Sodium Vapor Lamps (1.19 hw)  [ENDORSED]

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Kdhaines1B
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Joined: Fri Sep 26, 2014 2:02 pm

Sodium Vapor Lamps (1.19 hw)

Postby Kdhaines1B » Sun Oct 12, 2014 8:58 pm

Sodium vapor lamps, used for public lighting, emit yellow light of wavelength 589nm. 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

Please help on part b and c.

Quynh Nguyen 2O
Posts: 27
Joined: Fri Sep 26, 2014 2:02 pm

Re: Sodium Vapor Lamps (1.19 hw)

Postby Quynh Nguyen 2O » Mon Oct 13, 2014 1:03 am

b)

First, you convert mg to g
5.00 mg * (1g/1000mg) = 5.00 x 10^-3 g of Na

Since, it's asking for the how much energy is emitted by an excited SODIUM, you use the molar mass of Na

5.00x10^-3 g Na * (1 mol Na/ 23g Na) * (6.022 x10^23 atoms Na/ 1 g Na) = 1.31 x10^20 atoms of Na

(1.31 x10^20 atoms Na) * (3.37x 10^-19 J) = 44.1 J


c) this is the same process as b

1 mol Na * (6.022 x 10^23 atoms Na/ 1mol Na) = 6.022 x 10^23 atoms Na

(6.022 x 10^23 atoms Na) * (3.37 x 10^-19J) = 202941.4J or 2.03 x10^5 J

Quynh Nguyen 2O
Posts: 27
Joined: Fri Sep 26, 2014 2:02 pm

Re: Sodium Vapor Lamps (1.19 hw)  [ENDORSED]

Postby Quynh Nguyen 2O » Mon Oct 13, 2014 1:07 am

a)

wavelength = 589nm convert it to m, 5.89 x10^-7 m

you use the formula E= hc/wavelength

E = (6.626 x 10^-34)(2.99 x 10^8) / (5.89 x 10^-7)
= 3.37 x 10^-19 J

Lourick Bustamante 1B
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Joined: Wed Nov 16, 2016 3:02 am

Re: Sodium Vapor Lamps (1.19 hw)

Postby Lourick Bustamante 1B » Mon Oct 16, 2017 9:44 pm

Why is it that we use E = h(c/λ) instead of just E = hv?

Aman Sankineni 2L
Posts: 103
Joined: Fri Aug 30, 2019 12:17 am

Re: Sodium Vapor Lamps (1.19 hw)

Postby Aman Sankineni 2L » Tue Oct 15, 2019 11:56 pm

Quynh Nguyen 2O wrote:b)

First, you convert mg to g
5.00 mg * (1g/1000mg) = 5.00 x 10^-3 g of Na

Since, it's asking for the how much energy is emitted by an excited SODIUM, you use the molar mass of Na

5.00x10^-3 g Na * (1 mol Na/ 23g Na) * (6.022 x10^23 atoms Na/ 1 g Na) = 1.31 x10^20 atoms of Na

(1.31 x10^20 atoms Na) * (3.37x 10^-19 J) = 44.1 J


c) this is the same process as b

1 mol Na * (6.022 x 10^23 atoms Na/ 1mol Na) = 6.022 x 10^23 atoms Na

(6.022 x 10^23 atoms Na) * (3.37 x 10^-19J) = 202941.4J or 2.03 x10^5 J


Where did the 3.37 x 10^-19 J come from in this solution?

connie 2C
Posts: 106
Joined: Thu Jul 11, 2019 12:17 am

Re: Sodium Vapor Lamps (1.19 hw)

Postby connie 2C » Wed Oct 16, 2019 8:35 pm

Aman Sankineni 3E wrote:
Quynh Nguyen 2O wrote:b)

First, you convert mg to g
5.00 mg * (1g/1000mg) = 5.00 x 10^-3 g of Na

Since, it's asking for the how much energy is emitted by an excited SODIUM, you use the molar mass of Na

5.00x10^-3 g Na * (1 mol Na/ 23g Na) * (6.022 x10^23 atoms Na/ 1 g Na) = 1.31 x10^20 atoms of Na

(1.31 x10^20 atoms Na) * (3.37x 10^-19 J) = 44.1 J


c) this is the same process as b

1 mol Na * (6.022 x 10^23 atoms Na/ 1mol Na) = 6.022 x 10^23 atoms Na

(6.022 x 10^23 atoms Na) * (3.37 x 10^-19J) = 202941.4J or 2.03 x10^5 J


Where did the 3.37 x 10^-19 J come from in this solution?


the 3.37x10^-19 J comes from the answer to part a which is the energy emitted by an excited sodium atom when it generates a photon. to get that answer, look at the post above where someone else explains it!

Aman Sankineni 2L
Posts: 103
Joined: Fri Aug 30, 2019 12:17 am

Re: Sodium Vapor Lamps (1.19 hw)

Postby Aman Sankineni 2L » Sun Oct 20, 2019 2:56 pm

Ohhhh, just got it. Thank you!


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