jennilalala
Calculate the energy in J of 0.32 moles of photons whose frequency is 2.6 x 10^15 ?
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SheldonEinstein
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You must know the formula for calculating the energy... in this situation.
Do you know the formula?
aaronq
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E=hv
h is planks constant
v= frequency
this gives you the energy of 1 photon
so multiply this by the amount of photons you have:
amount of photons = avogadros constant x number of moles
SheldonEinstein
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that v is actually "new (pronounced as new)"
SheldonEinstein
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So @jennilalala with the help of : E = hv formula, can you find the energy by putting the values given ?
SheldonEinstein
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"v" --> \(\large{\color{blue}{\nu}}\) .... right @aaronq ?
jennilalala
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I am back sorry ahaah. yes i know the formulas. i just dont know what to do if it says ".32 moles of photon.."
SheldonEinstein
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Do you the planck's constant ?
jennilalala
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6.626x10^23 ?? not by heart...yet...
aaronq
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haha yep, v= "new"
jennilalala
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10^-34
SheldonEinstein
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OK so it is \[\large { h = 6.626 \times 10^{-34} m^{-1}}\]
SheldonEinstein
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oh wait it is m^2 kg/s
SheldonEinstein
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So right, now solve this :
\[\large{E = 6.626 \times 10^{-34} \times 2.6 \times 10^{15}}\]
What do you get?
jennilalala
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1.73x10^-18
SheldonEinstein
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My calculator says it as : \(\large{\color{blue}{ 17.2276 * 10^{-19}}}\) that is 1.72276 * 10^{-18}
SheldonEinstein
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OK so now .. this is the energy for \(1\) photon ... now we to calculate for 0.32 moles of photons.
SheldonEinstein
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See ... no. of photons = Avogadros's constant * no. of moles of photons...
SheldonEinstein
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You know avogadro's constant?
jennilalala
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6.022x10^23
SheldonEinstein
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Good! So now, solve this one :
\[\large{\color{red}{\mathbb{No. of Photons}} = \color{blue}{6.022 * {10}^{23}} \times \color{green}{0.32}}\]
jennilalala
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1.93x10^23
jennilalala
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then i multiply the two calculations?
SheldonEinstein
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Correct! So this is your No. of photons.
Now multiply this result with "1.73 * 10^{-18} "
jennilalala
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3.3x10^-5
SheldonEinstein
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Right!
jennilalala
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not negative typo lol
SheldonEinstein
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No problem.
SheldonEinstein
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So finally you have your correct way, I hope you got it ?
jennilalala
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Yes thank you so much!
SheldonEinstein
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You're welcome @jennilalala . keep up the good work.