- #1
Mechatron
- 38
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Is this equation equal to:
(e^(hf/kT)) - 1
or
e^( (hf/kT) - 1 )
http://s29.postimg.org/le6iqy3rb/exp.png
(e^(hf/kT)) - 1
or
e^( (hf/kT) - 1 )
http://s29.postimg.org/le6iqy3rb/exp.png
Last edited by a moderator:
First off, what you wrote is NOT an equation. An equation always has an = symbol in it.Mechatron said:Is this equation equal to:
(e^(hf/kT)) - 1
or
e^( (hf/kT) - 1 )
http://s29.postimg.org/le6iqy3rb/exp.png
Why did you put the -1 in the exponential? The parenthesis limit the argument of exp to hf/kT.Mark44 said:First off, what you wrote is NOT an equation. An equation always has an = symbol in it.
The image in the link is [exp(hf/kT) - 1].
What you have written is ambiguous, as what you probably meant is this:
$$e^{\frac{hf}{kT} - 1}$$
What you actually wrote, though, is this:
$$e^{\frac{hf}{k}T - 1}$$
The brackets - [] - around the entire expression are unnecessary.
The posted image, which doesn't have the -1 term, doesn't match the expressions in the first post. In the first post Mechatron asks about these expressions:DrClaude said:Why did you put the -1 in the exponential? The parenthesis limit the argument of exp to hf/kT.
DrClaude said:My guess is that Mechatron did not write that himself, but saw it in a book. It's most probably related to the Planck distribution (blackbody radiation). As economicsnerd said, the correct reading is
$$
e^{\beta h f} - 1 \mbox{ where } \beta = \frac{1}{kT}
$$
The additional bracket [] might be there because it is part of a greater equation.
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