HW 1.25 (b)


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Chau Nghiem 1K
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Joined: Fri Sep 25, 2015 3:00 am

HW 1.25 (b)

Postby Chau Nghiem 1K » Sun Oct 18, 2015 10:39 pm

Question: Sodium vapor lamps, used for public lighting, emit yellow light of wavelength 589 nm. How much energy is emitted by 5.00 mg of sodium atoms emitting light at this wavelength?

Could someone explain why it's possible that the units for energy is able to conveniently switch from just J to J/atoms in order to cancel out the necessary units in part (b)? Thank you.

ConnieDo3I
Posts: 31
Joined: Fri Sep 25, 2015 3:00 am

Re: HW 1.25 (b)

Postby ConnieDo3I » Sun Oct 18, 2015 11:01 pm

Hey!
So according to the solutions manual,
E = (5.00 x 10^-3 g Na / 22.99 g x mol^(-1)) x (6.022x10^23 atoms x mol^(-1)) x (3.37x10^(-19) J x atom^(-1)) = 44.1 J

If you follow the equation, the "atoms" unit should cancel out smoothly.
The following is the equation with only the units. It's probably easier to see how they cancel out this way.

g 1 atoms J
-- x ----------- x -------- x ------ = J
1 g mol^-1 mol atom

Hope this helps!

Angelica Dis 3k
Posts: 6
Joined: Fri Jun 23, 2017 11:40 am

Re: HW 1.25 (b)

Postby Angelica Dis 3k » Thu Oct 19, 2017 4:32 pm

Think about it this way...

In the question part A, we were looking for the energy of one photon given the wavelength. With this information we then could use E=hc/λ to get 3.37 x 10^-19. However, in part B we are given how much Sodium is present in mg. Now we are being asked to find the energy of multiple photons and not just one like in part A. In order to do so, we first convert mg of Na to g of Na to get 5.00 x 10^-3. From here we can divide by the atomic mass of Na, 22.99 g/mol -1. So far we have cancelled out grams and are left with mols but we need to find the number of atoms in order to solve for energy of photons. For this, we multiply by Avagadro's #, 6.023 x 10^23, and we are just left with atoms now! Our final answer is in atoms/Joules because we multiply our answer with our answer to part A to find how the total energy of photons from 5.00 mg of Na.

Hope this helps!


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