()
( )
314
2
1
HHH
e
3.01605u+1.00783u
4.00260u 931.5MeV/u
19.8MeV.
Emmmc
∆=
+ −
=
−
=
(b) The second step is to add energy to produce
3
HH
.
→
+
n
2
The energy needed is
( )
23
2
2
2.01410u+1.00867u 3.01605u 931.5MeV/u
6.26MeV.
n
Emm
m
c
=
−
=
(c) The third step:
2
H
→+
pn
, which — to make the electrons “balance” — may be
rewritten as
2
+
→
1
n
.
The work required is
( )( )
12
2
3
1.00783u 1.00867u
2.01410u 931.5MeV/u
2.23MeV.
n
m
c
=
+
−
=
(d) The total binding energy is
be
1
2
3
EE
E
E
∆
+
∆
+
∆
=
19.8MeV 6.26MeV
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This note was uploaded on 05/19/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Energy

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