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PHYSICS 6
Practice Exam 2 (2003)
SOLUTIONS
I.
1d ; 2b; 3c; 4a; 5d; 6d
II.
E
n
= 13.6/n
2
(eV)
a)
E(photon)= E
4
– E
2
= 13.6(1/4 – 1/16) eV = 2.6 eV
b)
E(photon) = h f
f = E(photon)/h = 2.6 eV/ 4.1
×
10
15
eV
•
sec = 6.3
×
10
14
sec
1
III. a) probability for one day to have precipitation
P = 11/38 = 0.29 or 29%
b) prob for no precipitation in one day = (1P) = 0.71 or 71%
So for 3 days have (1P)
3
= (0.71)
3
= 0.36
c) prob for ONLY ONCE in 3 days?
For (yesnono) it would be P
×
(1P)
2
. Same for (noyesno) and
(nonoyes).
So total prob for ONLY ONCE is
3
×
P
×
(1P)
2
=3
×
0.29
×
(0.71)
2
= 0.44
IV.
∆
x
∆
p
≥
h/4
π
So
∆
p(min) = h/(4
π
∆
x) = 6.6
×
10
34
J
•
sec / (4
π
×
10
7
m)
= 5.3
×
10
28
Kg
•
m/sec
(or units of
J
•
sec/m
since
1 J=1 Kg
•
m
2
/sec
2
)
V. For the essay the grade depends on evidence that you have read the play
as demonstrated by your use of appropriate citations to the text. Each essay
choice should refer to a section (or sections) of the play in which one or more
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This note was uploaded on 03/10/2008 for the course PHY 6 taught by Professor Garyr.goldstein during the Spring '06 term at Tufts.
 Spring '06
 GaryR.Goldstein
 Physics, Photon

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