## 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

kk! haha i stuck on this problem

2. NĐGK

the formula of wavelength is E=hc/λ??

3. telijahmed

the question now is how does the wavelength of an electron relate to the energy of light

4. telijahmed

|dw:1364196013585:dw|

5. rosewhite

it may related ,as I work on it with out the wavelength it still wrong

6. NĐGK

.__. this is too hard T^T

7. telijahmed

where m is the mass of the electron

8. telijahmed

i think i ll try that now

9. telijahmed

|dw:1364196120782:dw|

10. telijahmed

have you tried that @rosewhite

11. rosewhite

i think we need c of electron

12. rosewhite

yes, I used E=hc/λ and put c= 3*10^8 m/s

13. telijahmed

@rosewhite but you did not know E for light

14. telijahmed

if we use einsteins theory a new parameter comes in

15. telijahmed

which is the mass

16. telijahmed

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

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

okay the please try @rosewhite einsteins dual theory

19. rosewhite

how? @telijahmed

20. rosewhite

OK , I get C from the law and then get E and from it find lambda but still wrong !! @telijahmed

21. telijahmed

wow tough one eh @rosewhite sorry for wasting your time i am still working on it

22. rosewhite

Okey , if you got the answer, plz tell me @telijahmed

23. telijahmed

ok @rosewhite thanks for assisting

24. NĐGK

Can you tell me the answer if you got it? ;) thanks very much

25. Rainbow1980

anybody knows the answer? I can´t get it :(

26. ammanjordan

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

@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

@mikalika I am Sorry for that :-(

29. mikalika

@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

Thank you very much :-)

31. muratcankalem

32. JK1992

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

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

34. mafeg95

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

velocity, not speed sorry

36. JK1992

@MiniDino try mafeg's way and tell us :)

37. mafeg95

what would be n_i and n_f?? for rydberg's equation I mean

38. MiniDino

I tried nf = 3 and ni = 2 and nf = 2 and ni = 1, both answers given red

39. mafeg95

can you tell me step by step what you are doing? so that I can follow along better, cause I'm confused

40. JK1992

41. mafeg95

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

@mafeg95 ok :) then you are fine

43. nayon333

44. NĐGK

This court is really hard TT_TT i'm only 10th grade student

45. JK1992

could you try 3.121*10^-9 m ??

46. muratcankalem

3.121*10^-9 is WRONG

47. mafeg95

can someone try 4.25*10^-12 please?? I only have one subission left..

48. mafeg95

submission*

49. sodie69

4.25*10^-12 is wrong

50. Rainbow1980

wrong

51. monkeynero

Wrong

52. mafeg95

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

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

try 1.8746064 * 10^-6

55. muratcankalem

WRONG

56. nayon333

wrong

57. jamesmauk

wrong

58. fiorrr

That question is driving me nuts. Did anybody get the answer to number 9?

59. jamesmauk

no im suck on this one, 9 and number 1 if anyone can help?

60. nayon333

no, did u get answer of 9

61. fiorrr

which number one did you get? tungsten reaction? didnt get number 9 yet.

62. jamesmauk

i havent gotten any of the number 1's

63. fiorrr

i heard there are different questions for number 1: here's what i got: 0 35.28 52.336 15.38

64. mafeg95

Yeah, they use different elements. @fiorrr

65. jamesmauk

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

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

but can we focus on the question that they asked here?

68. mafeg95

Everybody is struggling with it...

69. nayon333

who has got K2TaF7 with Na?

70. MiniDino

@mafeg95 I almost give up, because I can't find the answer, I spent 6 times of my submission

71. nayon333

72. fiorrr

just opened a thread for #9

73. mafeg95

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

75. mafeg95

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

me too, if had the answer, i would give in this situation ....

77. mafeg95

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

i tried it more than 40 times to solve..............

79. telijahmed

there key is in the smallest energy transition

80. telijahmed

that statement i gave an answer and someone said wrong

81. nayon333

which one?

82. telijahmed

1.8746064 * 10^-6

83. nayon333

its wrong by the red signal

84. telijahmed

i am not to sure but i used|dw:1364235425854:dw|

85. maumay

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

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

check out the drawx here @maumay

88. fiorrr

1.64064*10^-7 is correct!

89. jamesmauk

now does anyone know how to do number 9?

90. nayon333

1.64064*10^-7 is correct

91. nayon333

anybody has 9 result.......

92. mafeg95

@maumay, you are a godsend! Thank you sooo much!

93. telijahmed

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

I T HAS NO RELATIONSHIP TO THE PROBLEM

95. telijahmed

I DID NOT USE THE RIGHT FORMULA BUT I AM COOL

96. nayon333

9 plz?

97. jamesmauk

i opened a threat for number 9

98. nayon333

get yet?

99. mikalika

I have different No 9 so I am opening too a new thread...

100. nayon333

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

who has it?

102. nayon333

velocity of 10^3 km/s . not 103

103. GoliS

is any one of u got answer of 1st problem (MgO)?????

104. nayon333

we have 7 min. only

105. NĐGK

Thanks so much

106. telijahmed

@GoliS i can walk you through

107. GoliS

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

okay now for MGO the kilograms is 0 because all og mgo is used up since the reaction goes to completion

109. GoliS

yupp

110. telijahmed

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

no that was wrong ..... :-(

112. GoliS

timed out

113. telijahmed

i dont understand

114. aaronq

wow some major cheating goin on here