# ELECTROMAGNETIC

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A. Wave Model.
Electromagnetic radiation constitutes the mode of energy propagation
for such phenomena as light waves , heat waves, radio waves, microwaves,
ultraviolet ray, x-rays ,and γ rays. These radiation are called
"electromagnetic " because they were first described , by Maxwell ,in terms
of oscillating electric and magnetic field.
As illustrated in figure (1) an electromagnetic wave can be represented
by the spatial variation in the intensities of an field (E) and a magnetic filed
(H), the fields being at right angles to each other at any given instant. Energy
is propagated with the speed of light (3x108 m/sec in vacuum) in the (Z)
direction . the relationship between wavelength (λ).frequency(ν) , and
velocity of propagation (c) is given by

c  v             (1)
In the above equation , c should be expressed in meters \second :λ , in
meters and ν, in cycles\second or hertz.

Figure(1): Graph showing electromagnetic wave at a given instant of time. E and H
are, respectively ,the peak amplitudes of electric and magnetic fields . The two fields
are perpendicular to each other

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Figure(2) : The electromagnetic spectrum. Ranges are approximate.

Figure (2) shows spectrum of electromagnetic radiations with wave lengths
ranging anywhere from 107 (radio waves) to10-13 (ultra-high energy x-
rays). Since wavelength and frequency are inversely related, the frequency
spectrum corresponding to the above range will be 3x10 1–3x1021
cycles/sec. It is interesting to note that only a very small portion of the
electromagnetic spectrum constitutes visible light bands .The wavelengths
of the wave to which the human eye responds range from 4x10 -7 (blue
light) to 7 x 10-7 m(red)
The wave nature of the electromagnetic radiation can be demonstrated
by experiments involving phenomena such as interference and diffraction
of light. Similar effects have been observed with x-rays using crystals
which possess interatomic spacing comparable to the x-ray wavelengths..
However, as the wavelengths becomes very small or the frequency
becomes very large, the dominant behavior of electromagnetic radiation
con only be explained by considering their particle or quantum nature.

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B. Quantum Model
In order to explain the results of certain experiment involving
interaction of action of radiation with matter, such as the photoelectric
effect and the Compton scattering, one has to consider electromagnetic
radiations as particles rather than waves . The amount of energy carried by
such a packet of energy, or photon. Is given by:
E  hv                    (2)
Where E is the energy (joules) carried by the photon, h is the Planck's
constant (6.62 x 10-19 J-sec), and ν is the frequency (cycles\second). By
combining equation 1.2 and 1.3, we have

hc
E                        (3)
λ

If E is to expressed in electron volt (eV) and λ in meters (m) ,then,
since 1eV=1.602 x 10-19 J,

6
1.24  10
E                           (4)
λ

The above equations indicate that as the wavelength becomes short or
the frequency becomes larger, the energy of the photon becomes greater.
This is also seen in figure (2)

2

Radioactivity may be described mathematically without regard to the
specific mechanisms of decay. The rate of decay (the number of decays per
unit time) of a radioactive sample is directly proportional to the number of
atoms present in the sample at the particular moment. This concept is
expressed mathematically as

ΔΝ
 λΝ             )1 (
Δt

Where ΔN/Δt is the rate of decay. The constant of proportionality λ is
called the decay constant. The minus sign indicates that the number of
parent nuclei and therefore the number decaying per unit time are
decreasing. The meaning of the decay constant is clear by rearranging
Equation (1) to

 ΔΝ/Ν
λ
Δt

The decay constant λ is the fractional rate of decay of the atoms. The
decay constant has units of inverse time (i.e, sec-1 , hr-1 , etc ). It has a
characteristic value for each nuclide and reflects its degree of instability.
A greater decay constant means a more unstable nuclide, i.e. , one that
undergoes decay more rapidly. The rate of decay is termed the quantity
“activity” which is defined as

 ΔΝ
Activity  A  λΝ 
Δt

The activity depends on both the decay constant of the atoms and the
number of atoms in a sample. A radioactive sample may have a high
activity because it is a very unstable material or because it is a moderately
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unstable material but is composed of many atoms. The traditional unit of
activity is the curie which is defined as
1 Ci = 3.7 x 1010 disintegrations per second ( dps)

The millicurie, microcurie, nanocurie, and picocurie are useful
divisions of the curie.

