1B #7

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Marisa Gaitan 2D
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1B #7

Postby Marisa Gaitan 2D » Thu Oct 15, 2020 3:16 pm

I'm a little confused on finding the energy emitted by 5.00mg of sodium atoms (Part B). I'm okay with parts A and C, but I think I might be overthinking B or messing up somehow.

Sodium vapor lamps, used for public lighting, emit yellow light of wavelength 589 nm. How much energy is emitted by (a) an excited sodium atom when it generates a photon; (b) 5.00 mg of sodium atoms emitting light at this wavelength; (c) 1.00 mol of sodium atoms emitting light at this wavelength?

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Re: 1B #7

Postby Chem_Mod » Thu Oct 15, 2020 3:25 pm

Hello! Begin by finding the energy using the equation E=hc/wavelength. You can then use dimensional analysis by converting the mass of sodium to moles and then to atoms and multiplying by the E value found to get final units of Joules. For part c, use a similar approach by converting 1 mol into atoms and multiplying by the energy found in part a!

Kaley Qin 1F
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Re: 1B #7

Postby Kaley Qin 1F » Thu Oct 15, 2020 3:30 pm

For part b, you have to first start by converting the 5.00 mg of Na to photons of Na. I converted 5.00 mg Na to moles of Na then to photons (there are 6.022 x 10^23 photons in a mol). Then I used the energy per photon I got in (a) multiplied by the number of photons I got to get the total energy emitted by 5.00mg Na.

Kaley Qin 1F
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Re: 1B #7

Postby Kaley Qin 1F » Thu Oct 15, 2020 3:30 pm

For part b, I first started by converting the 5.00 mg of Na to photons of Na. I converted 5.00 mg Na to moles of Na then to photons (there are 6.022 x 10^23 photons in a mol). Then I used the energy per photon I got in (a) multiplied by the number of photons I got to get the total energy emitted by 5.00mg Na.

Josh Chou 3K
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Re: 1B #7

Postby Josh Chou 3K » Thu Oct 15, 2020 3:31 pm

I believe you have to do some dimensional analysis to figure out how many atoms are in 5.00 mg of sodium --> (5.00 x 10^-3 g Na) x (1 mol Na/22.99 g Na) x (6.022 x 10^23 Na atoms / 1 mol Na). And then I think you multiply that result (# of atoms) your answer from part a (which you probably found using E = (hc)/λ)


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