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CCC Hoh Fuk Tong College
Second Term Examination, 2003 - 2004
Physics I
1.
(a)
(b)
- spectra is brighter
- each line is sharper (any 2)
- spreads out more
(2)
Since a sin = n
sin = n/a
For sin  1, we have n/a  1 or n 
a

1
n  3000009 = 5.6
600  10
Maximum order = 5
2.
(3)
yn
(c)
sin =
(d)
0.175
5.87  10-7
(1)
1  y 2n
0.354
5.9  10-7
0.532
5.91  10-7
(3)
(e)
mean  = 5.88  10-7 m
(1)
(f)
Third order
As it is further away, % error in measurement is less.
(2)
(a)
fo + fe = 100 cm
fo
fe = 9
(2)
Hence, fo = 90 cm and fe = 10 cm
(b)
(3)
objective
Scale = 1 : 10
100 cm
eyepiece
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(c)
minimum diameter of eyepiece fe
=f
diameter of objective
o
minimum diameter of eyepiece
(d)
(e)
3.
(a)
(2)
The eye-ring is the best position where the eye should be placed because it
collects as much light as possible from that passing through the telescope.
The diameter of the eye-ring = 0.44 cm
(3)
Advantage: It gives an erect image/It is shorter in length.
Disadvantage: The view angle is smaller/The image is dimmer.
(2)
(i)
To ensure that the elastic limit is not exceeded.
(1)
(ii)
slope of graph =
E
(iii)
(b)
10
= 4  90 = 0.444cm
(i)
(ii)
26.5 - 10.5
-1
-1
7.4 - 3 = 3.64 cm kg = 0.0364 m kg
F/A Fl Lgl
L
l
= e/l =
=
= e gA
eA
eA
1
l
1
2.00
= slope  g  A = 0.0364  10 
2


6.0  10  4 
4
9
-2
= 1.9  10 N m
(3)
F
15  10
= 5.3  108 N m-2
A= 
4 2
6.0  10 
4
1.9  10 9
E
v=
=
= 1300 m s-1
ρ
1100
d
10
t = v = 1300 = 0.0075 s
(2)
(3)
F
Breaking stress ( A ) remains unchanged, i.e. 5.3  108 N m-2
15 kg
4 = 2m
m = 1.875 kg
Therefore
(3)
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4.
(a)
(i)
(ii)
W
W
= -  P dV = - P (Vf – Vi) (for constant pressure)
= - (5  105)  (3  10-3 - 6  10-3)
= 1500 J
(2)
Vf
= -  P dV = - nRT ln(V ) (for constant temperature)
i
Vf
6 10-3
= - PB VB ln(V ) = - (5  105)  (3  10-3) ln(
)
3 10-3
i
= -1040 J
(2)
(b)
3
EA TA
Since E = 2 nRT, E = T
B
B
4500 1200
EB = 600 , EB = 2250 J
Since B and C are on the same isotherm, EC = EB = 2250 J
(c)
3
Since E = 2 nRT
3
At A: 4500 = 2 n (8.31)(1200)
n = 0.3 mole
(d)
(i)
(ii)
5.
(a)
(3)
(2)
U = W + Q
(2250 – 4500) = 1500 + Q
Q = -3750 J
U = W + Q
0 = -1040 + Q
Q = 1040 J
A
En = - n2 = -6.98
(4)
A
En+1 = - (n + 1)2 = -3.10
(n+1)2 6.98
n2 = 3.10  n = 2
A
Hence, - 22 = -6.98  A = 27.92
(b)
27.92
En = - n2
E1 = -27.92 eV
E2 = -6.98 eV
E3 = -3.10 eV
E4 = -1.745 eV……………
(3)
E4
E3
E2
E1
(2)
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(c)
E1,2 = -6.98 – (-27.92) = 20.94 eV  26.0 eV
E1,3 = -3.10 – (-27.92) = 24.82  26.0 eV
E1,4 = -1.745 – (-27.92) = 26.175  26.0 eV
The highest energy level = 3
Energy of bombarded electron = 26.0 – 24.82 = 1.18 eV
(d)
E =
(4)
hc

24.82  (1.6  10-19) =
(6.626 10-34) (3 108)

 = 5 10-8 m (500 Å)
It is ultra-violet.
(e)
(2)
Since an atom will only go to an excited state if it absorbs the right amount
of photon energy completely.
As there is no exact energy difference corresponding to the photon energy,
the atom will not be excited.
(2)
END OF PHYSICS I