Unformatted text preview: 20. (a) The ﬁrst step is to add energy to produce 4 He → p + 3 H, which – to make the electrons “balance”
– may be rewritten as 4 He → 1 H + 3 H. The energy needed is ∆E1 = (m3 H + m1 H − m4 He )c2 =
(3.01605 u + 1.00783 u − 4.00260 u)(931.5 MeV/u) = 19.8 MeV. The second step is to add energy to
produce 3 H → n + 2 H. The energy needed is ∆E2 = (m2 H + mn − m3 H )c2 = (2.01410 u+1.00867 u −
3.01605 u)(931.5 MeV/u) = 6.26 MeV. The third step: 2 H → p + n, which – to make the electrons
“balance” – may be rewritten as 2 H → 1 H + n. The work required is ∆E3 = (m1 H + mn − m2 H )c2 =
(1.00783 u + 1.00867 u − 2.01410 u)(931.5 MeV/u) = 2.23 MeV.
(b) The total binding energy is ∆Ebe = ∆E1 + ∆E2 + ∆E3 = 19.8 MeV + 6.26 MeV + 2.23 MeV =
28.3 MeV.
(c) The binding energy per nucleon is ∆Eben = ∆Ebe /A = 28.3 MeV/4 = 7.07 MeV. ...
View
Full
Document
This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Energy

Click to edit the document details