Monday 12 March 2012

Rutherford Α Particle Scattering


 


1.)  Beam Of Α Particle Allowed Fall On A Thin Foil Of Gold.

2.   Scattered Α Particle Were Observed Through A Roatating     Detector    Consisting Of Zinc- Sulphide Screen.
3.)  Scattered Α Particle Strik The Zinc Sulphide Screen Produce    Flashes;Ie Called “ Scientillations”.

Observation
A Graph Is Plotted Between The Sacttered Angle Θ And Number Of Α-Scattered Particle N(Θ) .. 
From This Graph We Observed That --
                                                 
1.)    Most Of The Α-Particle Pass Through The Foil And Suffer Small Scattering Angle .
2.)    The Α-Particle Defelected Backwards And Experience A Laeg Repulsive Force When It Goes Very Close To Central Part Of Atom Where The Whole  Mass Of Atom And Positive Charge Is Concentrated.
3.)    The Central Part Of Atom Is Called  “Nucleus”.
4.)    Size Of Nucleus Is About 10-¹5 M To 10-14 M.
5.)    Size Of Atom Is 10-10 M.
6.)    The Maginitued Of Force Is  F=1/4πє0*(Ze)2e/R2


Bhor`S Model Of Hydrogen Atom Postulates

1.)  An Electronin Atom Revolve In A Certain Stable Orbits With Out The Emission  Of  Radiant Energy .This Certain Stable Orbits Are Called Stationary Orbit .

2.) The Electron Revolves Arround The Nucleus Only In Those Orbits For Which The Angular Momentum Is Integral Multiple Of h  /2π;
  i.e.L=nh/2π

  L=mvr  
                                            Ln  =mvnrn
                                 Mvnrn=nh/2π              WHERE    rn= RADIUS OF nth ORBIT.
                                                                                                Vn=SPEED OF ELECTRON.
3.)When An Electron Jump From Higher Energy Level To Another Energy Level Or Vice-Versa A Photon Is Emmited Having Energy Equal To Energy Different Between Initial And Final States .
i.e. hυ=Ei-Ef

Ei=ENERGY OF INITIAL STATE.
                                                                         Ef=ENERGY OF FINAL STATE.


  TOTAL ENERGY OF THE ELECTRON IN STATIONARY STATE OF HYDROGEN ATOM IS GIVEN BY
                
   En=-13.6/n2*ev
ENERGY LEVEL DIAGRAM FOR HYDROGEN ATOM

The Energy Of State(N=1) E1=-13.6ev Is Called Ground State.
The Minimum Energy Required To Free A Electron Form The Ground State Of Hydrogen Atom Is 13.6ev.It Is Called Ionisation Energy Of Hydrogen Atom.
When An Electron Jump From n=2 TO n=1
E2-E1=-3.40ev-(-13.6ev)=10.2ev .
As The Excitation State Of Hydrogen Atom Increases(I.E.N Increases)The Value Of Minimum Energy Required To Free The Electron From excited Atom Decreases.
      i.e                                E3-E1=-1.51ev-(-13.6ev)=12.9ev.

SPECTRA LINES OF HYDROGEN ATOM


WAVE NUMBER  ὺ OF THE EMITEED PHOTONS IS GIVEN BY
 ὺ=R[1/nf2-1/ni2]     WHERE ni=QUANTUM NUMBER OF HIGHER ENERGY STATE
                                          nf=QUANTUM NUMBER OF LOWER ENERGY STATE
                                                    ὺ=WAVE NUMBER  
                                            R=RYDBERG CONSTANT
1. LYMAN SERIES:WHEN ELECTRON JUMP FROM ANY HIGHER ENERGY LAVEL TO FIRST LEVEL.
                                                       ὺ=R[1/12-1/ni2],
WHERE ni=2,3,4,…..
                                       THIS SERIES LIES IN ULTRA VIOLET REGION.
2.BALMER SERIES:
                                                             ὺ=R[1/22-1/ni2],
WHERE ni=3,4,5…
                                      THIS SERIES LIES IN THE VISIBLE REGION.
3.PASCHEN SERIES:
                                                           ὺ=R[1/32-1/ni2],
WHERE ni=4,5,6…
                                    THIS SERIES LIES IN THE INFRA RED REGION.




