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anonymous
 3 years 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?
anonymous
 3 years 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?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0kk! haha i stuck on this problem

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the formula of wavelength is E=hc/λ??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the question now is how does the wavelength of an electron relate to the energy of light

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1364196013585:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it may related ,as I work on it with out the wavelength it still wrong

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0.__. this is too hard T^T

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0where m is the mass of the electron

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think i ll try that now

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1364196120782:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0have you tried that @rosewhite

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think we need c of electron

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@rosewhite but you did not know E for light

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if we use einsteins theory a new parameter comes in

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0however 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
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.0okay the please try @rosewhite einsteins dual theory

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wow tough one eh @rosewhite sorry for wasting your time i am still working on it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okey , if you got the answer, plz tell me @telijahmed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok @rosewhite thanks for assisting

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Can you tell me the answer if you got it? ;) thanks very much

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0anybody knows the answer? I can´t get it :(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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 :)

anonymous
 3 years ago
Best 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@mikalika I am Sorry for that :(

anonymous
 3 years ago
Best 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you very much :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What is the answer?? I couldn't get! please help me

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well, 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
 3 years ago
Best 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 : (

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what 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
 3 years ago
Best ResponseYou've already chosen the best response.0velocity, not speed sorry

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@MiniDino try mafeg's way and tell us :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what would be n_i and n_f?? for rydberg's equation I mean

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'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
 3 years ago
Best ResponseYou've already chosen the best response.0@mafeg95 ok :) then you are fine

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0anybody got the answer....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This court is really hard TT_TT i'm only 10th grade student

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0could you try 3.121*10^9 m ??

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0arggggg! 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
 3 years ago
Best ResponseYou've already chosen the best response.0Im 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
 3 years ago
Best ResponseYou've already chosen the best response.0try 1.8746064 * 10^6

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That question is driving me nuts. Did anybody get the answer to number 9?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no, did u get answer of 9

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0which number one did you get? tungsten reaction? didnt get number 9 yet.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i havent gotten any of the number 1's

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, they use different elements. @fiorrr

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh 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:

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Personally, 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
 3 years ago
Best ResponseYou've already chosen the best response.0but can we focus on the question that they asked here?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Everybody is struggling with it...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0who has got K2TaF7 with Na?

anonymous
 3 years ago
Best 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0has anybody answer of 9?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0just opened a thread for #9

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.0plz tell us the answer

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well, 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
 3 years ago
Best ResponseYou've already chosen the best response.0me too, if had the answer, i would give in this situation ....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.0i tried it more than 40 times to solve..............

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0there key is in the smallest energy transition

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that statement i gave an answer and someone said wrong

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0its wrong by the red signal

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i am not to sure but i useddw:1364235425854:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hello, 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
 3 years ago
Best ResponseYou've already chosen the best response.0i 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
 3 years ago
Best ResponseYou've already chosen the best response.0check out the drawx here @maumay

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.01.64064*10^7 is correct!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now does anyone know how to do number 9?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.01.64064*10^7 is correct

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0anybody has 9 result.......

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@maumay, you are a godsend! Thank you sooo much!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we 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
 3 years ago
Best ResponseYou've already chosen the best response.0I T HAS NO RELATIONSHIP TO THE PROBLEM

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I DID NOT USE THE RIGHT FORMULA BUT I AM COOL

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i opened a threat for number 9

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I have different No 9 so I am opening too a new thread...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0A 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0velocity of 10^3 km/s . not 103

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@GoliS i can walk you through

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Magnesium 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay now for MGO the kilograms is 0 because all og mgo is used up since the reaction goes to completion

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now that means Si is the excess reagent so you need to find out how much of si consumed all of the mgo

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no that was wrong ..... :(

aaronq
 3 years ago
Best ResponseYou've already chosen the best response.1wow some major cheating goin on here
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