anonymous
  • anonymous
An electron of wavelength 1.74*10-10m strikes an atom of ionized helium (He+). What is the wavelength (m) of the light corresponding to the line in the emission spectrum with the smallest energy transition?
Chemistry
schrodinger
  • schrodinger
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anonymous
  • anonymous
kk! haha i stuck on this problem
anonymous
  • anonymous
the formula of wavelength is E=hc/λ??
anonymous
  • anonymous
the question now is how does the wavelength of an electron relate to the energy of light

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anonymous
  • anonymous
|dw:1364196013585:dw|
anonymous
  • anonymous
it may related ,as I work on it with out the wavelength it still wrong
anonymous
  • anonymous
.__. this is too hard T^T
anonymous
  • anonymous
where m is the mass of the electron
anonymous
  • anonymous
i think i ll try that now
anonymous
  • anonymous
|dw:1364196120782:dw|
anonymous
  • anonymous
have you tried that @rosewhite
anonymous
  • anonymous
i think we need c of electron
anonymous
  • anonymous
yes, I used E=hc/λ and put c= 3*10^8 m/s
anonymous
  • anonymous
@rosewhite but you did not know E for light
anonymous
  • anonymous
if we use einsteins theory a new parameter comes in
anonymous
  • anonymous
which is the mass
anonymous
  • anonymous
however if we use rhydbergs formula then we need to the VALUES OF n which for me should be 2 and 1 since it is helium
anonymous
  • anonymous
I think he said the lowest energy we can get that from the law E=-(Z^2/n^2)(me^2/8h^2E0^2) isnot it ?
anonymous
  • anonymous
okay the please try @rosewhite einsteins dual theory
anonymous
  • anonymous
anonymous
  • anonymous
OK , I get C from the law and then get E and from it find lambda but still wrong !! @telijahmed
anonymous
  • anonymous
wow tough one eh @rosewhite sorry for wasting your time i am still working on it
anonymous
  • anonymous
Okey , if you got the answer, plz tell me @telijahmed
anonymous
  • anonymous
ok @rosewhite thanks for assisting
anonymous
  • anonymous
Can you tell me the answer if you got it? ;) thanks very much
anonymous
  • anonymous
anybody knows the answer? I can´t get it :(
anonymous
  • anonymous
I got an answer of 5,4323575*10^(-18) but its my last chance and i would like to cross-check my result before check it out, so if anyone else try it and take a result just say it :-)
anonymous
  • anonymous
@ammanjordan I just checked the 5.4323575*10^-18 and I get a nice red x... so I don't suggest to check it
anonymous
  • anonymous
@mikalika I am Sorry for that :-(
anonymous
  • anonymous
@ammanjordan don't worry it was just my 2nd check. I just wanted to let you know in order to not loose your final check
anonymous
  • anonymous
Thank you very much :-)
anonymous
  • anonymous
What is the answer?? I couldn't get! please help me
anonymous
  • anonymous
well, as i see it, i thought we could use E=h*c/λ so that we find the energy of the electron and then apply this to the rydberg's equation and solve for λ
anonymous
  • anonymous
@JK1992 For this way my answer was 3.04*10^-8, and, unfortunately, it's not correct : (
anonymous
  • anonymous
what about using lambda = h/mv, and then using v instead of c? because c is for the speed of light, not for the speed of the electron...
anonymous
  • anonymous
velocity, not speed sorry
anonymous
  • anonymous
@MiniDino try mafeg's way and tell us :)
anonymous
  • anonymous
what would be n_i and n_f?? for rydberg's equation I mean
anonymous
  • anonymous
I tried nf = 3 and ni = 2 and nf = 2 and ni = 1, both answers given red
anonymous
  • anonymous
can you tell me step by step what you are doing? so that I can follow along better, cause I'm confused
anonymous
  • anonymous
@mafeg95 what is your level on chemistry???just curious cause you asked about ni and nf
anonymous
  • anonymous
I'm a junior in high school, and I'm taking AP chemistry, but I never fully got the spectral series, and we saw them on the first quarter :)
anonymous
  • anonymous
@mafeg95 ok :) then you are fine
anonymous
  • anonymous
anybody got the answer....
anonymous
  • anonymous
This court is really hard TT_TT i'm only 10th grade student
anonymous
  • anonymous
could you try 3.121*10^-9 m ??
anonymous
  • anonymous
3.121*10^-9 is WRONG
anonymous
  • anonymous
can someone try 4.25*10^-12 please?? I only have one subission left..
anonymous
  • anonymous
submission*
anonymous
  • anonymous
4.25*10^-12 is wrong
anonymous
  • anonymous
wrong
anonymous
  • anonymous
Wrong
anonymous
  • anonymous
arggggg! sorry, this is really stressful! If anyone is interested, I also asked this question in the physics section, if you guys want to follow that post too.
anonymous
  • anonymous
Im still working on this problem and can not get it! Did anyone get the 1st one about the antimony reaction that could help me?
anonymous
  • anonymous
try 1.8746064 * 10^-6
anonymous
  • anonymous
WRONG
anonymous
  • anonymous
wrong
anonymous
  • anonymous
wrong
anonymous
  • anonymous
That question is driving me nuts. Did anybody get the answer to number 9?
anonymous
  • anonymous
no im suck on this one, 9 and number 1 if anyone can help?
anonymous
  • anonymous
no, did u get answer of 9
anonymous
  • anonymous
which number one did you get? tungsten reaction? didnt get number 9 yet.