4.BRACKTT SERIES:
                                                             ὺ=R[1/42-1/ni2],
WHERE ni=5,6,7…
                                     THIS SERIES LIES IN THE INFRA RED REGION
5.PFUND SERIES:
                                                            ὺ=R[1/52-1/ni2],
   WHERE ni=6,7,8…
                                       THIS SERIES LIES IN THE INFRA RED REGION
NUCLEI

SIZE OF NUCLEUS:SIZE OF NUCLEUS IS DIRECTLY PROPORTIONAL TO ITS MASS NUMBER.
                                           4/3Π R3 α  A
                                          =>R3αA
                                          =>R αA1/3                             
                                          =>R =R0A1/3

(R=RADIUS OF NUCLEUS , R0 IS CONSTANT AND ITS VALUE IS 1.1*10-15m.)


ISOTOPES:ATOM HAVING DIFFERENT MASS NUMBER(A) AND HAVING SAME ATOMIC NUMBER(Z)
                           E.g.      92U235, 92U238

ISOBARS:ATOMS HAVING SAME MASS NUMBER(A)BUT DIFFERENT ATOMIC NUMBER (Z)
                          E.g.       24Cr54   26Fe54
ISOTONS:NUCLEI COANTAINING SAME NUMBER OF NEUTRONS ARE CALLED ISOTONS

                           E.g.17Cl37, 19K39
ATOMIC MASS UNIT:ONE u IS EQUAL TO 1/12th OF THE MASS OF AN ATOM OF 6C12 ISOTOPE OF CARBON.
                                      1u EQUIVALENT TO 931.5 mev
MASS DEFECT:
DIFFERENCE BETWEEN THE SUM OF MASSES OF PROTONS AND NEUTRON AND ACTUAL MASS OF NUCLES IS CALLED MASS DEFECT.IT IS DENOTED BY ∆M.


BINDING ENERGY:
IF A CERTAIN NUMBER OF NEUTRONS AND PROTONS ARE BROUGHT TOGETHER TO FORM A NUCLEUS,AS ENERGY Eb WILL REALSE IN THIS PROCESS.THIS ENERGY Eb IS CALLED THE BINDING ENERGY.
                              B.E. = ∆mc2
                              ∆M =Zmp+(A-Z)mn-MN(ZXA)
                              Zmp =MASS OF PROTONS.
                              (A-Z)mn=MASS OF NEUTRON.
                              MN(ZXA)=MASS OF NUCLEUS.

B.E.=[ Zmp+(A-Z)mn+ Zme-MN(ZXA)-me]C2          ( 931.5Mev)
B.E.=[Z(mp+me)+ (A-Z)mn-{mn(ZXA)+ me}] C2     (   931.5Mev)
B.E.=[ZmH+(A-Z)mn -m(ZXA)] C2                                      (  931.5Mev)

mH=MASS OF HYDROGEN ATOM.

B.E. PER  NUCLEOUS=     TOTAL B.E./A



GRAPH BETWEEN BINDING ENERGY PER NUCLEOUS Ebn VERSUS THE MASS NUMBER A


B.E./NUCLIOUS WITH MASS NUMBER A REVEALS THAT-

1.LIGHT NUCLI LIKE 1H1, 1H2, 1HHAVE SMALL VALUE OF B.E./A .

2.FOR MASS NUMBER RANGING FROM 2 TO 20 HAVE VERY SHARP PEAKS CORRESPONDING TO NUCLIE HAVING MASS NUMBER IN MULTIPLES OF FOUR .
E.g. 2He4,4Be8,6C12,8O16,10Ne20
PEAKS INDICATE THAT THESE NUCLIE ARE MORE STABLE THAN THE THE NUCLIE IN   THE NEIGHBOURHOOD.
3. FOR MASS NUMBERS RANGUNG FROM 30 TO 120B.E.CURVE HAS A BROAD   MAXIMUM.AVERAGE BE/A CORRESPONDING TO THIS RANG IS 8.5 Mev
PEAK VALUE IS   8.8Mev /A  (26Fe56)
ELEMENTS BELONGED TO THIN RANG ARE HIGHLY STABLE AND NON RADIOACTIVE.
4.AFTER A=120.B.E/A DECREASES GRADUALLY TO   7.6Mev/A    (92U238).THUS HEAVY NUCLIE ARE LESS STABLE.

