# "how to calculate Planck const at home"

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See in how many equations there is used h. You can reverse these equations.

It's basic mathematics - searching for unknown y, when there is known x... y=f(x)

Plug x to equation, and solve it, and you have y calculated.

But in pure math, values are usually meaningless. While in physics, they're measured quantities from real world.

E=h*c/wavelength

so after reversing equation we're receiving:

h=E*wavelength/c

E = Q*U

Q=I*t

so

E=I*t*U

so you have

h=I*t*U*wavelength/c

I - current from ampere meter

U - voltage from voltmeter

t - time from stopper

c - constant, but can be measured in experiment

wavelength - can be measured using optics equations..

For single electron Q=e, so equation simplifies to:

h=e*U*wavelength/c

When U is less than 1.9 V, and wavelength is 650 nm, red LED won't emit light.

If it's higher it's emitting light.

Similar for green,blue,ultraviolet diodes, but different voltages U will be needed.

Edited by Sensei
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h is the planck constant

Or height.

lol true enough

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lol true enough

Or height.

flash light????? ohhhhhhhhhhhhhhhhh, height, I see, Opps sorry..

See in how many equations there is used h. You can reverse these equations.

It's basic mathematics - searching for unknown y, when there is known x... y=f(x)

Plug x to equation, and solve it, and you have y calculated.

But in pure math, values are usually meaningless. While in physics, they're measured quantities from real world.

E=h*c/wavelength

so after reversing equation we're receiving:

h=E*wavelength/c

E = Q*U

Q=I*t

so

E=I*t*U

so you have

h=I*t*U*wavelength/c

I - current from ampere meter

U - voltage from voltmeter

t - time from stopper

c - constant, but can be measured in experiment

wavelength - can be measured using optics equations..

For single electron Q=e, so equation simplifies to:

h=e*U*wavelength/c

When U is less than 1.9 V, and wavelength is 650 nm, red LED won't emit light.

If it's higher it's emitting light.

Similar for green,blue,ultraviolet diodes, but different voltages U will be needed.

Copied this and will practice over and over again.

Just a question, can an exponent just be used like this in a calculator..

example:

4.768^-12

0000000000004.768

or

4.768^12

4.768000000000000

Or something like that.

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your decimal places are still wrong

4.768^10^-12 = 000000000004768

4.768*10^12 =4768000000000

where ever your decimal point is move it 12 spots to the left for -e

for positive exponent move over 12 spots to the right

1.0*10^1=10 take 1.0 move decimal 1 spot to right (1.0*101 =1.0*10=10)

1.0*10^-1=0.1 take 1.0 move decimal 1 spot left (1.0*10-1 =1.0/10 =0.1)

1.0*10^2=100 take 1.0 move decimal 2 spots right 1.0*10^2 =1.0*(10*10)

1.0*10^-2=0.01 take 1.0 move decimal 2 spots left 1.0*10-2 =1.0/(10*10) calculate the bracket first 1.0/100 =0.01

4.999*10^10 take 4.999 move decimal 10 spots right 4999000000.0

4.999*10^-10 take 4.999 move decimal 10 spots left. 0.0000000004999

this rule works only if your multiplying a number by 10x or 10-x

so take 4.999 *10^2 = 49.99

4.999 *10^-2 =0.04999

now do it the opposite way write 4999 in scientific notation count the number of decimal spots till you have 4.999

(when your coverting in this direction count in the opposite direction)

4999=4999.0 =4.999*10^3 count left x spots then write a positive x

write 0.000000499 in scientific notation 4.99*10^-7 count right x spots then write the -x value

let x=5

4.999*10x replace x with 5 so now you have 4.999*105 =499900.0

Edited by Mordred
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your decimal places are still wrong

4.768^10^-12 = 000000000004768

.....

At least he had a decimal point...

Before we have another thread dragged off topic - could Iwonder perhaps have a look through this

It is almost essential to any proper comprehension of science to be able to rapidly and accurately use scientific notation.

e is elementary charge when it is usually used in physics equation.

...

I would be willing to bet that e is more often Eulers number than elementary charge -- but I am just being pedantic. Really I wanted to high light this

You HAVE TO know which one is used at the moment..

Every scientists know it from context of equation.

This is the absolute crux of the matter. -

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I would be willing to bet that e is more often Eulers number than elementary charge -- but I am just being pedantic. Really I wanted to high light this

That depends on what you do. I am using e as elementary charge everyday, to go from Joules to eV unit and reverse..

e as Euler number is used while calculating decay rate using mean-life, f.e.

Edited by Sensei
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Imagine simple electronic circuit, battery with red LED (Light Emitting Diode).

Single red photon which has 650 nm wavelength has energy E=h*c/650nm = 3.056*10^-019 J

3.056*10^-19 J / 1.602*10^-19 = 1.9 eV (in electron volt unit)

Electron in electronic circuit that has voltage U has kinetic energy E=e*U

so if voltage in above circuit will be smaller than 1.9 V, there will be too little energy to emit red photon, and LED won't shine.

More or less...

