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NĐGK
 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?
NĐGK
 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?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

telijahmed
 one year ago
Best ResponseYou've already chosen the best response.2however 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

rosewhite
 one year 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 ?

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

rosewhite
 one year 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

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

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

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

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

Rainbow1980
 one year ago
Best ResponseYou've already chosen the best response.0anybody knows the answer? I can´t get it :(

ammanjordan
 one year 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 :)

mikalika
 one year 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

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

mikalika
 one year 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

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

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

JK1992
 one year 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

MiniDino
 one year 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 : (

mafeg95
 one year 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...

mafeg95
 one year ago
Best ResponseYou've already chosen the best response.0velocity, not speed sorry

JK1992
 one year ago
Best ResponseYou've already chosen the best response.0@MiniDino try mafeg's way and tell us :)

mafeg95
 one year ago
Best ResponseYou've already chosen the best response.0what would be n_i and n_f?? for rydberg's equation I mean

MiniDino
 one year 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

mafeg95
 one year 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

JK1992
 one year 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

mafeg95
 one year 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 :)

JK1992
 one year ago
Best ResponseYou've already chosen the best response.0@mafeg95 ok :) then you are fine

nayon333
 one year ago
Best ResponseYou've already chosen the best response.0anybody got the answer....

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

JK1992
 one year ago
Best ResponseYou've already chosen the best response.0could you try 3.121*10^9 m ??

muratcankalem
 one year ago
Best ResponseYou've already chosen the best response.03.121*10^9 is WRONG

mafeg95
 one year ago
Best ResponseYou've already chosen the best response.0can someone try 4.25*10^12 please?? I only have one subission left..

mafeg95
 one year 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.

jamesmauk
 one year 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?

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

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

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

nayon333
 one year ago
Best ResponseYou've already chosen the best response.0no, did u get answer of 9

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

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

fiorrr
 one year 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

mafeg95
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, they use different elements. @fiorrr

jamesmauk
 one year 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:

mikalika
 one year 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...

mafeg95
 one year ago
Best ResponseYou've already chosen the best response.0but can we focus on the question that they asked here?

mafeg95
 one year ago
Best ResponseYou've already chosen the best response.0Everybody is struggling with it...

nayon333
 one year ago
Best ResponseYou've already chosen the best response.0who has got K2TaF7 with Na?

MiniDino
 one year 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

nayon333
 one year ago
Best ResponseYou've already chosen the best response.0has anybody answer of 9?

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

mafeg95
 one year 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 :)

nayon333
 one year ago
Best ResponseYou've already chosen the best response.0plz tell us the answer

mafeg95
 one year 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! :)

nayon333
 one year ago
Best ResponseYou've already chosen the best response.0me too, if had the answer, i would give in this situation ....

mafeg95
 one year 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*

nayon333
 one year ago
Best ResponseYou've already chosen the best response.0i tried it more than 40 times to solve..............

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

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

nayon333
 one year ago
Best ResponseYou've already chosen the best response.0its wrong by the red signal

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

maumay
 one year ago
Best ResponseYou've already chosen the best response.6Hello, 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.

telijahmed
 one year ago
Best ResponseYou've already chosen the best response.2i 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

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

fiorrr
 one year ago
Best ResponseYou've already chosen the best response.01.64064*10^7 is correct!

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

nayon333
 one year ago
Best ResponseYou've already chosen the best response.01.64064*10^7 is correct

nayon333
 one year ago
Best ResponseYou've already chosen the best response.0anybody has 9 result.......

mafeg95
 one year ago
Best ResponseYou've already chosen the best response.0@maumay, you are a godsend! Thank you sooo much!

telijahmed
 one year ago
Best ResponseYou've already chosen the best response.2we 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

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

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

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

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

nayon333
 one year 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.

nayon333
 one year ago
Best ResponseYou've already chosen the best response.0velocity of 10^3 km/s . not 103

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

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

GoliS
 one year 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

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

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

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

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