How many phtons are emitted by green light?

[stickplot]

Homework Statement

How many photon/s are emitted by a green light (λ= 500 nm) 60 W incandescent bulb? Assume an efficiency of 2%.

Homework Equations

c= wavelength(f)
c=speed of light

E=hf

The Attempt at a Solution

im not sure which equation to use here, i don't know what to do with the 60 W and the efficiency of 2 percent.

 

[Andrew Mason]

Homework Statement

How many photon/s are emitted by a green light (λ= 500 nm) 60 W incandescent bulb? Assume an efficiency of 2%.

Homework Equations

c= wavelength(f)
c=speed of light

E=hf

The Attempt at a Solution

im not sure which equation to use here, i don't know what to do with the 60 W and the efficiency of 2 percent.

The question is actually a lot more complicated to answer correctly than the author of the question intended. I think he just wants you to determine the amount of light energy emitted by the bulb (ie. the total energy of all the photons emitted) and work out the energy of each photon. So:

1. What is the energy of one photon (λ= 500 nm)? (Hint: you need to use Planck's constant).

2. How much light energy is the lightbulb giving off (ie. what is the total energy of all the photons the bulb produces)?

3. What, then, is the number of photons?

[Note: the problem with this question is that an incandescent light emits a spectrum of light that is determined by the temperature of the filament. It cannot emit light of only one wavelength. To determine the proportion of green light in the range λ= 500 nm (+ or - 50 nm say) to the total energy requires an analysis of the temperature and energy distribution of the light.]

AM

 

[stickplot]

ok i took what you told me and tried it over.
E = hv
v=c/k
3x10^8/500=600,000
E=6.626 x 10^-34/600,000
1.1043333x10^-39
1.1043333x10^-39/50= .004 photons/s

im not sure if this is right .004 sounds very small

 

[stickplot]

1.1043333x10^-39 is total energy by photon
and .004 is total energy by light
and btw I am not sure if i have to include the 2% so help me out please
thanks

 

[mikelepore]

It sound to like the given 2 percent efficiency means that the light bulb consumes 60 watts, and it emits (0.02)(60) watts of light and (0.98)(60) watts of heat.

 

[stickplot]

alright so this is what i came up with correct me if I am wrong
p/(hc/wavelength)

6.626x10^-34/3x10^8= 1.989x10^-25
1.989x10^-25/500x10^-9= 3.978x10^-19

.02(60)= 1.2
1.2/3.978x10^-19= rate= 3x10^18 photon/s

 

[Andrew Mason]

alright so this is what i came up with correct me if I am wrong
p/(hc/wavelength)

6.626x10^-34/3x10^8= 1.989x10^-25
1.989x10^-25/500x10^-9= 3.978x10^-19

.02(60)= 1.2
1.2/3.978x10^-19= rate= 3x10^18 photon/s

I don't follow your work. Set out the formula for energy of a photon and then plug in the numbers.

Use [itex]E = h\nu = hc/\lambda[/itex]

where [itex]h = 6.626\times 10^{-34}; c=3 \times 10^8 m/sec; \lambda = 5 \times 10^-7 m.[/itex]

AM

 

[stickplot]

hv
6.626x10^-34(3x10^8)= 1.9878x10^-25

1.9878x10^-25/5x10^-7= 3.9756x10^-19

E=3.9756x10^-19

ok i used the formula that you showed me, but why does it come out to a negative number? and after this do i divide P/E= to get the number of photons that occurred on the effect for the 60 W lightbulb? (60/3.9756x10^-19)??

 

[Andrew Mason]

hv
6.626x10^-34(3x10^8)= 1.9878x10^-25

1.9878x10^-25/5x10^-7= 3.9756x10^-19

E=3.9756x10^-19

ok i used the formula that you showed me, but why does it come out to a negative number? and after this do i divide P/E= to get the number of photons that occurred on the effect for the 60 W lightbulb? (60/3.9756x10^-19)??

E=3.9756x10^-19 J for ONE photon. The total energy is .012 x 60 W = .72 J/sec

What is the total number of photons/sec?

[tex]E_{total} = nE_{photon}[/tex]

You have to work out n (per second).

AM

 

[stickplot]

o ok.
.72 J/sec= n(3.9756x10^-19 J)
n= 1.8 photons per second? how does that look?
and how did you get .012?
did you mean to put .02?

 

[Andrew Mason]

o ok.
.72 J/sec= n(3.9756x10^-19 J)
n= 1.8 photons per second? how does that look?

You are missing the 10^18

and how did you get .012?
did you mean to put .02?

Sorry. Going by memory. It is 2% not 1.2%. It is 1.2 watts. (.02 x 60).

So it is n = 1.2/3.9756x10^-19= 3 x 10^18 photons / sec

AM