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Physics Problem Solutions

1) The document discusses the calculation of binding energy and specific binding energy for two nuclei: 26Fe56 and 55Cs133. 2) It is found that 26Fe56 has a binding energy of 478.98 MeV and specific binding energy of 8.553 MeV, while 55Cs133 has a binding energy of 1090.95 MeV and specific binding energy of 8.203 MeV. 3) Since 26Fe56 has a higher specific binding energy, it is determined to be the more stable nucleus.

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31 views4 pages

Physics Problem Solutions

1) The document discusses the calculation of binding energy and specific binding energy for two nuclei: 26Fe56 and 55Cs133. 2) It is found that 26Fe56 has a binding energy of 478.98 MeV and specific binding energy of 8.553 MeV, while 55Cs133 has a binding energy of 1090.95 MeV and specific binding energy of 8.203 MeV. 3) Since 26Fe56 has a higher specific binding energy, it is determined to be the more stable nucleus.

Uploaded by

Prince Vegeta
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Set-5

1 q
33. Formula: V 
4 0 r
Point inbetween
(5x108C) A B (3x108C)

C O
x 16x

Let O be a point at a distance ‘x’ from A (between the charges) where the
potential is zero.
9 5x10 8 
At O, potential due to 5x10 C, VA  9 x10 
8


 x 
  3x10 8 
At O, potential due to 3x108C, VB  9 x10 9  

 16  x 
Net potential at O is zero if
VA  VB  0
8
9 5x10  9  3x10 8 
   
9 x10 
x   9 x10  16  x   0
   
 5x10 8   3x10 8 
9 x10 9    9 x10 9 
  16  x 

 x   
5 3
  805x=3x  80=8x
x 16  x
x=10cm from A in between the charges.
Point outside:
(5x108C) (3x108C)
A B

C O
16+x
x
Let O be a point at a distance ‘x’ from B (outside the charges) where the
potential is zero.
1 q
We have V 
4 0 r
 5x10 8 
At O, potential due to 5x108C, VA  9 x10 9  

 16  x 
9  3x10 8 
8
At O, potential due to 3x10 C, VB  9 x10   
x 
 
Net potential at O is zero if VA  VB  0
 5x10 8    3x10 8 
9 x10 9    9 x10 9 
 
0

 16  x   x 

 5x10 8  3x10 8
9 x10 9    9 x10 9 
  x 

 16  x   
5 3
  5x=48+3x  2x=48
16  x x
X=24cm from B and 40cm from A outside the charges.
34. Formula: V=IR, Node rule and Loop Rules.

The circuit diagram, various nodes and currents through various branches are
shown. Applying KVL to the loop ABDA (traversing clockwise)
E+IR=0
2I2 + 10Ig –8I1 =0
–8I1 +2I2 + 10Ig =0..........(1)

Applying KVL to the loop BCDB


(traversing clockwise) B
I4=I2Ig
4(I2Ig)  4(I1+Ig)10Ig=0
2
4I24Ig  4I14Ig10Ig=0 Ig 4
4I1+4I218Ig = 0............(2) I2
A  10 C
Applying KVL to the loop I1
ABCEA
8 4
(traversing clockwise) I
5+ 2I2+4(I2Ig) = 0 I3=I1+Ig
5+ 2I2+4(I2Ig) = 0 I=I1+I2 D
I
0 + 6I2  4Ig = 5.............(3) E
Multiplying equation (1) by 2
–16I1 +4I2 + 20Ig =0
writing equation (2) again 5V
4I1 + 4I2  18Ig = 0
Subtracting,
–12I1 +0 + 38Ig =0...........(4)
Multiplying equation (1) by 3
–24I1 +6I2 + 30Ig =0 writing equation (3) again
0 + 6I2  4Ig = 5
Subtracting, –24I1 +0 + 34Ig =5...............(5)
Equation (4) x2  –24I1 +0 + 76Ig = 0
Solving, 42Ig = 5
5
Ig   0.119A
42
35. Given power factor Cos = 1 or =0O.
1 1
Hence XL=XC or o  or  o 
LC 2 LC
(i) At resonance frequency, power factor is unity. Hence resonance frequency is
1
o   15.9Hz
2x3.142 200 x10 3 x500 x10 6
V V 100
(ii) Current, I rms  rms  rms   10A
Z R 10
Current amplitude Io  I rms 2  10 2  14.14A
o L 2 o L 2 x 3.142 x15.9 x 200 x10 3
(iii) Q factor = Q=   2
R R 10
 o L 1 L
Also note that Q   
 R R C
D
36. Formula: fringe width,  
d

width 20 fringes 8x10 3


   0.4mm
20 20
xD 3 546 x10 9 x1.25
  0.4x10 
d d
9
546 x10 x1.25
d  1.706mm
0.4x10 3
Distance between slits is d=1.706mm.
Fringe width if 594nm is used,
xD 594 x10 9 x1.25
   0.435mm
d 1.706 x10 3
Width of 20 fringes = 20=20x0.435=8.7mm

56
37. 26Fe contains 26 protons and 30 neutrons. Formula for Mass defect is

Δm  Zm P  A  Z m n   M

m=[26x1.00728 + 30x1.00867] – 55.9349

m=[26.18928 + 30.260] – 55.9349

m=[56.4493 - 55.9349]

m=0.51448u

Hence binding energy of 26Fe56 is


B.E.= m x931MeV=0.51448x931x106ev=478.98MeV

B.E. 478 .98


Specific binding energy, E    8.553MeV
A 56
133
55Cs contains 55 protons and 78 neutrons. Formula for Mass defect is

Δm  Zm P  A  Z m n   M

m=[55x1.00728 + 78x1.00867] –132.9049

m=[55.4004 + 78.6726] – 132.9049

m=[134.07666 – 132.9049]

m=1.1718u

Hence binding energy of 55Cs133 is

B.E.= m x931MeV=1.1718x931x106ev=1090.95MeV

B.E. 1090 .95


Specific binding energy, E    8.203MeV
A 56

Specific binding energy of 26Fe56 is greater than specific binding energy of 55Cs133.

Hence 26Fe56 is more stable.

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