- The direct voltage of a diode is not the material bandgap that may relate with the wavelength. Silicon has 1.12eV bandgap but a silicon diode direct voltage is rather 0.55V to 0.7V (more if losses are big).

- Many Led radiate from dopant levels deep in the bandgap to obtain the desired colour, typically the green and yellow ones made of GaP. Then the photon energy is significantly less than the bandgap.

- The direct voltage does not convert in electron kinetic energy. A few microwave components achieve 0.5eV or 1eV electron kinetic energy, so do avalanche components as well. In about all others, the carriers' kinetic energy is small, under the 26meV thermal energy at RT.

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See section "Colors and materials" in article

http://en.wikipedia.org/wiki/Light-emitting_diode

There is table with data how drop of voltage on element is related to wavelength and color of photon emitted by LED.

ps. Iwonderaboutthings had already trouble understanding all these informations, without going into such details.

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Sensei - just a note - Iwonderaboutthings complained once too often that we hadn't banned him when he had demanded that we should; in the end we obliged.

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Maybe he will find peace now in his heart..

He appeared more and more pissed off.

Perhaps we were giving him too much data..

Edited by Sensei
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See section "Colors and materials" in article

http://en.wikipedia.org/wiki/Light-emitting_diode

There is table with data how drop of voltage on element is related to wavelength and color of photon emitted by LED.

And if you had made the computation, you would have seen that it doesn't fit. The direct voltage is not the bandgap.

Also, by choosing the dopant, GaP with the same direct voltage produces red, orange, yellow or green light.

There is some link between voltage and wavelength, but too inaccurate to get the first mantissa position of Planck's constant.

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And if you had made the computation, you would have seen that it doesn't fit.

Did you do calcs? Or even better do experiment in real world electronic circuit?

Article says about wavelength = 760 nm

E = h*c/760 nm = 2.61374448971061E-019 J / e = 1.6315508675 eV

Article's voltage drop = 1.63 V (does match)

Article says about wavelength = 610 nm

E = h*c/610 nm = 3.25646854455747E-019 J / e = 2.0327519005 eV

Article's voltage drop = 2.03 V (does match)

Article says about wavelength = 590 nm

E = h*c/590 nm = 3.36685730877976E-019 J / e = 2.1016587446 eV

Article's voltage drop = 2.1 V (does match)

And so on so on with the rest, with little exceptions (f.e. white led with colorful plastic cover)

I didn't do just calcs in thread.

I have checked it a long time ago before, using colorful LEDs to check it by myself.

Before even seeing that table on wikipedia (that I noticed three days ago).

And drop of voltage on element is in many cases matching photon energy emitted by diode (h*f=e*U) (+ little lost).

Plug diode to breadboard, attach cables to laboratory power supply, with smooth adjustment of current and voltage, set I=10 mA, and start with U=0 V, and spin it until diode is starting emitting light. Compare with wavelength of emitted photon. Done.

Edited by Sensei
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Sensei, you've derailed.

I used Leds possibly before you were born. I started playing with electronics at 10.

My first profession was engineer for semiconductors, including III-V compounds and optoelectronics - so yes, I've already done such calculations, and more than once.

Read a bit and learn that GaP makes red, orange, yellow and green diodes, all with the same direct voltage. Or better, experiment it: I did, obviously you didn't. GaP alone disproves any fixed relationship between the direct voltage and the wavelength.

9 months ago you still believed that energy was quantized in multiples of h, here on this forum. Come on!

Go on learning, try to separate very accurately what you know from what you suppose, and don't replace knowledge by loud claims.

Edited by Enthalpy
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!

Moderator Note

Civility, please. Let's ease up on the personal angles and talk about some science.

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Here's a datasheet for a series of LED diodes

http://www.tme.eu/de/Document/c20126608a1d3a620ecfe31f04c5b3c7/L-1002YD.pdf

- Some are just GaP, others have a GaAsP heterojunction on GaP (p. 1)

- The typical direct voltage varies between 2.0V (for red) and 2.25V (red as well) passing by 2.1V (yellow) and 2.2V (green) (p. 2)

- The center wavelength varies from 700nm to 565nm (p. 3)

The lower voltage corresponds to the heterojunction: yellow and high-eficiency red. The homojunction red and green take a higher voltage. Especially, bright red (700nm) requires a higher voltage (2.25V) than yellow (590nm and 2.1V), because this relates to the bandgaps and junction, but not directly with the emitted wavelength.

What is not in Wiki's diagram about Leds is that they often use deep dopant levels to make the light, GaP especially so. Then the wavelength doesn't relate simply with the bandgap. Heterojunctions are one other complication and are common.

What is in Wiki's diagram is that the direct voltage of a diode is smaller than the material's bandgap.

Let's take the bright red example: 2.25V direct voltage, but 700nm make only 1.77eV photon energy because of deep dopant.

Or take this green Led based on gallium nitride:

http://datasheet.octopart.com/IF-E93-Industrial-Fiberoptics-datasheet-63924.pdf

- Typical forward voltage 3.5V

- 520nm are 2.4eV photon energy, because a deep dopant makes the green light.

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