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An electron of wavelength 1.74*1010m 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?
 one year ago
 one year ago
An electron of wavelength 1.74*1010m 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?
 one year ago
 one year ago

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NĐGKBest ResponseYou've already chosen the best response.0
kk! haha i stuck on this problem
 one year ago

NĐGKBest ResponseYou've already chosen the best response.0
the formula of wavelength is E=hc/λ??
 one year ago

telijahmedBest ResponseYou've already chosen the best response.2
the question now is how does the wavelength of an electron relate to the energy of light
 one year ago

telijahmedBest ResponseYou've already chosen the best response.2
dw:1364196013585:dw
 one year ago

rosewhiteBest ResponseYou've already chosen the best response.0
it may related ,as I work on it with out the wavelength it still wrong
 one year ago

NĐGKBest ResponseYou've already chosen the best response.0
.__. this is too hard T^T
 one year ago

telijahmedBest ResponseYou've already chosen the best response.2
where m is the mass of the electron
 one year ago

telijahmedBest ResponseYou've already chosen the best response.2
i think i ll try that now
 one year ago

telijahmedBest ResponseYou've already chosen the best response.2
dw:1364196120782:dw
 one year ago

telijahmedBest ResponseYou've already chosen the best response.2
have you tried that @rosewhite
 one year ago

rosewhiteBest ResponseYou've already chosen the best response.0
i think we need c of electron
 one year ago

rosewhiteBest ResponseYou've already chosen the best response.0
yes, I used E=hc/λ and put c= 3*10^8 m/s
 one year ago

telijahmedBest ResponseYou've already chosen the best response.2
@rosewhite but you did not know E for light
 one year ago

telijahmedBest ResponseYou've already chosen the best response.2
if we use einsteins theory a new parameter comes in
 one year ago

telijahmedBest ResponseYou've already chosen the best response.2
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
 one year ago

rosewhiteBest ResponseYou've already chosen the best response.0
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 ?
 one year ago

telijahmedBest ResponseYou've already chosen the best response.2
okay the please try @rosewhite einsteins dual theory
 one year ago

rosewhiteBest ResponseYou've already chosen the best response.0
OK , I get C from the law and then get E and from it find lambda but still wrong !! @telijahmed
 one year ago

telijahmedBest ResponseYou've already chosen the best response.2
wow tough one eh @rosewhite sorry for wasting your time i am still working on it
 one year ago

rosewhiteBest ResponseYou've already chosen the best response.0
Okey , if you got the answer, plz tell me @telijahmed
 one year ago

telijahmedBest ResponseYou've already chosen the best response.2
ok @rosewhite thanks for assisting
 one year ago

NĐGKBest ResponseYou've already chosen the best response.0
Can you tell me the answer if you got it? ;) thanks very much
 one year ago

Rainbow1980Best ResponseYou've already chosen the best response.0
anybody knows the answer? I can´t get it :(
 one year ago

ammanjordanBest ResponseYou've already chosen the best response.0
I got an answer of 5,4323575*10^(18) but its my last chance and i would like to crosscheck my result before check it out, so if anyone else try it and take a result just say it :)
 one year ago

mikalikaBest ResponseYou've already chosen the best response.0
@ammanjordan I just checked the 5.4323575*10^18 and I get a nice red x... so I don't suggest to check it
 one year ago

ammanjordanBest ResponseYou've already chosen the best response.0
@mikalika I am Sorry for that :(
 one year ago

mikalikaBest ResponseYou've already chosen the best response.0
@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
 one year ago

ammanjordanBest ResponseYou've already chosen the best response.0
Thank you very much :)
 one year ago

muratcankalemBest ResponseYou've already chosen the best response.0
What is the answer?? I couldn't get! please help me
 one year ago

JK1992Best ResponseYou've already chosen the best response.0
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
 one year ago