STABILITY OF A NUCLEUS DEPEND UPON ----
1.High Valu Of B.E/A.

2.Even-Even Nucleus(Even Z And Even A).It Is Found To Be More Stable,Even If It Has Comparatively Low Value Of B.E/A. An Even –Odd Nucleus Or Odd Even Nucleus Is Lesser Stable And Odd –Odd  Nucleus Is Least Stable.

3. Nuclei Having Higher Or Lower Value Of Neutron To Proton Ratio Than The Stable Nuclie Are Also Un Stable.

MAIN FEATURE OF THE GRAPH B.E/A VERSUS  A
1. The Binding Energy Per Nucleos Ebn Is Practically Constant.The Curv Has A Maximum Of About 8.75mevfor A=56 And Has A Value Of 7.6 Mev For A=238.
2.  Ebn Is Lower For Both  Light Nuclei  A<30 And Havy Nuclei A>170.



FOLLOWING RESULTS  FROM THESE TWO OBSERVATION IS UNDER –

1.The Force Is Attractive And Sufficiently Strong To Produce a Binding Of A Few Mev Per Nucleon.
2.The Binding Energy In The Range 3o<A<17o Is A
3.A Very Heavy Nucleus Has Lower Binding Energy Per Nucleon.This Impalies Energy Would Be Realse In This Process Of Fission..
4.The Binding Energy Per Nucleous Of Fused Havier Nuclei Is More Than The Binding Energy Per Nucleon, Of The Light Nuclei.This Implies That Energy Would Be Released In The Process Of Fusion.
         
  RADIOACTIVITY

Radio Activity: Radioactivity Is Phenomenon In Which A Heavy Nucleus Distintegrate Itself .The Element Exhibiting This Property Is Called Radioactive Element . Thair Nuclei Are Unstable Nuclei  E.G. Radium , Thorium, Actinium , Polarium.

Redioactive Decay:Three Type Of Radioactive Decay Occur In Nature
1.α-DECAY –α-DECAY IN WHICH HELIUM NUCLUES    2He4   IS EMITED
2. β-DECAY –IN β – DECAY ELECTRON OR POSITORN ARE EMITTED  .
3. γ-DECAY – IN WHICH HIGH ENERGY PHOTON ARE EMITTED .

LAW OF REDIOACTIVE  DECAY
-         When A Radioactive Substances Under Goes Α; Β  And  Γ Decay  It Is Observed That Number Of Nucliei Undergoing The Decay Per Unit Time Is Proportional To Total Number Of Nuclei Is Present .
-         In Sample
.         If N Is Number Of Nuclei In The Sample And Δn Under Goes Decay In Time Δt,Then 
                                   ΔN/Δt   α  -N
                               
                               =>dN/dt   α  -N         
   .                       –    =>dN/dt=-λN   
WHERE λ IS CALLED DECAY CONSTANT OR DISINTERGRATION CONSTANT
HERE    -ve SIGN  SHOWS   THAT ONE OF ORIGINAL  NUCLEI,ΔN HAVE DECAYED.      
                                          =>dN/N=-λdt
                              NoN dN/N=-λ ∫0tdt
=> NoNloge N= -λ[t]ot = -λt

=> logeN- logeN0 = - λt

logeN/N0=-λt
                                       
                                                             N/N0=e-λt


      N= N0   e-λt
 
 

                                   
                                   


    4.   DISINTEGRATE CONSTANT
                 N=N0e-λt       ,IF t=1/λ
                  N=N0e-λ*1/λ
                  N=N0e-1  =  N=N0/e
                  λ=1/t