 1mCi =10-3 Ci = 3.7 x 107 dps

 1µCi = 10-6 Ci = 3.7 x 104 dps

 1nCi = 10-9 Ci = 3.7 x 101 dps

 1pCi = 10-6 Ci = 3.7 x 10-2 dps

One curie was originally defined as the rate of decay of 1 g of radium.
Although recent measurements have established the decay rate as 3.61×10 10
dps for radium, the curie has retained its original numerical definition. The
Becquerel is the modern unit of activity. The Becquerel is defined as

1 Bq = 1 dps = 2.7 × 10-11 Ci

 Example (1)
a.     TI has a decay constant of 9.49×10-3 hr-1 . Find the activity in becquerels
201
81

of a sample containing 1010 atoms
From Equation 1
Α  λΝ

4
                               
 9.4910  3 hr 1  1010 atoms 
                               
                                  
3,600sec/hr

atoms
 2.64 104
sec

 2.64 104 Bq
b. How many atoms of 11 C with a decay constant of 2.08 hr-1 would be required
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to obtain the same activity as the sample in part a ?

Α
Ν
λ
atoms          sec 
2.64  104          3,600     
sec          hr 

(2.08 1    )
hr

= 4.57 107 atoms
More atoms of 201 TI than of 11 C are required to obtain the same activity
81            6

because the difference in decay constants.

Decay Equations and Half-Life

Equation (1) can be solved for the number of atoms of parent N
existing at any time t in the sample (Fig 3) the solution is

N  N e  λt           (2)
0

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Where Nο is the number of atoms originally present and e is the
exponential quantity 2.7183. After multiplying both sides of the equation
by the decay constant, this expression can be written as

Α  Α e  λt       (3)
0

where A is the activity remaining after time t, and A is the original
activity . The number of atoms N* decaying in an interval is N o – N (See
Fig 3)

Ν*  Ν (1  e λ t )   )4(
0

Figure(3):Mathematics of radioactive decay with the parent decaying to a
stable daughter.

The physical half-life T1/2 of a radioactive nuclide is the time required
for decay of half of the atoms in a sample of the nuclide.
In Equation (2),   Ν   1
2   Ν0   when t =T1/2. By substitution in Equation (2)

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 λΤ1
1   Ν
   e             2   (5 )
2 Ν0

1
 
LN     λΤ
2       1/2
 

0.693
Τ    
1/2      λ

Where 0.693 is the ln 2 (natural or Napierian logarithm of 2):
ln2   0.693
Τ1/2      
λ      λ

Illustrated in Figure (4) is the percentage of the original activity of a
radioactive sample described as a function of time expressed in unites of
physical half-life. By replotting these data semilogarithmically (activity on a
logarithmic axis, time on a linear axis), a straight line can be obtained, as
shown in Figure (5).

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Figure (4): The percentage of the original activity of a radioactive sample is
expressed as a function

FIG (5): Semilogarithmic graph of the data in Figure (4)
Example (2)
The physical half-live is 1.7 hours for 113mIn
a. A sample of 113mIn has a mass of 2 µg. How many                    113m
In atoms are
present in the sample?
                                          
                                                         
 No.of          No.of
grams          atoms                            

Number of atoms Nο =                          
                 gram atomic

mass 

No.ofg
gram atomic mass

                                       
                  23 atoms                      
 210- 6  6.0210                               
         
=            
                  gram atomic

mass 

113 g
gram atomic mass
8
= 1.07  10
16

b. How many 113mIn atoms remain after 4 hours have elapsed?

Number of atoms remaining =                    N  N e  λt
0

Since λ = 0.693 / T1/2
                          
                          
                          
  0.693                    t
       Τ                  
         1                
Ν Ν e                      2              
0

 0.693       
                               t
  1.071016 atoms e        1.7hr 
                 
                 

  1.071016 atoms 0.196
                              
                 
                              

 2.10 1015 atoms remaining

c. What is the activity of the sample when t = 4.0 hours?

0.693
λ Ν                   Ν   A=
Τ
1
2

       15     
0.693 2.110 atoms 
              
                                  
            
1.7hr  3,600sec 
         hr 

 2.4 1011dps

In units of activity A = 2.4 x 1011 Bq

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d. specific activity is defined as the activity per unit mass of a radioactive
sample. What is the specific activity of the 113mIn sample after 4 hours?

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2.410 Bq            Bq
Specific activity             1.2  1017
6               g
210 g

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