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Mathematics 9 Online
OpenStudy (anonymous):

find the color of light whose photon has 4.75x10^-19 J of energy

OpenStudy (anonymous):

@marcoduuuh

OpenStudy (anonymous):

@TuringTest

OpenStudy (anonymous):

Woaaah. I don't know this. I sowwy. :(

OpenStudy (anonymous):

k no problem marco :)

OpenStudy (anonymous):

@phi

OpenStudy (phi):

do you know the equation for the energy of a photon? \[ E= \hbar \omega \]

OpenStudy (phi):

if you can find the wavelength or the frequency, then you can use a table http://en.wikipedia.org/wiki/Color

OpenStudy (anonymous):

i have a table for the colors just need to know how to find the wavelength since thats all the info i have

OpenStudy (phi):

it is not h it is h bar http://en.wikipedia.org/wiki/Planck_constant

OpenStudy (anonymous):

i know what the h is so thats good enough lol.....but how do i find the wavelength with just the energy??

OpenStudy (phi):

\[ \omega= 2\pi \nu \] \(\nu\) is the frequency

OpenStudy (phi):

yes, you may know h but your equation is wrong it is either \[ E= h \nu \] or \[ E= \hbar \omega\]

OpenStudy (anonymous):

oohhh...wait i think i got it \[E=h\times\nu\]\[4.75x10^{-19}=6.626x10^{-34}\times\nu\]\[\nu\approx 7.17x10^{14}\]the you do\[c=\lambda\times\nu\]\[3.00x10^8=\lambda\times7.17x10^{14}\]\[\lambda\approx4.18x10^{-7}\]

OpenStudy (phi):

yes, that looks good

OpenStudy (anonymous):

so then you convert that to nanometers and get\[\lambda\approx418.41~nm\]

OpenStudy (anonymous):

and thats Violet

OpenStudy (anonymous):

I ROCK T(^.^T)

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