1. Jul 12, 2012

### Johnahh

there is a question in it as follows;
The second is defined as 9,192,631,770 periods of radiation from a caesium-133 atom, The metere is defined as the distance travelled by light in 1/299792458 of a second. How many wavelengths of caesium-133 radiation used to define the second is this?

So I started with 1/299792458 this equals 3.3x10^-9
then 9,192,631,770 / 3.3x10^-9 but obviously when dividing powers you minus them and a minus and minus make a plus this gives the answer 2.8x10^18.
I must be doing something wrong as this is a higher answer than the initial amount of radiation in a second?
This is not a homework question but not being able to work it out aggravates me!

2. Jul 12, 2012

### mikeph

1 second is equivalent to 9192631770 cycles
frequency = [9192631770 cycles]/[1 second] = 9192631770 s^-1

1 metre is travelled in 1/299792458 seconds
speed = [1 metre]/[1/299792458 seconds] = 299792458 ms^-1

the wavelength is speed/frequency = [299792458 ms^-1]/[9192631770 s^-1] = 0.0326 m

Think that's right... going from words to numbers is not my strong point

3. Jul 12, 2012

### Johnahh

Thanks for the reply but your answer is in metres? I am after how much radiation is emitted from a caesium-133 atom in the time it takes light to travel 1 metre. which is 3.3x10^-9 s

4. Jul 12, 2012

### Simon Bridge

If you have a period T, then the number of periods in S seconds is S/T - you did it ... different.
You divided the number of periods in one second - which is 1/T - by the number of seconds S ... when embarking on something like this is is useful to be able to visualize what you are talking about somehow. Thus MickyW's reply. It also helps to do the algebra before you sub in the numbers.

The wavelength of the key spectral line is the speed of light times the period or $\lambda = cT$ for:
299792458(m.s-1)/9192631770(s-1)=0.032612m

You appeared to be trying to calculate how many periods of the caesium line fit in 1m of travel... so, since the wavelength is the distance gone in one period, which is 0.03m, how many oscillations fit into a whole meter?

Thus MickyW did give you the answer you wanted and in a way easy to visualize.
OR you could just correct the original calculation as per the first sentence (above) :)

Note1: of course if you really mean "how much radiation" then that is a whole different kettle of piranhas.
Note2: even when it is not homework, the answer won't be just spoon-fed to you. That would be disrespectful to you and besides: you learn better when you do it for yourself.

Last edited: Jul 12, 2012
5. Jul 12, 2012

### Johnahh

This is the part I didnt figure out, one caesium wavelength is 0.03m therefore 1/0.03m = 33.33....
My apologies mikeyW you were in fact correct just I didn't work out the next step as I am not very familiar with wavelength equations.

I stated this as there is a special sub forum for homework and I did not want the post to get moved there :)
Thank you both.

6. Jul 13, 2012

### Simon Bridge

Good on yer mate :)