Reduced mass to atomic mass units conversion help please

In summary, the conversation is about understanding how to convert the reduced mass into atomic mass units. The equation for reduced mass is $m_1m_2/(m_1+m_2)$ and for two similar masses, it becomes $m^2 / 2m = 1/2 * m$. To convert to atomic mass units, the book uses the value of 1/2u, which is obtained by substituting the mass of 4u for $m_4$ in the equation.
  • #1
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Hello

I am trying to understand how to write the reduced mass into atomic mass units but i am confused how it was done.

The equation is

$$m_1m_2/(m_1+m_2)$$

For two similar masses in my particular case i have:

$$m^2 / 2m = 1/2 * m$$

Then to convert to atomic mass units, the book says it gives:
1604877616274.png


But how did they get that value? I had $$1/2 * u$$My conversion must be failing me, i am not sure how they got that value..
 
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  • #2
So the book has ##m_4 = 4u## while you have ##m_4 = u## ?

Hard to tell which is which on PF since ##m_4## hasn't been introduced here. Is it described somehow in the book ?
 
  • #3
@BvU it was representing a $$_2^4 He$$

Don't know how to write it inline with my sentence, anyway it does seem I forgot that it would be 4u.
 
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Likes BvU
  • #4
All clear ! (the subscript 4 was a giveaway :wink: )

Inline ##\LaTeX## by enclosing it in double #: ##\LaTeX##
 

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