UV vs IR Light Bulb Power Comparison

[georgeh]

An ultraviolet light bulb, emitting at 400nm and an infared light, emitting at 700nm are both rated at 400 W.
a. Which bulb radiates more photons at the greater rate?
I thought about it long and hard and only thing i can think of
is to use the rleationship: v=lambda*f
where v would be C- the speed of light
solve for the frequency and we know the wavelength for each bulb.
b. How many more photons does it generate per second than the other bulb?
I was thinking of using the following relationship:
E=hc/Lamda
Determine the energy, and we know that Power=Energy/time
solve for the time?

 

[neutrino]

An ultraviolet light bulb, emitting at 400nm and an infared light, emitting at 700nm are both rated at 400 W.
a. Which bulb radiates more photons at the greater rate?
I thought about it long and hard and only thing i can think of
is to use the rleationship: v=lambda*f
where v would be C- the speed of light
solve for the frequency and we know the wavelength for each bulb.
...
E=hc/Lamda
Determine the energy, and we know that Power=Energy/time
solve for the time?

Yes, you have to use both these relations, excpet for a minor addition to the latter : $$E = nh\nu$$ since you are calculating the energy due to a stream of photons rather than a single one.

 

[andrevdh]

Each photon is a little packet of energy that is emitted by the source. The energy of each photon can be calculated with
$$E=hf$$
which comes to
$$E=\frac{hc}{\lambda}$$
Shorter wavelength photons therefore carry more energy per packet than longer wavelength photons.

The power of the light bulb express the rate at which it emits energy, or the total amount of energy emitted by it per second. This energy is made up by the stream of photons, let's say $n$ of them emitted per second by the light bulb. Since the shorter wavelength photons have more energy per photon less of them will be needed to make up the $400\ J$ per second energy than the longer wavelength light bulb. You can solve the problem by equating the emitted energies of the two light bulbs which will give you the fractional relationship between the required amounts of photons of the light bulbs.

 

[thiotimoline]

An ultraviolet light bulb, emitting at 400nm and an infared light, emitting at 700nm are both rated at 400 W.
a. Which bulb radiates more photons at the greater rate?
I thought about it long and hard and only thing i can think of
is to use the rleationship: v=lambda*f
where v would be C- the speed of light
solve for the frequency and we know the wavelength for each bulb.
b. How many more photons does it generate per second than the other bulb?
I was thinking of using the following relationship:
E=hc/Lamda
Determine the energy, and we know that Power=Energy/time
solve for the time?

The frequency determines the strength of the radiation, ie the higher the frequency the more energy the radiation carries. The power watt determines the intensity of the radiation, ie the brighter the more intensified. For your question, both are shining at 400W, so both are emitting photons at the same rate, with one of them emitting more energetic photons.

 

[Hootenanny]

For your question, both are shining at 400W, so both are emitting photons at the same rate, with one of them emitting more energetic photons.

You may want to check that one. Both sources are of the same power and therefore emit energy at the same rate. The two sources emit photons of differing energies and therefore must be emitting photons at different rates.

 

[thiotimoline]

You may want to check that one. Both sources are of the same power and therefore emit energy at the same rate. The two sources emit photons of differing energies and therefore must be emitting photons at different rates.

Negative. I think you should check yours instead, coz I remember that frequency and intensity are two different things. As I mentioned, frequency determines the energy of the radiation while the intensity (power) determines the rate at which photons are being emitted. For the question, for eg both radiation are emitting at n photons per unit time but with one of them being more energetic.

 

[nrqed]

Originally Posted by Hootenanny
You may want to check that one. Both sources are of the same power and therefore emit energy at the same rate. The two sources emit photons of differing energies and therefore must be emitting photons at different rates

...

Negative. I think you should check yours instead, coz I remember that frequency and intensity are two different things. As I mentioned, frequency determines the energy of the radiation while the intensity (power) determines the rate at which photons are being emitted. For the question, for eg both radiation are emitting at n photons per unit time but with one of them being more energetic.

I think that there is a misunderstanding here. What Hoot said is correct.

Intensity does not *only* determine the rate at which photon are being emitted. It does, but that's not the only variable.

Intensity is basically the power output of the source over 4 pi r^2 (for a point source). The power output of the source is the number of photon emitted per unit time multiplied by the energy of each photon . So there are *two* variables here. The rate of emission of photons AND the energy of each photon.

To answer the OP: the 400 nm photons are more energetic. Since the two lightbulbs have the same power output, they emit the same total energy per second. Since each 400 nm photon is more energetic than each 700 nm photon, this means that the 400 W bulb emitting at 400 nm emits fewwer photons per second than the 400 W bulb emitting at 700 nm.
I am in agreement with Andrevdh.

Patrick

 

[SonnieInn]

v(frequency) = c/λ
c = 299792458 m/s (exact)​

400 nm => 749.481145 THz
700 nm => 426.27494 THz

E = hv
h = 6.62606896 x10-34 Joule seconds (+/- 0.00000033 x10-34 Js)​

Thus, E = ch/λ

400 nm => 749.481145 THz => 496.6113751 x10-21 Joules (per photon)
700 nm => 426.27494 THz => 283.777928629 x10-21 Joules (per photon)

P = 400 Watts = 400 Joules per second
photons per second = P / (J/photon)
Thus, photon emission rate = P/E = Watts x λ / (c h)​

400 nano meters => 805.45879546 billion billion photons per second
700 nano meters => 1409.55289206 billion billion photons per second

REFERENCES
speed of light (c) =
http://physics.nist.gov/cgi-bin/cuu/Value?c|search_for=speed+of+light

Planck constant (h) =
http://physics.nist.gov/cgi-bin/cuu/Value?h|search_for=plancks+constant

Related Source
http://hyperphysics.phy-astr.gsu.edu/HBASE/mod2.html