Homework Question 1B7b


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Emma Scholes 1L
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Joined: Fri Sep 28, 2018 12:18 am

Homework Question 1B7b

Postby Emma Scholes 1L » Sun Oct 07, 2018 2:55 pm

Sodium vapor lamps emit yellow light of wavelength 589nm. How much energy is emitted by 5.00 mg of sodium atoms?

Mindy Kim 4C
Posts: 65
Joined: Fri Sep 28, 2018 12:25 am

Re: Homework Question 1B7b

Postby Mindy Kim 4C » Sun Oct 07, 2018 8:47 pm

First, you want to convert the 5.00 mg of sodium atoms to how many atoms of Na there are. You can do this through the following:

(5.00 mg Na)(1 g Na/1000 mg Na)(1 mol Na/22.99 g Na)(6.022x10^23 atoms/1 mol Na) which equals 1.3097 x 10^20 atoms of Na.

Then, to find the energy emitted by 5.00 mg of sodium atoms, you can multiply the atoms of Na by the energy emitted by one Na atom (the answer of part a).

(1.3097x10^20 atoms)(3.37x10^-19 J/atom) = 44.1 J

Hope this helped!

Quinn_Simpson_3D
Posts: 12
Joined: Fri Sep 28, 2018 12:26 am

Re: Homework Question 1B7b

Postby Quinn_Simpson_3D » Wed Oct 10, 2018 10:11 am

In order to solve this problem, you need to find the amount of atoms of Na. To do this convert the 5.00 mg to .005 g. Then divide by 22.99 g.mol^-1 (Na molar Mass) to find the number of moles, which is 2.17 x 10^-4. You then multiply by Avogadro's constant (6.022 x 10^23) to find the number of atoms, which turns out to be 1.31 x 10^20 atoms Na. In order to find the energy of the sodium emitted, we need to use the equation E=hv, where h is Planck's constant and v is the frequency. However, we only have the wavlength. Luckily, we can use c=v(lambda) to find the frequency. c is the speed of light or 3.00 x 10^8. The wavelength is in nanometers, so we must multiply by 1 x 10-9 to convert to meters, making the wavelength 5.89 x 10^-7 m. We use c/lambda = v to find the frequency, which results in 5.09 x 10^14 Hz. We can now use E=hv to find the energy of a single atom, which is 3.37 x 10^-19 (6.63 x 10^-34 x (5.09 x 10^14)). We can then multiply the found energy of a single atom to the amount of atoms we have (1.31 x 10^20), which results in the energy being 44.1 J.


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