## HW Help 1.25

$\lambda=\frac{h}{p}$

Moderators: Chem_Mod, Chem_Admin

Janelle Magaling 3L
Posts: 11
Joined: Thu Oct 04, 2018 12:16 am

### HW Help 1.25

Hi can someone help me with this question?

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

I solved (a) already, but got stuck on (b). What equation should I use considering the given units?

Jane Burgan 1C
Posts: 73
Joined: Fri Sep 28, 2018 12:15 am

### Re: HW Help 1.25

Since you're given the amount of sodium in mg, convert this amount into moles by dividing 5.00 mg by the molar mass of sodium. Once you have the amount of sodium in moles, you can multiply this number by avogrado's number to get the number of sodium atoms in 5.00 mg. Finally, multiply your answer by the energy you got in part a. This will give you the amount of energy emitted by 5.00 mg of sodium.

Destiny Diaz 4D
Posts: 51
Joined: Fri Sep 28, 2018 12:28 am

### Re: HW Help 1.25

I followed the process suggested and am not getting the answer offered in the solutions from the book, my answer is off by a few decimal points and was wondering if you could break down the process a bit more just so I can understand conceptually. Thank you so much <3

Jeremy_Guiman2E
Posts: 82
Joined: Fri Sep 28, 2018 12:29 am

### Re: HW Help 1.25

Essentially, your set up would look like this: E = (5.00 x 10-3 g Na) (1 mol Na / 22.99 g Na) (6.022 x 1023 atoms/mol) (3.37 x 10-19 J/atom) = 44.1 J

The 3.37 x 10-19 J/atom comes from your answer in part a. Hope this helps!

Destiny Diaz 4D
Posts: 51
Joined: Fri Sep 28, 2018 12:28 am

### Re: HW Help 1.25

jguiman4H wrote:Essentially, your set up would look like this: E = (5.00 x 10-3 g Na) (1 mol Na / 22.99 g Na) (6.022 x 1023 atoms/mol) (3.37 x 10-19 J/atom) = 44.1 J

The 3.37 x 10-19 J/atom comes from your answer in part a. Hope this helps!

Yes thank you <3

Return to “DeBroglie Equation”

### Who is online

Users browsing this forum: No registered users and 1 guest