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

In summary, after attempting to solve a problem involving the deuteron and using the reduced mass, binding energy, and distance, it was determined that the value of Kappa is 5939.86 and the result of -Kappa*R is 8.90*10^-13. However, there may have been some errors in the calculation of units and the value of fm was attempted to be converted into cm.
  • #1
Labboi
5
0
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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  • #2
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.
 
  • #3
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?
 
  • #4
Charles Link said:
fm=1.0 E-15 m
I disagree. One fm is ##10^{-6}~\rm nm##.
 
  • #5
When you put in for the mass, ## m=E(in \, MeV)/c^2 ##.
 
  • #6
vela said:
I disagree. One fm is ##10^{-6}~\rm nm##.
Then, in fact, you agree with @Charles Link!
 
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  • #7
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
 
  • #8
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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  • #9
I did, just forgot it in the equation, sorry.
 
  • #10
Thank you, I must have messed up my parentheses, it's been a long night.
 

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

What is the importance of using units in scientific calculations?

Units are essential in scientific calculations because they provide a standard way to measure and compare quantities. Without units, it would be impossible to accurately communicate and reproduce experimental results. Additionally, units help to ensure that calculations are performed correctly and that the final answer is in the correct form.

What is a normalized function?

A normalized function is a mathematical function that has been scaled or adjusted in such a way that it has a maximum value of 1. Normalization is often used in data analysis to compare different datasets or to remove the effects of varying scales.

Why is it important to solve a normalized function?

Solving a normalized function can provide valuable insights into the underlying patterns and relationships in a dataset. It can also help to simplify complex data and make it easier to compare and interpret.

How do units affect the solution of a normalized function?

The choice of units can affect the solution of a normalized function as it can impact the scale and magnitude of the data. For example, using different units for temperature (e.g. Celsius vs. Fahrenheit) can result in different normalized values and potentially alter the interpretation of the data.

What are some common mistakes when dealing with units in scientific calculations?

Some common mistakes when dealing with units in scientific calculations include forgetting to convert between units, using inconsistent units for different quantities, and not paying attention to the units in the final answer. It is important to always double-check units and ensure they are consistent throughout the calculation process.

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