## energy required to remove an electron [ENDORSED]

$c=\lambda v$

hojae_lee_1C
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### energy required to remove an electron

I am having a tough time calculating the energy required to remove an electron from a sodium atom. The given information includes the ejected electron's velocity v = 6.61 x 10^5 m/s and sodium's work function = 150.6 kJ/mol. How do I approach this question?

Paul Wong1B
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### Re: energy required to remove an electron

The energy required to remove an electron is the value of sodium's work function, so the value would be 150.6 kj/mol.

Joe Rich 1D
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### Re: energy required to remove an electron

It should also be said that the question in the post-assessment asks for the energy required to remove 1 electron, but the work function is given in kJ per mol of electrons. In order to convert this number into the energy (in Joules) to remove a single electron, we first multiply 150.6 kJ by 1000 to get the energy in Joules, which is 1.506 x 10^5 J. We then divide the energy in Joules by 6.02 x 10^23 (number of electrons in one mol) to get a work function of 2.502 x 10^-19 J/electron.

Sarah_Wilen
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### Re: energy required to remove an electron

Hope the image of the worked out problem attached helps. I'll walk through my brain jumble:

1) To solve for the KE of the ejected electron, use the equation on the right side of the diagram because this is the amount of KE of the electrons. This equation is E$E_{KE}=\frac{1}{2}m_{electron}\times v_{electron}^{2}$
Now, all you have to do is take your known quantities (velocity, mass of electron constant), then plug and chug. You should get $1.99\times 10^{-19}J$

2) To solve for the energy needed to remove an e- from the metal, remember that this energy is the amount of energy the photon needs to meet or exceed to remove the e-, so wouldn't it make sense for this energy to be the threshold energy (work function)? Yes, it does. You are actually given the threshold energy, but it is in KJ. Convert the quantity into Joules by a unit conversion (1000 Joules in 1 KJ). You should get $1.506\times 10^{5}J$

3) Finally, we need to find the frequency of the incident light. You already know that $E_{photon}-\phi =E_{KE}$ Using this equation, rearrange to solve for the energy of the photon (because we know the threshold and KE of photon already). Plug and chug to solve for the energy. Now that we know the energy, to find the frequency of the light, use the equation $E=h\nu$ Rearrange the equation to solve for frequency. Plug and chug, then you're done! Yippee! :)

https://ibb.co/ciCKXk
This is the image^ Idk how to post it. LMK if the URL fails me

**EDIT**
Divide the energy to remove the metal (threshold) by Avogadro's number because it is the energy per electron. Then, use that number instead of the mess I wrote to plug and chug to find the frequency.
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Last edited by Sarah_Wilen on Thu Jun 29, 2017 10:30 am, edited 1 time in total.

Xinye_Jiang_1A
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### Re: energy required to remove an electron  [ENDORSED]

Since the work function is given by J/mol, is it possible that we need to devide 1.506*10^5 by 6.022*20^23 to get the work function for one electron per atom?

Sarah_Wilen wrote:Hope the image of the worked out problem attached helps. I'll walk through my brain jumble:

1) To solve for the KE of the ejected electron, use the equation on the right side of the diagram because this is the amount of KE of the electrons. This equation is E$E_{KE}=\frac{1}{2}m_{electron}\times v_{electron}^{2}$
Now, all you have to do is take your known quantities (velocity, mass of electron constant), then plug and chug. You should get $1.99\times 10^{-19}J$
506*
2) To solve for the energy needed to remove an e- from the metal, remember that this energy is the amount of energy the photon needs to meet or exceed to remove the e-, so wouldn't it make sense for this energy to be the threshold energy (work function)? Yes, it does. You are actually given the threshold energy, but it is in KJ. Convert the quantity into Joules by a unit conversion (1000 Joules in 1 KJ). You should get $1.506\times 10^{5}J$

3) Finally, we need to find the frequency of the incident light. You already know that $E_{photon}-\phi =E_{KE}$ Using this equation, rearrange to solve for the energy of the photon (because we know the threshold and KE of photon already). Plug and chug to solve for the energy. Now that we know the energy, to find the frequency of the light, use the equation $E=h\nu$ Rearrange the equation to solve for frequency. Plug and chug, then you're done! Yippee! :)

https://ibb.co/ciCKXk
This is the image^ Idk how to post it. LMK if the URL fails me

Sarah_Wilen
Posts: 62
Joined: Fri Jun 23, 2017 11:39 am
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### Re: energy required to remove an electron

You are exactly right! I didn't catch that thank you so much. Yes, divide 1.506*10^5 by 6.02*10^23 to get the function for one electron per atom. Then, use that number instead of whatever I wrote down to get the final answer to solve for frequency. Sorry for the confusion! Thanks

Xinye_Jiang_1A wrote:Since the work function is given by J/mol, is it possible that we need to devide 1.506*10^5 by 6.022*20^23 to get the work function for one electron per atom?

Sarah_Wilen wrote:Hope the image of the worked out problem attached helps. I'll walk through my brain jumble:

1) To solve for the KE of the ejected electron, use the equation on the right side of the diagram because this is the amount of KE of the electrons. This equation is E$E_{KE}=\frac{1}{2}m_{electron}\times v_{electron}^{2}$
Now, all you have to do is take your known quantities (velocity, mass of electron constant), then plug and chug. You should get $1.99\times 10^{-19}J$
506*
2) To solve for the energy needed to remove an e- from the metal, remember that this energy is the amount of energy the photon needs to meet or exceed to remove the e-, so wouldn't it make sense for this energy to be the threshold energy (work function)? Yes, it does. You are actually given the threshold energy, but it is in KJ. Convert the quantity into Joules by a unit conversion (1000 Joules in 1 KJ). You should get $1.506\times 10^{5}J$

3) Finally, we need to find the frequency of the incident light. You already know that $E_{photon}-\phi =E_{KE}$ Using this equation, rearrange to solve for the energy of the photon (because we know the threshold and KE of photon already). Plug and chug to solve for the energy. Now that we know the energy, to find the frequency of the light, use the equation $E=h\nu$ Rearrange the equation to solve for frequency. Plug and chug, then you're done! Yippee! :)

https://ibb.co/ciCKXk
This is the image^ Idk how to post it. LMK if the URL fails me

Chem_Mod
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### Re: energy required to remove an electron

A lot of good chemistry here!

Great to see students working collaboratively and helping each other.

taywebb
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Joined: Fri Sep 28, 2018 12:15 am

### Re: energy required to remove an electron

Yeah if you do the full problem out (by dividing Avogadro's number) you end up with 2.501 x 10^-19 J of energy as the work function, which after plugging it in to get the frequency of incident light gets you 6.78 x 10^14 Hz!

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