How Do You Calculate Neutron Wavelength?

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SUMMARY

The discussion centers on calculating neutron wavelength using the formula λ^2 = h^2/m^2v^2. Participants clarify the placement of variables, specifically the confusion between mass (m) and meters (m). The correct interpretation is that 'm' in the formula represents mass, which should be in the numerator, while the unit 'm' refers to meters. The conversation highlights the importance of proper notation and understanding in physics, especially for those revisiting the subject after a long time.

PREREQUISITES
  • Understanding of quantum mechanics concepts
  • Familiarity with the Planck constant (h)
  • Knowledge of kinetic energy in electron volts
  • Basic algebraic manipulation of equations
NEXT STEPS
  • Study the derivation of the de Broglie wavelength formula
  • Learn about the implications of quantum mechanics on particle behavior
  • Explore the relationship between energy and wavelength in quantum physics
  • Review common notation in physics to avoid confusion
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Students of quantum mechanics, physicists revisiting foundational concepts, and educators teaching advanced physics topics.

jjson775
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Homework Statement
Show that the wavelength of a neutron is (2.86x10-11/√K)m where K is the kinetic energy in electron volts.
Relevant Equations
λ=h/mv
λ^2=h^2/m^2v^2
=h^2/2/(m^2v^2/2)
=h^2/2/(Km)
=h^2/2Km
 
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Is there a question here ?

As an urgent hint: use brackets where necessary !
 
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BvU said:
Is there a question here ?

As an urgent hint: use brackets where necessary !
There are 2 brackets missing in the homework statement. They look OK to me in the attempted solution. I can't get to the answer (λ in terms of K and m) from my last line. m should be in the numerator.
 
jjson775 said:
m should be in the numerator.
You mean denominator.
 
kuruman said:
You mean denominator.
 
No, numerator is correct. Actually, when I looked again at the homework statement, the brackets are OK.
 
jjson775 said:
No, numerator is correct. Actually, when I looked again at the homework statement, the brackets are OK.
The mass in the numerator is not correct.
##K=\dfrac{p^2}{2m}~\Rightarrow~p=\sqrt{2mK}##.

##\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2mK}}.##
 
m is shown in the numerator in the textbook. See attached file.
 

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  • #10
See Problem 3
 

Attachments

  • 7B238C36-30C8-4F9F-81B1-9BA3605D85B3.jpeg
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  • #11
That "m" stands for "meters" not "mass". In other words, if you enter a number for the kinetic energy in electron volts, take the square root and divide the result into 2.86 x 10-11, the result will be in meters. I dislike mixed unit formulas of this kind exactly because of the confusion they sow as we have seen in this case.

Do you know how to proceed now?
 
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  • #12
jjson775 said:
See Problem 3
The variable ##m## is usually typeset in an italic font whereas the unit m is not.
 
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  • #13
vela said:
The variable ##m## is usually typeset in an italic font whereas the unit m is not.
Unfortunately the textbook page where the formula appears was posted a few replies after the confusion had already set in.
 
  • #14
I am actually past this point in the book and knew I was hung up on something silly with this problem, which was the case. I mixed up the m for mass with meters. A mental block. I am a retired engineer, 78 years old, seeing modern physics for the first time. Past special relativity and now in QM. Thanks.
 
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  • #15
jjson775 said:
I am actually past this point in the book and knew I was hung up on something silly with this problem, which was the case. I mixed up the m for mass with meters. A mental block. I am a retired engineer, 78 years old, seeing modern physics for the first time. Past special relativity and now in QM. Thanks.
Good luck with your endeavor. We are here to help.
 

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