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Author Topic: Ionization Energy of Carbon  (Read 2232 times)

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bravoghost

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Ionization Energy of Carbon
« on: March 19, 2011, 02:36:20 PM »

I found that it takes 11.26 eV to liberate the first electron off a neutral carbon atom. I needed to calculate the wavelength, and my final answer was around 15 nm. I watched someone do the same problem online, and they got roughly 150 nm. Who was right?
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opti384

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Re: Ionization Energy of Carbon
« Reply #1 on: March 19, 2011, 04:17:21 PM »

Can you show your process?

bravoghost

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Re: Ionization Energy of Carbon
« Reply #2 on: March 27, 2011, 02:14:51 PM »

So I did the problem without such liberal rounding, and it came out to about 1.11*10^-7.

11.26 eV = 1.6*10^-19 C

Thus, 11.26 x 1.6*10^-19 = 1.8*10^-18 C (the energy of the ionization)

E=hf
1.8*10^-18 = 6.63*10^-34 x f
f = 2.71*10^15

v=λf
3*10^8 = λ x 2.71*10^15
λ = 1.11*10^-7 m

but that isn't 110 nm, right? If we put that number in nanometers, it'd be .011 nm, right??
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rabolisk

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Re: Ionization Energy of Carbon
« Reply #3 on: March 27, 2011, 02:35:22 PM »

So I did the problem without such liberal rounding, and it came out to about 1.11*10^-7.

11.26 eV = 1.6*10^-19 C

This makes no sense.

Thus, 11.26 x 1.6*10^-19 = 1.8*10^-18 C (the energy of the ionization)

Neither does this.

E=hf
1.8*10^-18 = 6.63*10^-34 x f
f = 2.71*10^15

v=λf
3*10^8 = λ x 2.71*10^15
λ = 1.11*10^-7 m

Nevertheless, this is correct.

1 nanometer is 10-9 meter.
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