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NĐGK

  • one year ago

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?

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  1. NĐGK
    • one year ago
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    kk! haha i stuck on this problem

  2. NĐGK
    • one year ago
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    the formula of wavelength is E=hc/λ??

  3. telijahmed
    • one year ago
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    the question now is how does the wavelength of an electron relate to the energy of light

  4. telijahmed
    • one year ago
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    |dw:1364196013585:dw|

  5. rosewhite
    • one year ago
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    it may related ,as I work on it with out the wavelength it still wrong

  6. NĐGK
    • one year ago
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    .__. this is too hard T^T

  7. telijahmed
    • one year ago
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    where m is the mass of the electron

  8. telijahmed
    • one year ago
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    i think i ll try that now

  9. telijahmed
    • one year ago
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    |dw:1364196120782:dw|

  10. telijahmed
    • one year ago
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    have you tried that @rosewhite

  11. rosewhite
    • one year ago
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    i think we need c of electron

  12. rosewhite
    • one year ago
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    yes, I used E=hc/λ and put c= 3*10^8 m/s

  13. telijahmed
    • one year ago
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    @rosewhite but you did not know E for light

  14. telijahmed
    • one year ago
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    if we use einsteins theory a new parameter comes in

  15. telijahmed
    • one year ago
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    which is the mass

  16. telijahmed
    • one year ago
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    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

  17. rosewhite
    • one year ago
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    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 ?

  18. telijahmed
    • one year ago
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    okay the please try @rosewhite einsteins dual theory

  19. rosewhite
    • one year ago
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    how? @telijahmed

  20. rosewhite
    • one year ago
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    OK , I get C from the law and then get E and from it find lambda but still wrong !! @telijahmed

  21. telijahmed
    • one year ago
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    wow tough one eh @rosewhite sorry for wasting your time i am still working on it

  22. rosewhite
    • one year ago
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    Okey , if you got the answer, plz tell me @telijahmed

  23. telijahmed
    • one year ago
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    ok @rosewhite thanks for assisting

  24. NĐGK
    • one year ago
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    Can you tell me the answer if you got it? ;) thanks very much

  25. Rainbow1980
    • one year ago
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    anybody knows the answer? I can´t get it :(

  26. ammanjordan
    • one year ago
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    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 :-)

  27. mikalika
    • one year ago
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    @ammanjordan I just checked the 5.4323575*10^-18 and I get a nice red x... so I don't suggest to check it

  28. ammanjordan
    • one year ago
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    @mikalika I am Sorry for that :-(

  29. mikalika
    • one year ago
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    @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

  30. ammanjordan
    • one year ago
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    Thank you very much :-)

  31. muratcankalem
    • one year ago
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    What is the answer?? I couldn't get! please help me

  32. JK1992
    • one year ago
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    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

  33. MiniDino
    • one year ago
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    @JK1992 For this way my answer was 3.04*10^-8, and, unfortunately, it's not correct : (

  34. mafeg95
    • one year ago
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    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...

  35. mafeg95
    • one year ago
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    velocity, not speed sorry

  36. JK1992
    • one year ago
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    @MiniDino try mafeg's way and tell us :)

  37. mafeg95
    • one year ago
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    what would be n_i and n_f?? for rydberg's equation I mean

  38. MiniDino
    • one year ago
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    I tried nf = 3 and ni = 2 and nf = 2 and ni = 1, both answers given red

  39. mafeg95
    • one year ago
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    can you tell me step by step what you are doing? so that I can follow along better, cause I'm confused

  40. JK1992
    • one year ago
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    @mafeg95 what is your level on chemistry???just curious cause you asked about ni and nf

  41. mafeg95
    • one year ago
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    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 :)

  42. JK1992
    • one year ago
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    @mafeg95 ok :) then you are fine

  43. nayon333
    • one year ago
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    anybody got the answer....

  44. NĐGK
    • one year ago
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    This court is really hard TT_TT i'm only 10th grade student

  45. JK1992
    • one year ago
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    could you try 3.121*10^-9 m ??

  46. muratcankalem
    • one year ago
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    3.121*10^-9 is WRONG

  47. mafeg95
    • one year ago
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    can someone try 4.25*10^-12 please?? I only have one subission left..

  48. mafeg95
    • one year ago
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    submission*

  49. sodie69
    • one year ago
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    4.25*10^-12 is wrong

  50. Rainbow1980
    • one year ago
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    wrong

  51. monkeynero
    • one year ago
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    Wrong

  52. mafeg95
    • one year ago
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    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.

  53. jamesmauk
    • one year ago
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    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?