anonymous
  • anonymous
i havent gotten any of the number 1's
anonymous
  • anonymous
i heard there are different questions for number 1: here's what i got: 0 35.28 52.336 15.38
anonymous
  • anonymous
Yeah, they use different elements. @fiorrr
anonymous
  • anonymous
oh i didnt know that i got this one Antimony (Sb) is produced by reacting Antimony Sulfide (Sb2S3) with elemental Iron (Fe). The by-product is Iron Sulfide (FeS). 67.0 kg of Fe is reacted 1.0 kg Sb2S3. Assume the reaction goes to completion. After the the reaction has reached completion, the reactor contents at the completion of this reaction will be the following: Enter the number of kg of Sb2S3 remaining: Enter the number of kg of Fe remaining: Enter the number of kg of FeS present: Enter the number of kg of Sb present:
anonymous
  • anonymous
Personally, I have magnesium... and I am struggling with No 9 for so many hrs that I don't think that I will have the time to go back to the No1...
anonymous
  • anonymous
but can we focus on the question that they asked here?
anonymous
  • anonymous
Everybody is struggling with it...
anonymous
  • anonymous
who has got K2TaF7 with Na?
anonymous
  • anonymous
@mafeg95 I almost give up, because I can't find the answer, I spent 6 times of my submission
anonymous
  • anonymous
has anybody answer of 9?
anonymous
  • anonymous
just opened a thread for #9
anonymous
  • anonymous
I know me too @MiniDino... But in about 30 minutes, I'm going to go to a chemistry teacher in my school to see if she can help me figure it out, and I'll post my procedure if get it right :)
anonymous
  • anonymous
plz tell us the answer
anonymous
  • anonymous
well, personally, I'm against of sharing answers, but since time is short, I'll make an exception and share the answer as soon as I see the precious green check! :)
anonymous
  • anonymous
me too, if had the answer, i would give in this situation ....
anonymous
  • anonymous
I can't promise you guys that I'll get it right though... This is the 5th chemistry teacher that I've asked this question, and one of them had a PhD in chemistry... So it's a pretty hard question * sigh*
anonymous
  • anonymous
i tried it more than 40 times to solve..............
anonymous
  • anonymous
there key is in the smallest energy transition
anonymous
  • anonymous
that statement i gave an answer and someone said wrong
anonymous
  • anonymous
which one?
anonymous
  • anonymous
1.8746064 * 10^-6
anonymous
  • anonymous
its wrong by the red signal
anonymous
  • anonymous
i am not to sure but i used|dw:1364235425854:dw|
anonymous
  • anonymous
Hello, guys. The final formula to solve this problem is the following: K * z^2 * (1/n_f^2 - n_i^2) = c*h/l where K = 2.17*10^-18 J (ground state energy) z = 2 (helium) n_i, n_f - initial and final states. They're asking for smallest energy transition. That would be 2->3 c = 3*10^8 m/s h = 6.62*10^-34 Js Solving for l: 2.1789*10^-18 * 2^2 * (1/2^2 - 1/3^2) = (6.62*10^-34 * 3e8)/l We're having nice little photon flying out with wavelength of 1.64064*10^-7 m.
anonymous
  • anonymous
i used this other method but i dont like the algebra here so much so maybe that is why it is wrong let me show you what i did
anonymous
  • anonymous
check out the drawx here @maumay
anonymous
  • anonymous
1.64064*10^-7 is correct!
anonymous
  • anonymous
now does anyone know how to do number 9?
anonymous
  • anonymous
1.64064*10^-7 is correct
anonymous
  • anonymous
anybody has 9 result.......
anonymous
  • anonymous
@maumay, you are a godsend! Thank you sooo much!
anonymous
  • anonymous
we think alike @maumay i thought that the wavelength for the ELECTRON WAS A DECOY AND I DID NOT USE IT IN MY MOST SURE ATTEMPT
anonymous
  • anonymous
I T HAS NO RELATIONSHIP TO THE PROBLEM
anonymous
  • anonymous
I DID NOT USE THE RIGHT FORMULA BUT I AM COOL
anonymous
  • anonymous
9 plz?
anonymous
  • anonymous
i opened a threat for number 9
anonymous
  • anonymous
get yet?
anonymous
  • anonymous
I have different No 9 so I am opening too a new thread...
anonymous
  • anonymous
A lithium ion with a single orbiting electron is subjected to ultraviolet radiation (λ = 3.711e-8 m). On interaction of this radiation with the lithium ion, the radiative energy is transferred to the electron, which is ejected from orbit as a result. As a free electron, it exhibits a velocity of 103 km/s. Given this information, determine the principal quantum number (n) of the electron in lithium prior to its interaction with the UV radiation.
anonymous
  • anonymous
who has it?
anonymous
  • anonymous
velocity of 10^3 km/s . not 103
anonymous
  • anonymous
is any one of u got answer of 1st problem (MgO)?????
anonymous
  • anonymous
we have 7 min. only
anonymous
  • anonymous
Thanks so much
anonymous
  • anonymous
@GoliS i can walk you through
anonymous
  • anonymous
Magnesium can be produced by reaction of magnesium oxide (MgO) with silicon (Si). The by-product of the reaction is magnesium orthosilicate (Mg2SiO4). A reactor is charged with 260.0 kg of MgO and 170.0 kg Si
anonymous
  • anonymous
okay now for MGO the kilograms is 0 because all og mgo is used up since the reaction goes to completion
anonymous
  • anonymous
yupp
anonymous
  • anonymous
now that means Si is the excess reagent so you need to find out how much of si consumed all of the mgo
anonymous
  • anonymous
no that was wrong ..... :-(
anonymous
  • anonymous
timed out
anonymous
  • anonymous
i dont understand
aaronq
  • aaronq
wow some major cheating goin on here

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