Homework Help: Finding momentum of a photon/ Finding the mass of a object

1. Mar 20, 2017

LionLieOn

1. The problem statement, all variables and given/known data
So me and my friend were comparing our homework and we noticed that although we had the same answers, we both had different ways of answer them.

So I was just wondering who's work, my friend or mine, is the most correct?

Here's the questions.

1. Calculate the momentum of a 140 eV photon.
2. A certain microscopic object has a wavelength of 8.4 X 10-14 m and a speed of 1.2 X 105 m/s. What is the mass of the object?

2. Relevant equations

3. The attempt at a solution

My work:

140 eV Convert into joules:
E=(1.60 X 10-19)(140)
E= 2.24 X 10-17 J

Finding momentum:

p=E/c

p=(2.24 X 10-17) / ( 3.0 X 108 )

p= 7.466666667 X 10-26
Rounded,
p= 7.5 X 10-26 kgm/s

-------------------------------------------------

My friends work:

140 eV Convert into joules:
E=(1.60 X 10-19)(140)
E= 2.24 X 10-17 J

First find the wavelength:

E=hc/ (Lambda symbol)

(Lambda symbol)= hc/E

(Lambda symbol)= 8.9 X 10-9 m

Now find momentum:

p=h/ (Lambda symbol)

p=(6.63 X 10-34) / (8.9 X 10-9 )

p= 7.5 X 10 -26 kgm/s
----------------------------------------------------------------------------------
Second question.

My work:

(Lambda symbol) = h/mv

m=h/ (Lambda symbol)v

m= (6.63 X 10-34)/ (8.4 X 10-14) (1.2 X 105)

m= 6.577380952 X 10-26)
Rounded,
m= 6.6 X 10-26) kg
---------------------------------------------------------
my friends work:

Finding momentum of the object.

p= h/ (Lambda Symbol)

p= (6.63 X 10-34) / (8.4 X 10-14)

p= 7.89 X 10-21 kgm/s

Now find the mass of the object

p=mv

m=p/v

m= 7.89 X 10-21 / (1.2 X 105)

m= 6.577 X 10-26

Rounded,
m= 6.6 X 10-26 kg

2. Mar 20, 2017

PeroK

For 1) I would say the photon has a momentum of $140 eV/c$.

3. Mar 20, 2017

LionLieOn

Ohh did I get the unit wrong for the momentum of a photon?

4. Mar 20, 2017

PeroK

If you measure energy in $eV$ it would seem appropriate to measure momentum in $eV/c$.

5. Mar 20, 2017

LionLieOn

Hmmm what if I converted 140eV into joules? Would I still use eV/c or put it as kgm/s ?

6. Mar 20, 2017

TJGilb

Your units aren't wrong per se, it's just strange that you would choose to convert to Joules instead of staying in eV which would be much easier in this case.

7. Mar 20, 2017

PeroK

If you want the momentum in SI units just convert energy to Joules and divide by $c$. No need for any rigmarole.

8. Mar 20, 2017

TJGilb

As to which approach is more correct, both are perfectly fine. You'll find that problems in physics tend to have a great number of viable approaches. It's just a matter of which makes the most sense to you.

9. Mar 20, 2017

LionLieOn

Thank you guys very much!