Download Example 27-3 The Binding Energy of 4He

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Metastable inner-shell molecular state wikipedia , lookup

Nuclear fission wikipedia , lookup

Nuclear fusion wikipedia , lookup

Valley of stability wikipedia , lookup

Nuclear drip line wikipedia , lookup

Nuclear binding energy wikipedia , lookup

Atomic nucleus wikipedia , lookup

Transcript
Example 27-3 The Binding Energy of 4He
Previously, we determined the binding energy per nucleon in the 4He nucleus using the mass of the nucleus. Use the values
from Appendix C to determine the binding energy per nucleon (in MeV>c 2) in the 4He nucleus using the atomic mass of
4
He.
Set Up
We’ll use Equation 27-2 and the values
given in Appendix C for the neutron mass,
the atomic mass of 1H, and the atomic
mass of 4He.
Binding energy of a nucleus:
EB = (Nmn + Zm1H 2 matom)c2
(27-2)
binding energy
of 4He =
rest energy of
rest energy of
rest energy of
+
–
two 1H atoms
two neutrons
a 4He atom
+
+
+
+
Solve
Calculate the binding energy of the 4He
nucleus, which has two neutrons (N = 2)
and two protons (Z = 2).
From Appendix C,
neutron mass = mn = 1.008665 u
atomic mass of 1H = m1H = 1.007825 u
atomic mass of 4He = m4He = 4.002602 u
Substitute these into Equation 27-2, with matom = m4He:
E B = 1211.008665 u2 + 211.007825 u2 - 4.002602 u2c 2
= 0.030378 uc2
Since 1 u = 931.494 MeV>c 2,
E B = 0.030378 uc 2 a
931.494 MeV>c 2
1u
= 28.297 MeV
The binding energy per nucleon equals the
binding energy of the nucleus divided by
the number of nucleons in the nucleus.
b
The 4He nucleus has four nucleons (two neutrons and two protons), so
the binding energy per nucleon is
EB
28.297 MeV
=
= 7.0742 MeV
A
4
Reflect
Previously, we calculated E B >A = 7.1 MeV to two significant figures using the mass of the 4He nucleus; our new calculation is consistent with this.
Why would we do this kind of calculation using atomic masses rather than nuclear masses? The reason is that in
general, the masses of neutral atoms have been well measured, but precise measurements of the masses of atomic nuclei
in isolation are difficult to obtain.
Ch27_example.indd 3
9/4/13 12:54 PM