  54. telijahmed
    • one year ago
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    try 1.8746064 * 10^-6

  55. muratcankalem
    • one year ago
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    WRONG

  56. nayon333
    • one year ago
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    wrong

  57. jamesmauk
    • one year ago
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    wrong

  58. fiorrr
    • one year ago
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    That question is driving me nuts. Did anybody get the answer to number 9?

  59. jamesmauk
    • one year ago
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    no im suck on this one, 9 and number 1 if anyone can help?

  60. nayon333
    • one year ago
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    no, did u get answer of 9

  61. fiorrr
    • one year ago
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    which number one did you get? tungsten reaction? didnt get number 9 yet.

  62. jamesmauk
    • one year ago
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    i havent gotten any of the number 1's

  63. fiorrr
    • one year ago
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    i heard there are different questions for number 1: here's what i got: 0 35.28 52.336 15.38

  64. mafeg95
    • one year ago
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    Yeah, they use different elements. @fiorrr

  65. jamesmauk
    • one year ago
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    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:

  66. mikalika
    • one year ago
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    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...

  67. mafeg95
    • one year ago
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    but can we focus on the question that they asked here?

  68. mafeg95
    • one year ago
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    Everybody is struggling with it...

  69. nayon333
    • one year ago
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    who has got K2TaF7 with Na?

  70. MiniDino
    • one year ago
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    @mafeg95 I almost give up, because I can't find the answer, I spent 6 times of my submission

  71. nayon333
    • one year ago
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    has anybody answer of 9?

  72. fiorrr
    • one year ago
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    just opened a thread for #9

  73. mafeg95
    • one year ago
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    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 :)

  74. nayon333
    • one year ago
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    plz tell us the answer

  75. mafeg95
    • one year ago
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    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! :)

  76. nayon333
    • one year ago
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    me too, if had the answer, i would give in this situation ....

  77. mafeg95
    • one year ago
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    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*

  78. nayon333
    • one year ago
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    i tried it more than 40 times to solve..............

  79. telijahmed
    • one year ago
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    there key is in the smallest energy transition

  80. telijahmed
    • one year ago
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    that statement i gave an answer and someone said wrong

  81. nayon333
    • one year ago
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    which one?

  82. telijahmed
    • one year ago
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    1.8746064 * 10^-6

  83. nayon333
    • one year ago
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    its wrong by the red signal

  84. telijahmed
    • one year ago
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    i am not to sure but i used|dw:1364235425854:dw|

  85. maumay
    • one year ago
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    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.

  86. telijahmed
    • one year ago
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    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

  87. telijahmed
    • one year ago
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    check out the drawx here @maumay

  88. fiorrr
    • one year ago
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    1.64064*10^-7 is correct!

  89. jamesmauk
    • one year ago
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    now does anyone know how to do number 9?

  90. nayon333
    • one year ago
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    1.64064*10^-7 is correct

  91. nayon333
    • one year ago
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    anybody has 9 result.......

  92. mafeg95
    • one year ago
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    @maumay, you are a godsend! Thank you sooo much!

  93. telijahmed
    • one year ago
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    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

  94. telijahmed
    • one year ago
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    I T HAS NO RELATIONSHIP TO THE PROBLEM

  95. telijahmed
    • one year ago
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    I DID NOT USE THE RIGHT FORMULA BUT I AM COOL

  96. nayon333
    • one year ago
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    9 plz?

  97. jamesmauk
    • one year ago
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    i opened a threat for number 9

  98. nayon333
    • one year ago
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    get yet?

  99. mikalika
    • one year ago
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    I have different No 9 so I am opening too a new thread...

  100. nayon333
    • one year ago
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    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.

  101. nayon333
    • one year ago
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    who has it?

  102. nayon333
    • one year ago
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    velocity of 10^3 km/s . not 103

  103. GoliS
    • one year ago
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    is any one of u got answer of 1st problem (MgO)?????

  104. nayon333
    • one year ago
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    we have 7 min. only

  105. NĐGK
    • one year ago
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    Thanks so much

  106. telijahmed
    • one year ago
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    @GoliS i can walk you through

  107. GoliS
    • one year ago
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    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

  108. telijahmed
    • one year ago
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    okay now for MGO the kilograms is 0 because all og mgo is used up since the reaction goes to completion

  109. GoliS
    • one year ago
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    yupp

  110. telijahmed
    • one year ago
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    now that means Si is the excess reagent so you need to find out how much of si consumed all of the mgo

  111. GoliS
    • one year ago
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    no that was wrong ..... :-(

  112. GoliS
    • one year ago
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    timed out

  113. telijahmed
    • one year ago
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    i dont understand

  114. aaronq
    • one year ago
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    wow some major cheating goin on here

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