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rosedewittbukater

  • one year ago

Using the conversion factor from eV (electron volts) to joules, determine which energy line for an electron dropping from one energy level to another would show a line at 435 nm in an emission spectrum. Use the formulas and the energy level information below to answer the question.

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  1. rosedewittbukater
    • one year ago
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    Needed constants: 1.00eV = 1.6 * 10^-19 J c = 3.00 * 10^8 m/s h = 6.63 * 10^-34 J*s Energy Level Values: E6: E= -0.378 eV E5: E= -0.544 eV E4: E= -0.850 eV E3: E= -1.51 eV E2: E= -3.403V

  2. rosedewittbukater
    • one year ago
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    @chmvijay @wio do you think one of you could help me out on this problem?

  3. rosedewittbukater
    • one year ago
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    @Luigi0210 do you know how to solve this Physics problem?

  4. Luigi0210
    • one year ago
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    I don't know physics, sorry.

  5. chmvijay
    • one year ago
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    |dw:1434429760324:dw| You know h,c and lamda (435nm) energy value comes in joules so use energy and eV conversion value and find the value in eV that is 1.6 * 10^-19 J =1eV Energy u got from above= 1eV* 1.6 * 10^-19 J /Energy did u get it ?

  6. chmvijay
    • one year ago
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    @rosedewittbukater u got it ?

  7. rosedewittbukater
    • one year ago
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    @chmvijay no I'm still pretty lost

  8. rosedewittbukater
    • one year ago
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    E = (3.00 * 10^8 m/s)(6.63 * 10^-34 J*s)/435 nm Is that it?

  9. rosedewittbukater
    • one year ago
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    I'm really confused and can't figure out what I'm supposed to be doing.

  10. rosedewittbukater
    • one year ago
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    I tried to understand how other people were explaining it and got this: 435 nm = 435 * 10^-9 m E = (6.63 * 10^-34 J*s) (3.00 * 10^8 m/s)/(435 * 10^-9 m) E = 4.57 * 10^-19 J (4.57 * 10^-19 J) (1 eV/1.60 * 10^-19) = 2.86 eV E5 - E2 = -0.544 - (-3.403) = 2.86 eV

  11. chmvijay
    • one year ago
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    yaa its right indeed RIGHT:)

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