Problem with units? (trying to solve a normalized function)

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The discussion revolves around solving a deuteron problem involving the reduced mass, binding energy, and the conversion of units. The user is struggling with the calculations, particularly with the value of Kappa and the correct application of units. There are disagreements on the conversion of femtometers to nanometers and the correct mass calculation using energy and speed of light. Participants emphasize the importance of including units in calculations and the need for careful arithmetic. The conversation highlights common pitfalls in physics calculations and the significance of unit consistency.
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Homework Statement
I'm trying to solve a normalized function and cannot seem to get the right answer.
Relevant Equations
Kappa=sqrt(2m(EB)/hbar^2))
-kappa*R=-0.34
I'm trying to solve for this in a deuteron problem. But can't seem to get the right answer.
The reduced mass of the deuteron is 469.4 MeV, the binding energy Eb is 2.226 MeV and R = 1.5fm.

Using hbar = 6.5817x10^-16 eV.s

I get Kappa = sqrt((2(469.4)*2.226)/(6.5817*10^-22)^2) = 6.94*10^16

-kappa*(1.5*10^-13) = 10418.

I feel like I'm messing up units or something.
 
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fm=1.0 E-15 m, but your arithmetic has other errors as well. Google of the mass of the deuteron also is in disagreement.
 
Labboi said:
I feel like I'm messing up units or something.
Like not including them on your answer? Have you tried including the units and working out the end result?
 
Charles Link said:
fm=1.0 E-15 m
I disagree. One fm is ##10^{-6}~\rm nm##.
 
When you put in for the mass, ## m=E(in \, MeV)/c^2 ##.
 
vela said:
I disagree. One fm is ##10^{-6}~\rm nm##.
Then, in fact, you agree with @Charles Link!
 
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Kappa = Sqrt((2(469.4MeV/3.0*10^8m.s^-2)*2.226Mev)/(6.5817*10^-22MeV.s)^2 = 5939.86

-Kappa*R = 5939.86*1.5*10^-13 = 8.90*10^-13

I'm trying to convert the fm into cm
 
You need to take ##c ## to the second power. In addition ## \hbar ## is to the second power inside the square root. You need to do the arithmetic carefully.
 
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I did, just forgot it in the equation, sorry.
 
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Thank you, I must have messed up my parentheses, it's been a long night.
 
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