Finding momentum of a photon/ Finding the mass of a object

AI Thread Summary
Both participants in the discussion calculated the momentum of a 140 eV photon correctly, using different methods. One used the energy-momentum relation directly, while the other derived the wavelength first before calculating momentum. For the second question regarding the mass of a microscopic object, both approaches yielded the same mass value, demonstrating the validity of their methods. The conversation highlighted that both methods are acceptable, and the choice often depends on personal preference and clarity. Ultimately, the discussion emphasizes that multiple approaches can lead to correct answers in physics problems.
LionLieOn
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Homework Statement


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?

Homework Equations

The Attempt at a Solution



Answers for the first question.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
 
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For 1) I would say the photon has a momentum of ##140 eV/c##.
 
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PeroK said:
For 1) I would say the photon has a momentum of ##140 eV/c##.

Ohh did I get the unit wrong for the momentum of a photon?
 
LionLieOn said:
Ohh did I get the unit wrong for the momentum of a photon?
If you measure energy in ##eV## it would seem appropriate to measure momentum in ##eV/c##.
 
PeroK said:
If you measure energy in ##eV## it would seem appropriate to measure momentum in ##eV/c##.

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

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

If you want the momentum in SI units just convert energy to Joules and divide by ##c##. No need for any rigmarole.
 
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.
 
PeroK said:
If you want the momentum in SI units just convert energy to Joules and divide by ##c##. No need for any rigmarole.
TJGilb said:
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.

Thank you guys very much!
 
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