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Need simple chem question answered!

TecHNooB

TecHNooB

Diamond Member
The maximum kinetic energy (in J) of an electron removed from a Li atom by light of 6.51 x 1015 s-1 frequency is:

A. 4.67 x 10-18 Joules.
B. 4.31 x 10-18 Joules.
C. 3.95 x 10-18 Joules.
D. 0.365 x 10-18 Joules.
E. none of these

I tried using bohr's model for H to solve this... either it doesn't work, or I'm not using it right. I'm dumb 🙁
 
Be sure you get all your units right. Convert frequency to wavelength, then get energy joules. If it matches one of the answers that should be it. But.... I'm not sure if the nucleus matters in this case - I'm assuming you're doing freshman chem rather than some advanced p chem or something.

If you're doing advanced stuff you might have to consider the orbitals/energy states available for the three electrons - the first two would be identical (I can't see why they wouldn't be), the third different, so there's only two states to consider. Be warned that this is QED stuff, so it's a looooong way away from my area of expertise and I'm open to correction.
 
Oh this is freshman chem alright. Friggin Bohr's model... never works for me. If someone gets the answer, please leave an explanation. Thnx!
 
hmm ok,

the KE of an ejected electron will equal the energy of the photon (hf) minus the work function for lithium, which, according to google, is 2.9eV.

2.9 eV is equivalent to 4.64e-19 J

so the KE should be

(6.6e-34)(6.51e15) - 4.64e-19 = about 3.9e-18 J

haha, so i guess my first guess was slightly off.
 
I don't know why they make you do this crap. You don't even begin to learn chemistry until you get into organic. I love organic chem even though i hated general chem.
 
Originally posted by: Shawn
I don't know why they make you do this crap. You don't even begin to learn chemistry until you get into organic. I love organic chem even though i hated general chem.
Click to expand...

this is more physics than chem.

but now that i think about it... everything is physics, really.
 
Steven, the answer you got was close, but not close enough. I need something that's just dead on. Perhaps this question will help get things going. I got this wrong as well; same mistake I'm sure.

It takes 495.8 kJ/mol of energy to ionize sodium. What frequency (s-1) and wavelength (nm) of light would be required to ionize sodium?


frequency (s-1) wavelength (nm)
(a) 7.48 x 10^38 241
(b) 1.24 x 10^15 2.41 x 104
(c) 1.24 x 10^15 241
(d) 7.48 x 10^38 3.95 x 10-22
(e) none of the above

The numbers after the exponentials are wavelengths in nm. AT won't add the extra spaces.
 
Originally posted by: TecHNooB
Steven, the answer you got was close, but not close enough. I need something that's just dead on. Perhaps this question will help get things going. I got this wrong as well; same mistake I'm sure.

It takes 495.8 kJ/mol of energy to ionize sodium. What frequency (s-1) and wavelength (nm) of light would be required to ionize sodium?


frequency (s-1) wavelength (nm)
(a) 7.48 x 10^38 241
(b) 1.24 x 10^15 2.41 x 104
(c) 1.24 x 10^15 241
(d) 7.48 x 10^38 3.95 x 10-22
(e) none of the above

The numbers after the exponentials are wavelengths in nm. AT won't add the extra spaces.
Click to expand...

my first instinct tells me this is a trick question. shouldn't the freq*wavelength of light always be equal to 3e8 m/s ?

choice (C) multiplies to be just about 3e8, so my first guess is to go with that

*EDIT* man I wish I still had Powertoy Calculator (doesn't work with Vista 🙁 )
 
I think there is more than one choice that comes close to the speed of light. However, the way I did it the first time through (the wrong way) was using E = hv, and finding wavelength through that.

And yes, the answer is C. Does anyone know why? T___T
 
Originally posted by: TecHNooB
I think there is more than one choice that comes close to the speed of light. However, the way I did it the first time through (the wrong way) was using E = hv, and finding wavelength through that.

And yes, the answer is C. Does anyone know why? T___T
Click to expand...

only choice (C) gives you the right units and numbers for speed of light.

choice (B) gives a speed of light that's too high - 3e10

*EDIT* ahh, i see what you mean, lemme think
 
Originally posted by: StevenYoo
Originally posted by: TecHNooB
I think there is more than one choice that comes close to the speed of light. However, the way I did it the first time through (the wrong way) was using E = hv, and finding wavelength through that.

And yes, the answer is C. Does anyone know why? T___T
Click to expand...

only choice (C) gives you the right units and numbers for speed of light.

choice (B) gives a speed of light that's too high - 3e10

*EDIT* ahh, i see what you mean, lemme think
Click to expand...

😀

 
both (C) and (D) yield the speed of light, but I dunno if there exists electromagnetic radiation of frequency >10^22 Hz

only reasoning I can think of right now
 
Ok, I asked a friend of mine and he worked it out backwards.

(2.18x10^-18) / 6 - hv. That equals the answer exactly. But neither of us know where the 6 came from. It's exactly 6 though.
 
OK. so.

495.8 kJ / mole = 8.23e-19 J per atom. This should equal the energy of the photon required to ionize.

E = hf
8.23e-19 J = (6.63e-34) (f)

f = 1.24e15

this frequency corresponds to choice (C)
 
Originally posted by: StevenYoo
OK. so.

495.8 kJ / mole = 8.23e-19 J per atom. This should equal the energy of the photon required to ionize.

E = hf
8.23e-19 J = (6.63e-34) (f)

f = 1.24e15

this frequency corresponds to choice (C)
Click to expand...

OH YEA. I forgot about that. So this problem doesn't have any correlation with the previous one I posted?
 
Originally posted by: TecHNooB
Originally posted by: StevenYoo
OK. so.

495.8 kJ / mole = 8.23e-19 J per atom. This should equal the energy of the photon required to ionize.

E = hf
8.23e-19 J = (6.63e-34) (f)

f = 1.24e15

this frequency corresponds to choice (C)
Click to expand...

OH YEA. I forgot about that. So this problem doesn't have any correlation with the previous one I posted?
Click to expand...

guess not
 
yeah i was gonna say.. E = hf. or otherwise, you could look at the work function (cant remember if that's the right term) for an electron in lithium 😀
 
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