MiniDinoBest ResponseYou've already chosen the best response.0
@JK1992 For this way my answer was 3.04*10^8, and, unfortunately, it's not correct : (
 one year ago

mafeg95Best ResponseYou've already chosen the best response.0
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...
 one year ago

mafeg95Best ResponseYou've already chosen the best response.0
velocity, not speed sorry
 one year ago

JK1992Best ResponseYou've already chosen the best response.0
@MiniDino try mafeg's way and tell us :)
 one year ago

mafeg95Best ResponseYou've already chosen the best response.0
what would be n_i and n_f?? for rydberg's equation I mean
 one year ago

MiniDinoBest ResponseYou've already chosen the best response.0
I tried nf = 3 and ni = 2 and nf = 2 and ni = 1, both answers given red
 one year ago

mafeg95Best ResponseYou've already chosen the best response.0
can you tell me step by step what you are doing? so that I can follow along better, cause I'm confused
 one year ago

JK1992Best ResponseYou've already chosen the best response.0
@mafeg95 what is your level on chemistry???just curious cause you asked about ni and nf
 one year ago

mafeg95Best ResponseYou've already chosen the best response.0
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 :)
 one year ago

JK1992Best ResponseYou've already chosen the best response.0
@mafeg95 ok :) then you are fine
 one year ago

nayon333Best ResponseYou've already chosen the best response.0
anybody got the answer....
 one year ago

NĐGKBest ResponseYou've already chosen the best response.0
This court is really hard TT_TT i'm only 10th grade student
 one year ago

JK1992Best ResponseYou've already chosen the best response.0
could you try 3.121*10^9 m ??
 one year ago

muratcankalemBest ResponseYou've already chosen the best response.0
3.121*10^9 is WRONG
 one year ago

mafeg95Best ResponseYou've already chosen the best response.0
can someone try 4.25*10^12 please?? I only have one subission left..
 one year ago

mafeg95Best ResponseYou've already chosen the best response.0
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.
 one year ago

jamesmaukBest ResponseYou've already chosen the best response.0
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?
 one year ago

telijahmedBest ResponseYou've already chosen the best response.2
try 1.8746064 * 10^6
 one year ago

fiorrrBest ResponseYou've already chosen the best response.0
That question is driving me nuts. Did anybody get the answer to number 9?
 one year ago

jamesmaukBest ResponseYou've already chosen the best response.0
no im suck on this one, 9 and number 1 if anyone can help?
 one year ago

nayon333Best ResponseYou've already chosen the best response.0
no, did u get answer of 9
 one year ago

fiorrrBest ResponseYou've already chosen the best response.0
which number one did you get? tungsten reaction? didnt get number 9 yet.
 one year ago

jamesmaukBest ResponseYou've already chosen the best response.0
i havent gotten any of the number 1's
 one year ago

fiorrrBest ResponseYou've already chosen the best response.0
i heard there are different questions for number 1: here's what i got: 0 35.28 52.336 15.38
 one year ago

mafeg95Best ResponseYou've already chosen the best response.0
Yeah, they use different elements. @fiorrr
 one year ago

jamesmaukBest ResponseYou've already chosen the best response.0
oh i didnt know that i got this one Antimony (Sb) is produced by reacting Antimony Sulfide (Sb2S3) with elemental Iron (Fe). The byproduct 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:
 one year ago

mikalikaBest ResponseYou've already chosen the best response.0
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...
 one year ago

mafeg95Best ResponseYou've already chosen the best response.0
but can we focus on the question that they asked here?
 one year ago

mafeg95Best ResponseYou've already chosen the best response.0
Everybody is struggling with it...
 one year ago

nayon333Best ResponseYou've already chosen the best response.0
who has got K2TaF7 with Na?
 one year ago

