macho_investigator
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Can "separation" or "radiation" energy be calculated from mass excess (in keV)?
This book has newly discovered or updated atomic masses. 2016-2017.
See NUBASE2016 pdf.
Taken from this page: https://www-nds.iaea.org/amdc/
Now as you see they have 2 versions of data. 2012 and 2016.
They also have 2 variants: AME and NUBASE. I am more interested in NUBASE.
Now they already have separation energies, they call it reaction energies, for AME. But I wanna calculate NUBASE ones. But for NUBASE they only gave
mass excess, half-life, decay mode, and excitation energy data.
Can I somehow know separation or radiation energies just by knowing already provided data in NUBASE2016?
Note: separation energy of something is reaction energy of something else
For example: neutron separation energy of He4 is same as reaction energy of He3 + n.
When I say reaction energy maybe it is same as fusion result energy.
There are:
neutron separation energy
proton separation energy
alpha separation energy
2 protons separation energy...
as you can see we are playing with balls here, just like some game, we could think of million separation energies.
But neutron and proton separation energies is what i wanna know.
Binding energy is energy released when nucleus is made from nucleons (electrons and protons), correct?
It's negative of course, because it's released, or we need to add it to separate atom into nucleons.
Separation energy is energy needed to separate "something" from atom?
Like neutron...
How can we know separation energy?
That pdf has table that goes until end of document...it has values.
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ShadowWeirdo
Reincarnation of troll PhDChemist
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Table III. Nuclear-reaction and separation energies
EXPLANATION OF TABLE
We present, for all nuclides for which such data can be derived, separation energies (in keV) of particles (or groups of particles)
and nuclear-reaction energies obtained as the following combinations of atomic masses (see accompanying diagram):
Q(β − ) = M(A,Z)−M(A,Z+1) (in Part I) (a)
Q(2β − ) = M(A,Z)−M(A,Z+2) (b)
Q(4β − ) = M(A,Z)−M(A,Z+4) (c)
Q(β − n) = M(A,Z)−M(A−1,Z+1)−n (d)
S(n) = − M(A,Z)+M(A−1,Z)+n (e)
S(p) = − M(A,Z)+M(A−1,Z−1)+ 1 H (f)
Q(εp) = M(A,Z)−M(A−1,Z−2)− 1 H (g)
S(2n) = − M(A,Z)+M(A−2,Z)+2n (h)
Q(d,α) = M(A,Z)−M(A−2,Z−1)− 2 H− 4 He (i)
S(2p) = − M(A,Z)+M(A−2,Z−2)+2 1 H (j)
Q(p,α) = M(A,Z)−M(A−3,Z−1)− 4 He+p (k)
Q(n,α) = M(A,Z)−M(A−3,Z−2)− 4 He+n (l)
Q(α) = M(A,Z)−M(A−4,Z−2)− 4 He (m)
Other reaction energies can be derived from the given data with the help of the following relations:
Q(γ,p) = − S(p)
Q(γ,n) = − S(n)
Q(γ,2p) = − S(2p)
Q(γ,pn) = Q(d,α) − 26071.0939±0.0005
Q(γ,d) = Q(d,α) − 23846.5279±0.0002
Q(γ,2n) = − S(2n)
Q(γ,t) = Q(p,α) − 19813.8649±0.0003
Q(γ, 3 He) = Q(n,α) − 20577.6194±0.0005
Q(γ,α) = Q(α)
Q(p,n) = Q(β − ) − 782.3465±0.0005
Q(p,2p) = − S(p)
Q(p,pn) = − S(n)
Q(p,d) = − S(n) + 2224.5660±0.0004
Q(p,2n) = Q(β − n) − 782.3465±0.0005
Q(p,t) = − S(2n) + 8481.7949±0.0009
Q(p, 3 He) = Q(d,α) − 18353.0535±0.0003
Q(n,2p) = Q(εp) + 782.3465±0.0005
Q(n,np) = − S(p)
Q(n,d) = − S(p) + 2224.5660±0.0004
Q(n,2n) = − S(n)
Q(n,t) = Q(d,α) − 17589.2989±0.0005
Q(n, 3 He) = − S(2p) + 7718.0404±0.0005
Q(d,pn) = 0 − 2224.5660±0.0004
Q(d,t) = − S(n) + 6257.2290±0.0005
Q(d, 3 He) = − S(p) + 5493.4744±0.0001
Q( 3 He,t) = Q(β − ) − 18.5920±0.0001
Q( 3 He,α) = − S(n) + 20577.6194±0.0005
Q(t,α) = − S(p) + 19813.8649±0.0003
https://www-nds.iaea.org/amdc/ame2016/AME2016-b.pdf
[Edited on 22-1-2018 by ShadowWeirdo]
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