This preview shows pages 1–2. Sign up to view the full content.
Homework set # 21
Problems: 26.16, 22; 27.9, 10
Solutions:
26.16
(a)
2
22
7
8
13
1 MeV
1.67 10
kg
3.00 10 m s
939 MeV
1.60 10
J
R
Em
c
(b)
2
2
3
2
1
939 MeV
3.01 10 MeV
3.01 GeV
1
0.950
R
R
E
c
E
vc
(c)
33
939 MeV
2.07 10 MeV
2.07 GeV
R
KE
E
E
26.22
Let
1
m
be the mass of the fragment moving at
1
0.868
, and
2
m
be the mass
moving at
2
0.987
.
From conservation of massenergy,
2
12
1
0.868
1
0.987
mc
E
mc
giving
27
2.01
6.22
3.34 10
kg
mm
(1)
The momenta of the two fragments must add to zero, so the magnitudes must be
equal, giving
pp
or
111
2 22
mv
. This yields
2.01
0.868
6.22
0.987
, or
3.52
(2)
Substituting equation (2) into (1) gives
27
2
7.07
6.22
3.34 10
kg
m
, or
28
2
2.51 10
kg
m
Equation (2) then yields
28
28
1
3.52 2.51 10
kg
8.84 10
kg
m
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document 27.9
(a)
From the photoelectric effect equation, the work function is
max
hc
KE
, or
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 05/12/2011 for the course PHYS 1112 taught by Professor Seaton during the Spring '08 term at University of Georgia Athens.
 Spring '08
 SEATON
 Work

Click to edit the document details