How do I calculate watelength of photon and energy contained by photons?

Question

Ok, so the question goes 'A photon has a frequency (v) of 9.73x10^14Hz. Use the relationships E=hv and c= (gamma sign - can't do it on here!) v to calculate the wavelength of the photon and the energy contained by 1mol of such photons. To what region of the electromagnetic spectrum does this photon belong?
c=2.9979x10^8 m s-1 h=6.626x10^34 J s-1 Na = 6.022x10^23 mol-1


Please please don't just tell me the answer because I already have that, I need to know How to do it. Also if you know any website that can explain all this stuff that would be a Great answer too! Thanks:)

I said gamma sign...duh I meant lambda!! And it's 'wavelength' not watelength...:s

Answer 1

To find wavelenght
wavelenght = c/v
= (2.9979x10^8) / 9.73x10^14

To find energy
Energy = (plancks constant) x Frequency
= (6.626x10^34) x (9.73 x10^14)

Answer 2

You have the frequency, and Hz is 1/second.
So if you multiply the frequency you were given by Plank's constant (h), then you will get the energy of that photon in Joules.
If you have one mole of photon, you multiply that energy by 6.022E23 (which is the number of particle per mole)

As for the wavelength, you have the speed of light, in m/s. Multiply the frequency by the speed of light, (the "Hz" multiplied by "second" cancel each other) and you have an answer in metre for the wavelength.

Note this: this is an important and VERY useful tool, that is called unit analysis. If you need to have a length, which should be expressed in "m" then look at what you have that is expressed in m times whatever, and find the whatever than will cancel it by multiplying or dividing. You will always have a consistent unit, the only thing that could be off is if there is a proportional constant added (a pure number, with no dimension).
You do not need a website, all you need is in that last paragraph.

Answer 3

Hmmmmm.....
The wavelength is worked out by using c= lamda v
Rearranging gives: lamda = c/v
Avogadro's number has units (particles) per mol, so i assume you are trying to work out the energy of 6.022 x 10^23 photons.
So the answer's 6.022 x 10^23 x energy of a photon.
Energy of a photon is calculated using E= hv.
The region of the EM spectrum is worked out by looking at a textbook or website (try http://www.electro-optical.com/bb_rad/emspect.htm)and looking at the wavelength and which region it falls under: radio, micro, infra-red, visible, ultra violet, x-ray, gamma. The shorter the wavelength, the higher the energy.

Answer 4

Okay, change the subject of the speed of light equation to make the wavelength the subject:

lambda = c/v

Now, you have been given the value for c to use and you have been given v.

For the energy in one mol of the photons, first calculate the energy of one photon, given by hxv = ? (I am using x for "times")

I mol is a quantity which you have been given as 6.022x10^23

So the energy in one mol of these photons is:

6.022x10^23 x h x v

6.022x10^23 x 6.626x10^34 x 9.73x10^14

Try searching in www.wikipedia.org or www.howstuffworks.com

Answer 5

it really is a thanks to unravel it: we've: h=6.626*10^-34 j.s (plank consistent) v=102.3 MHz first 102.3 MHz to Hz the end result will be 102.3*10^6 Hz then: locate E=capacity E=(h)(v) E=(6.626*10^-34 j.s)(102.3*10^6 Hz) =6.818*10^-26 J