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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?

  • one year ago
  • one year ago

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  1. NĐGK
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    kk! haha i stuck on this problem

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

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

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

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

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

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

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

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

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

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

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

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

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

    • one year ago
  15. telijahmed
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    which is the mass

    • one year ago
  16. telijahmed
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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

    • one year ago
  17. rosewhite
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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 ?

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

    • one year ago
  19. rosewhite
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    how? @telijahmed

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

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

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

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

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

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

    • one year ago
  26. ammanjordan
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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 :-)

    • one year ago
  27. mikalika
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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

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

    • one year ago
  29. mikalika
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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

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

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

    • one year ago
  32. JK1992
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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

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

    • one year ago
  34. mafeg95
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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...

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

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

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

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

    • one year ago
  39. mafeg95
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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

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

    • one year ago
  41. mafeg95
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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 :)

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

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

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

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

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

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

    • one year ago
  48. mafeg95
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    submission*

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

    • one year ago
  50. Rainbow1980
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    wrong

    • one year ago
  51. monkeynero
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    Wrong

    • one year ago
  52. mafeg95
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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.

    • one year ago
  53. jamesmauk
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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?

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

    • one year ago
  55. muratcankalem
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    WRONG

    • one year ago
  56. nayon333
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    wrong

    • one year ago
  57. jamesmauk
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    wrong

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

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

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

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

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

    • one year ago
  63. fiorrr
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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

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

    • one year ago
  65. jamesmauk
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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:

    • one year ago
  66. mikalika
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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...

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

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

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

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

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

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

    • one year ago
  73. mafeg95
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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 :)

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

    • one year ago
  75. mafeg95
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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! :)

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

    • one year ago
  77. mafeg95
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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*

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

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

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

    • one year ago
  81. nayon333
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    which one?

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

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

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

    • one year ago
  85. maumay
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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.

    • one year ago
  86. telijahmed
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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

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

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

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

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

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

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

    • one year ago
  93. telijahmed
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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

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

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

    • one year ago
  96. nayon333
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    9 plz?

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

    • one year ago
  98. nayon333
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    get yet?

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

    • one year ago
  100. nayon333
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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.

    • one year ago
  101. nayon333
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    who has it?

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

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

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

    • one year ago
  105. NĐGK
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    Thanks so much

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

    • one year ago
  107. GoliS
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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

    • one year ago
  108. telijahmed
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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

    • one year ago
  109. GoliS
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    yupp

    • one year ago
  110. telijahmed
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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

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

    • one year ago
  112. GoliS
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    timed out

    • one year ago
  113. telijahmed
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    i dont understand

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

    • one year ago
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