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NĐGK
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?
kk! haha i stuck on this problem
the formula of wavelength is E=hc/λ??
the question now is how does the wavelength of an electron relate to the energy of light
|dw:1364196013585:dw|
it may related ,as I work on it with out the wavelength it still wrong
where m is the mass of the electron
i think i ll try that now
|dw:1364196120782:dw|
have you tried that @rosewhite
i think we need c of electron
yes, I used E=hc/λ and put c= 3*10^8 m/s
@rosewhite but you did not know E for light
if we use einsteins theory a new parameter comes in
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
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 ?
okay the please try @rosewhite einsteins dual theory
OK , I get C from the law and then get E and from it find lambda but still wrong !! @telijahmed
wow tough one eh @rosewhite sorry for wasting your time i am still working on it
Okey , if you got the answer, plz tell me @telijahmed
ok @rosewhite thanks for assisting
Can you tell me the answer if you got it? ;) thanks very much
anybody knows the answer? I can´t get it :(
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 :-)
@ammanjordan I just checked the 5.4323575*10^-18 and I get a nice red x... so I don't suggest to check it
@mikalika I am Sorry for that :-(
@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
Thank you very much :-)
What is the answer?? I couldn't get! please help me
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
@JK1992 For this way my answer was 3.04*10^-8, and, unfortunately, it's not correct : (
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...
velocity, not speed sorry
@MiniDino try mafeg's way and tell us :)
what would be n_i and n_f?? for rydberg's equation I mean
I tried nf = 3 and ni = 2 and nf = 2 and ni = 1, both answers given red
can you tell me step by step what you are doing? so that I can follow along better, cause I'm confused
@mafeg95 what is your level on chemistry???just curious cause you asked about ni and nf
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 :)
@mafeg95 ok :) then you are fine
anybody got the answer....
This court is really hard TT_TT i'm only 10th grade student
could you try 3.121*10^-9 m ??
3.121*10^-9 is WRONG
can someone try 4.25*10^-12 please?? I only have one subission left..
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.
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?
try 1.8746064 * 10^-6
That question is driving me nuts. Did anybody get the answer to number 9?
no im suck on this one, 9 and number 1 if anyone can help?
no, did u get answer of 9
which number one did you get? tungsten reaction? didnt get number 9 yet.
i havent gotten any of the number 1's
i heard there are different questions for number 1: here's what i got: 0 35.28 52.336 15.38
Yeah, they use different elements. @fiorrr
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:
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...
but can we focus on the question that they asked here?
Everybody is struggling with it...
who has got K2TaF7 with Na?
@mafeg95 I almost give up, because I can't find the answer, I spent 6 times of my submission
has anybody answer of 9?
just opened a thread for #9
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 :)
plz tell us the answer
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! :)
me too, if had the answer, i would give in this situation ....
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*
i tried it more than 40 times to solve..............
there key is in the smallest energy transition
that statement i gave an answer and someone said wrong
its wrong by the red signal
i am not to sure but i used|dw:1364235425854:dw|
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.
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
check out the drawx here @maumay
1.64064*10^-7 is correct!
now does anyone know how to do number 9?
1.64064*10^-7 is correct
anybody has 9 result.......
@maumay, you are a godsend! Thank you sooo much!
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
I T HAS NO RELATIONSHIP TO THE PROBLEM
I DID NOT USE THE RIGHT FORMULA BUT I AM COOL
i opened a threat for number 9
I have different No 9 so I am opening too a new thread...
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.
velocity of 10^3 km/s . not 103
is any one of u got answer of 1st problem (MgO)?????
@GoliS i can walk you through
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
okay now for MGO the kilograms is 0 because all og mgo is used up since the reaction goes to completion
now that means Si is the excess reagent so you need to find out how much of si consumed all of the mgo
no that was wrong ..... :-(
wow some major cheating goin on here