy in a mole of photons and divide it by the number of photons in a mole of photons.

    It's really important that you do sanity checks. Pengwuino asked, "Does 1.656x10^29 Joules of energy PER atom sound even vaguely reasonable?" THINK about that. 1.656x10^29 is a really BIG number. A photon is the smallest possible particle of light, so it should have a really SMALL energy. 2.2x10^18 is smaller than 1.656x10^29, but it's still way, way too big. You expect 10^-something for the energy. You should immediately realize, getting numbers like that, that something is wrong.
     
  10. Jul 27, 2011 #9
    I really still don't get it. I re-did the math switching around the division of the avagadro's number. I put the 275000 in the calculator first and I got 4.566E10^-19. When I divide with the hc I'm getting 4.35 m and when I convert to nm I still get 4.35 nm and not 435
     
  11. Jul 27, 2011 #10
    Well, at least now you're close. You've just got to find where you let the decimal point slip by a couple of places. Everyone makes those mistakes -- just go back and track it down.

    Could you explain this remark?

    How can you convert 4.35 m to nm and get 4.35? 4.35 m = 4,350,000,000 nm.
     
  12. Jul 27, 2011 #11

    SteamKing

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    What is the numerical value of the product hc equal to when you do your calculation?
     
  13. Jul 27, 2011 #12

    Pengwuino

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    You're just a computation error away from the problem.

    However, to answer your original question, Avagadros number is actually [itex]6.022x10^23 {{atoms}\over{mole}}[/itex]. It's the number of atoms in 1 mole of something. That's why your answer came out incorrectly and your dimensional analysis made no sense.

    Remember, whenever you use a conversion, you need to know exactly what it means.
     
  14. Jul 28, 2011 #13
    I really still don't get. Was just reading about it in the textbook and I still don't understand. The texbook has this:

    "One mole of anything is 6.022E10^23 units of that thing."
    Does that basically mean just divide?

    Here's another thing, 1 mol = 6.0221421E10^23

    So If I used that as a conversion factor to convert 275,000 J/mol to get the mol to cancel out, wouldn't I set it up like this:

    .......................6.022E10^23
    275,000 J/mol -------------------
    .........................1 mol

    That's why I was multiplying in the first place. I really still don't get this. Why do I do it like this instead? :

    275,000 J/mol
    ----------------------
    6.022E10^23/mol

    The book seems like it's also teaching something else about converting between number of atoms and number of moles. Is the problem here that I'm not converting between moles and atoms in my particular problem but rather I just need to get the mol to cancel out? Is that why I have to divide?
     
  15. Jul 28, 2011 #14
    Oh sorry. Problem solved. I finally got 435.348226 and due to significant figures it comes to 435 nm. I got the right answer this time around. Thanks.
     
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