MiniDinoBest ResponseYou've already chosen the best response.0
@mafeg95 I almost give up, because I can't find the answer, I spent 6 times of my submission
 one year ago

nayon333Best ResponseYou've already chosen the best response.0
has anybody answer of 9?
 one year ago

fiorrrBest ResponseYou've already chosen the best response.0
just opened a thread for #9
 one year ago

mafeg95Best ResponseYou've already chosen the best response.0
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 :)
 one year ago

nayon333Best ResponseYou've already chosen the best response.0
plz tell us the answer
 one year ago

mafeg95Best ResponseYou've already chosen the best response.0
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! :)
 one year ago

nayon333Best ResponseYou've already chosen the best response.0
me too, if had the answer, i would give in this situation ....
 one year ago

mafeg95Best ResponseYou've already chosen the best response.0
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*
 one year ago

nayon333Best ResponseYou've already chosen the best response.0
i tried it more than 40 times to solve..............
 one year ago

telijahmedBest ResponseYou've already chosen the best response.2
there key is in the smallest energy transition
 one year ago

telijahmedBest ResponseYou've already chosen the best response.2
that statement i gave an answer and someone said wrong
 one year ago

nayon333Best ResponseYou've already chosen the best response.0
its wrong by the red signal
 one year ago

telijahmedBest ResponseYou've already chosen the best response.2
i am not to sure but i useddw:1364235425854:dw
 one year ago

maumayBest ResponseYou've already chosen the best response.6
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.
 one year ago

telijahmedBest ResponseYou've already chosen the best response.2
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
 one year ago

telijahmedBest ResponseYou've already chosen the best response.2
check out the drawx here @maumay
 one year ago

fiorrrBest ResponseYou've already chosen the best response.0
1.64064*10^7 is correct!
 one year ago

jamesmaukBest ResponseYou've already chosen the best response.0
now does anyone know how to do number 9?
 one year ago

nayon333Best ResponseYou've already chosen the best response.0
1.64064*10^7 is correct
 one year ago

nayon333Best ResponseYou've already chosen the best response.0
anybody has 9 result.......
 one year ago

mafeg95Best ResponseYou've already chosen the best response.0
@maumay, you are a godsend! Thank you sooo much!
 one year ago

telijahmedBest ResponseYou've already chosen the best response.2
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
 one year ago

telijahmedBest ResponseYou've already chosen the best response.2
I T HAS NO RELATIONSHIP TO THE PROBLEM
 one year ago

telijahmedBest ResponseYou've already chosen the best response.2
I DID NOT USE THE RIGHT FORMULA BUT I AM COOL
 one year ago

jamesmaukBest ResponseYou've already chosen the best response.0
i opened a threat for number 9
 one year ago

mikalikaBest ResponseYou've already chosen the best response.0
I have different No 9 so I am opening too a new thread...
 one year ago

nayon333Best ResponseYou've already chosen the best response.0
A lithium ion with a single orbiting electron is subjected to ultraviolet radiation (λ = 3.711e8 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.
 one year ago

nayon333Best ResponseYou've already chosen the best response.0
velocity of 10^3 km/s . not 103
 one year ago

GoliSBest ResponseYou've already chosen the best response.0
is any one of u got answer of 1st problem (MgO)?????
 one year ago

telijahmedBest ResponseYou've already chosen the best response.2
@GoliS i can walk you through
 one year ago

GoliSBest ResponseYou've already chosen the best response.0
Magnesium can be produced by reaction of magnesium oxide (MgO) with silicon (Si). The byproduct of the reaction is magnesium orthosilicate (Mg2SiO4). A reactor is charged with 260.0 kg of MgO and 170.0 kg Si
 one year ago

telijahmedBest ResponseYou've already chosen the best response.2
okay now for MGO the kilograms is 0 because all og mgo is used up since the reaction goes to completion
 one year ago

telijahmedBest ResponseYou've already chosen the best response.2
now that means Si is the excess reagent so you need to find out how much of si consumed all of the mgo
 one year ago

GoliSBest ResponseYou've already chosen the best response.0
no that was wrong ..... :(
 one year ago

aaronqBest ResponseYou've already chosen the best response.1
wow some major cheating goin on here
 one year ago
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