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NĐGK

  • 2 years 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
    • 2 years ago
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    kk! haha i stuck on this problem

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

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

  4. telijahmed
    • 2 years ago
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    |dw:1364196013585:dw|

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

  6. NĐGK
    • 2 years ago
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    .__. this is too hard T^T

  7. telijahmed
    • 2 years ago
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    where m is the mass of the electron

  8. telijahmed
    • 2 years ago
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    i think i ll try that now

  9. telijahmed
    • 2 years ago
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    |dw:1364196120782:dw|

  10. telijahmed
    • 2 years ago
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    have you tried that @rosewhite

  11. rosewhite
    • 2 years ago
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    i think we need c of electron

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

  13. telijahmed
    • 2 years ago
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    @rosewhite but you did not know E for light

  14. telijahmed
    • 2 years ago
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    if we use einsteins theory a new parameter comes in

  15. telijahmed
    • 2 years ago
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    which is the mass

  16. telijahmed
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    okay the please try @rosewhite einsteins dual theory

  19. rosewhite
    • 2 years ago
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    how? @telijahmed

  20. rosewhite
    • 2 years 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
    • 2 years ago
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    wow tough one eh @rosewhite sorry for wasting your time i am still working on it

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

  23. telijahmed
    • 2 years ago
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    ok @rosewhite thanks for assisting

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

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

  26. ammanjordan
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    @mikalika I am Sorry for that :-(

  29. mikalika
    • 2 years 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
    • 2 years ago
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    Thank you very much :-)

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

  32. JK1992
    • 2 years 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
    • 2 years ago
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    @JK1992 For this way my answer was 3.04*10^-8, and, unfortunately, it's not correct : (

  34. mafeg95
    • 2 years 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
    • 2 years ago
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    velocity, not speed sorry

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

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

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

  39. mafeg95
    • 2 years 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
    • 2 years ago
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    @mafeg95 what is your level on chemistry???just curious cause you asked about ni and nf

  41. mafeg95
    • 2 years 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
    • 2 years ago
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    @mafeg95 ok :) then you are fine

  43. nayon333
    • 2 years ago
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    anybody got the answer....

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

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

  46. muratcankalem
    • 2 years ago
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    3.121*10^-9 is WRONG

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

  48. mafeg95
    • 2 years ago
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    submission*

  49. sodie69
    • 2 years ago
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    4.25*10^-12 is wrong

  50. Rainbow1980
    • 2 years ago
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    wrong

  51. monkeynero
    • 2 years ago
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    Wrong

  52. mafeg95
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    try 1.8746064 * 10^-6

  55. muratcankalem
    • 2 years ago
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    WRONG

  56. nayon333
    • 2 years ago
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    wrong

  57. jamesmauk
    • 2 years ago
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    wrong

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

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

  60. nayon333
    • 2 years ago
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    no, did u get answer of 9

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

  62. jamesmauk
    • 2 years ago
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    i havent gotten any of the number 1's

  63. fiorrr
    • 2 years 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
    • 2 years ago
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    Yeah, they use different elements. @fiorrr

  65. jamesmauk
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    but can we focus on the question that they asked here?

  68. mafeg95
    • 2 years ago
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    Everybody is struggling with it...

  69. nayon333
    • 2 years ago
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    who has got K2TaF7 with Na?

  70. MiniDino
    • 2 years 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
    • 2 years ago
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    has anybody answer of 9?

  72. fiorrr
    • 2 years ago
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    just opened a thread for #9

  73. mafeg95
    • 2 years 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
    • 2 years ago
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    plz tell us the answer

  75. mafeg95
    • 2 years 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
    • 2 years ago
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    me too, if had the answer, i would give in this situation ....

  77. mafeg95
    • 2 years 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
    • 2 years ago
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    i tried it more than 40 times to solve..............

  79. telijahmed
    • 2 years ago
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    there key is in the smallest energy transition

  80. telijahmed
    • 2 years ago
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    that statement i gave an answer and someone said wrong

  81. nayon333
    • 2 years ago
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    which one?

  82. telijahmed
    • 2 years ago
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    1.8746064 * 10^-6

  83. nayon333
    • 2 years ago
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    its wrong by the red signal

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

  85. maumay
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    check out the drawx here @maumay

  88. fiorrr
    • 2 years ago
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    1.64064*10^-7 is correct!

  89. jamesmauk
    • 2 years ago
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    now does anyone know how to do number 9?

  90. nayon333
    • 2 years ago
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    1.64064*10^-7 is correct

  91. nayon333
    • 2 years ago
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    anybody has 9 result.......

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

  93. telijahmed
    • 2 years 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
    • 2 years ago
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    I T HAS NO RELATIONSHIP TO THE PROBLEM

  95. telijahmed
    • 2 years ago
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    I DID NOT USE THE RIGHT FORMULA BUT I AM COOL

  96. nayon333
    • 2 years ago
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    9 plz?

  97. jamesmauk
    • 2 years ago
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    i opened a threat for number 9

  98. nayon333
    • 2 years ago
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    get yet?

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

  100. nayon333
    • 2 years 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
    • 2 years ago
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    who has it?

  102. nayon333
    • 2 years ago
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    velocity of 10^3 km/s . not 103

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

  104. nayon333
    • 2 years ago
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    we have 7 min. only

  105. NĐGK
    • 2 years ago
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    Thanks so much

  106. telijahmed
    • 2 years ago
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    @GoliS i can walk you through

  107. GoliS
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    yupp

  110. telijahmed
    • 2 years 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
    • 2 years ago
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    no that was wrong ..... :-(

  112. GoliS
    • 2 years ago
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    timed out

  113. telijahmed
    • 2 years ago
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    i dont understand

  114. aaronq
    • 2 years ago
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    wow some major cheating goin on here

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