# B Measurement standard for joules = Momentum in kilograms

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1. Nov 18, 2016

### Camdenl

Basically, I found the energy of a photon using its frequency ((6.8*10^15) and E=(3.6496394*10^-16)) and so I then used "p=E/c" to find the momentum and got 1.2173887*10^-24 however I don't know the measurement of the momentum. Is there a better equation to use to find momentum, and what would the measurements be?

(edit) I'd like to know the momentum in kilogram-meters per second.

2. Nov 18, 2016

### BvU

Hi Camden,

What are the units of the E you found ? And what are then the units of p if you use c = 3 108 m/s a (as it seems you do) ?

And, eh, if you have 2 digits of $f$, you should end up with 2 (perhaps 3) digits of p, not 8

3. Nov 18, 2016

### Camdenl

The "E" was in joules, and I would like to find out the momentum "p" in kilogram-meter per second.

4. Nov 18, 2016

### ZapperZ

Staff Emeritus
Two things:

1. Pay attention to significant figures! It is meaningless to write all those numbers up to 7 decimal places.

2. If you keep all your units in SI (i.e. energy in Joules, c in m/s), then your momentum will also be in SI units that you are looking for. Remember that "Joules" is equal to kg.m2/s2.

Zz.

5. Nov 18, 2016

### BvU

So, Camden, is it clear to you now that you had p in the units you wanted all along ?

6. Nov 18, 2016

### Cutter Ketch

First, just out of curiosity I checked the wavelength. This is 44 nm, a pretty hard x-ray. That's fine, I just thought I'd check and make sure that is what you intended.

You say you took this frequency and found the energy. How? E = h f. These aren't just numbers. E has units. h has units. f has units. You could use completely different units for h and get a completely different number for E. The numbers would mean the same energy if they are written with their units. Completely different numbers meaning the same thing with units should underscore the point that the numbers don't mean anything without units. You must use units.

So I'm looking for the units you might have used for h. In SI units of J-s h is 6.626E-34.

6.8E15 / s * 6.626E-34 J - s = 4.5E-18 J

So your energy doesn't seem to be Joules. In fact I don't see any obvious choice of units for h that would produce your number. Where did that energy number come from?

So a J = kg m^2 / s^2

E / c = p so

(4.5E-18 kg m^2 / s^2) / ( 2.997E8 m/s) = 1.5E-26 kg m/s

in the units you requested

7. Nov 18, 2016

### BvU

[oh boy, it didn't occur to me to check if Camden computed $h\nu$ properly .... kudos CK!]

8. Nov 18, 2016

### Camdenl

The numbers don't have 7 decimal spots, they have 24, they're shortened. For example, 1.2173887*10^-24 is equal to 0.000000000000000012173887. The whole thing's a decimal.

9. Nov 18, 2016

### Camdenl

Thanks, I'm fairly new to stuff like this since I'm learning out of personal interest.

10. Nov 18, 2016

### ZapperZ

Staff Emeritus
Since you are doing your own personal lesson, I suggest you look up the meaning of "significant figures".

Zz.

11. Nov 19, 2016

### Camdenl

Oh, I got it, I just didn't understand. I'm pretty new to this stuff, sorry

12. Nov 21, 2016

### Camdenl

I'm sorry, but I checked for 44 nm on a wavelength chart, and it came under uv light. Did I make some mistake, or what?

13. Nov 21, 2016

### BvU

Wavelength is fine: C = 3 10 8 m/s divided by frequency 6.8 1015 1/s gives 4.4 10-8 m or 44 nm.
An a chart like this ranges it as far uv, very soft x-ray.
(Same chart shows an energy between 10 and 100 eV)

Now we would like to now how you found your E :

did you use a formula ? which ?

14. Nov 21, 2016

### Camdenl

I adopted the frequency from "http://www.csun.edu/~jte35633/worksheets/Chemistry/5-2PlancksEq.pdf" and then attempted to use the formula available on that site. The probable reason for why I got my E would likely be because of my novice in this subject. Also, could you explain to me how I could use nm and convert it so I could use it in an equation?

Last edited: Nov 21, 2016
15. Nov 21, 2016

### BvU

Ok, I see the frequency 6.8 1015 1/s in question 1. They give $\ h$ = 6.63 10-34 J s .

What do you get when you apply $\ E=h\nu\$ ?

Nanometers are 10-9 m. So 4.4 10-8 m / 10-9 (m/nm) = 44 nm.

And conversely: wavelength in nm x 10-9 (m/nm) = wavelength in m

16. Nov 21, 2016

### Cutter Ketch

Ah, yes. I was completely off base calling that X-ray. My typing out ran my brain. My apologies.

17. Nov 21, 2016

### Camdenl

So I grabbed plancks constant for joules(6.626070040(81)×10^−34) and multiplied it by the frequency(6.8×10^15) using the google calculator it came out at 3.6496394e-16. Also, thanks for the info on nanometers.

18. Nov 21, 2016

### jbriggs444

If you multiply 6.8 times 6.6, what do you get? If you subtract 15 from 34, what do you get?

Always do the easy sanity checks.

19. Nov 21, 2016

### Camdenl

Ok, thanks.

20. Nov 25, 2016

### BvU

The google calculator (in my case I just type it in on the search bar) tells you what it does to calculate the result:

6.626070040 * (81) * (10^−(34)) * 6.8 * (10^15) just above the 'result' 3.6496394e-16​

6.626×10^−34 * 6.8×10^15 works a lot better: 4.50568e-18​

So it uses e notation for the result. Funny enough it doesn't like same notation for input

6.626e−34 * 6.8e15 is read as (6.626 * e) − (34 * 6.8e15) = -2.312e+17​

And I (seasoned Fortran programmer) found no way to enter a negative exponent of 10 using this e notation. (anyone ?)

Someone should tell those guys to clean up their act. School kids (and lots of others) blindly copy the nonsensical result !

21. Nov 25, 2016

### Camdenl

Thanks for the info!!