HW 1.B.7 (b and c)


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WYacob_2C
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Joined: Sat Jul 20, 2019 12:16 am

HW 1.B.7 (b and c)

Postby WYacob_2C » Fri Oct 18, 2019 12:38 pm

1.B.7 says "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?"

I understood part (a) but could someone take me through their thought process to get through part (b) or (c). Thank you

Hiba Alnajjar_2C
Posts: 108
Joined: Fri Aug 09, 2019 12:17 am

Re: HW 1.B.7 (b and c)

Postby Hiba Alnajjar_2C » Fri Oct 18, 2019 12:45 pm

For parts b and c, you essentially do dimensional analysis, just as we did in the high school review unit. Because your answer to part a gives you the amount of energy emitted per atom, you can use dimensional analysis to convert from mg to g to mol to atoms. After finding the number of atoms of Na, you would multiply that number by your answer to part a to find the amount of energy emitted in total. Similarly, for part c, you would go from mol to atoms and then multiply that number by your answer to part a. If you're unsure if you're doing it correctly, it may help to double check whether or not your units cancelled out: you should be left with J at the end. Hope this helps!

Amir Bayat
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Re: HW 1.B.7 (b and c)

Postby Amir Bayat » Fri Oct 18, 2019 12:49 pm

For part (b) we convert the mg of Na to moles of Na and then multiply by Avogadro's number to get the number of atoms of Na. We then multiply by the energy in part (a) to get how much energy is emitted by 5 mg of Na.

For part (c) we multiply the energy from part (a) by Avogadro's number to get the light energy emitted by one mole of Na atoms.

Miriam Villarreal 1J
Posts: 105
Joined: Sat Aug 17, 2019 12:16 am

Re: HW 1.B.7 (b and c)

Postby Miriam Villarreal 1J » Fri Oct 18, 2019 12:58 pm

For part (b) you covert the 5.00mg into .005g Na and divide it by its molar mass of 23g/mol and multiply the answer you get by 6.02 x 10^23 atoms which should result in 1.31x10^20 atoms which you then multiply by the energy obtained in part a and you should get 44.1J.

For part (c) you take the 1.00 mol of Na atoms multiply by 6.02x10^23, multiply by the 3.37x10^-19J (energy obtained in part a). The answer in the textbook is given in Kj so just make sure to convert.


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