5.     HALF LIFE AND MEAN LIFE:

Half Life:Half Life Is The Time In Which One-Half Of The Radioactive       Substance Is Dis Intigrated.
IT IS DENOTED BY T1/2
   t= T1/2  ; N= N0/2
                 ·.·  N=N0e-λt
               => N0/2=N0e-λ*1/λ= N0 e- λT½                          ·.·t=T½
               =>1/2= e- λT½ =1/ eλT½
                2= eλT½
               TAKING log ON BOTH SIDE
                  loge2  =  logeeλT½
                          =λT½ logee
     3.303* loge102=λT½
                   λT½  =3.303*0.3010=0.693
                                    T½ = 0.693/λ
  IMPORTANT
    t=    T½  =T
    N= N0/2
               IN ANOTHER HALF LIFE       (i.e.AFTER 2 HALF LIFE t=2T)
                N=1/2*N0/2  = N0/4  = N(1/2)2
             AFTER ANOTHER HALF LIFE(i.e.AFTER 3 HALF LIFE t=3T)
            N= N0/4*1/2 = N0/8= N0(1/2)3         AND SO ON
   
                              FOR n HALF LIFE

                           ·.·  N = N0 (1/2)n       
                           ·.·t=nT                            =>n=t/T


          N=N0(1/2)n= N0(1/2)t/T
 
 

          =>

                              T=TOTAL TIME OF n HALF LIFE
                         n=n NUMBER OF HALF LIFE

                               t = total time of  n half life 
                                n = no. of half life

Average life or mean life: AVERAGE LIFE OF A RADIOACTIVE ELEMENT IS RECIPROCAL OF THE DECAY CONSTANT OF THE ELEMENT
                                i.e. τ = 1/ λ

τ = SYMBOL OF AVERAGE  LIFE OR MEAN LIFE.
 => τ= 1/0.693*T  , ·.·λ=0.693/T

=> τ=T/0.693       , T=HALF LIFE

=> τ=1.44T        , λ=DECAY CONSTANT.

THUS AVERAGE LIFE OF A RADIO SUBSTANCE IS 1.44 TIMES THE HALF LIFE OF ELEMENT.

RADIO-ACTIVE DECAY

ALPHA DECAY-
  IT IS THE PHENOMENON OF EMISSION OF AN α PARTICLE  FROM A radioactive NUCLEUS. When a nucleus EMITS AN α –PARTICLE , ITS MASS NUMBER   DECREASE BY 4 AND CHARGE NUMBER (ATOMIC NUMBER ) DECREASE BY 2.
  ZXA→(Z-2)Y(A-4)+2He4+Q

92U23890Th234+2He4+Q

X=u,Y=Th.X=PARENT ELECTRON,Y=DAUGHTER ELECTRON.

-DECAY:
               IT IS THE PHENOMENON OF EMISSION OF AN ELECTRON FROM A RADIOACTIVE NUCLEUS .
              
                                          ZXA→(Z+1)Y-1e0+   Q
  IN -DECAY:-   A NUCLEUS SPONTANOUSLY EMITS AN ELECTRON -DECAY)OR A POSISOTRON (β+DECAY)

   EX.OF β- DECAY -
                                             15P3216S32+-1e0+ύ       -1e0=ELECTRON , ύ=ANTI NUTRINO
        EX.OF β+   DECAY                           
                                             11Na2210Ne22++1e0+ύ                 +1e0=POSITRON , ύ=NUTRINO

DURING β- DECAY  n/P RATIO DECREASES


γ-DECAY: PHENEMENON OF EMISSION OF γ RAYS PHOTON FROM A RADIOACTIVE NUCLEUS IS CALLED γ-DECAY

β-DECAY OF 27Co60 NUCLEUS FOLLOWED BY EMISSION OF TWO γ-RAYS FROM DEEXCITATION OF THE DAUGHTER NUCLEUS 28Ni60






NUMERICALS
Q1.      THE VALUE OF GROUND STATE ENERGY OF HYDROGEN ATOM IS -13·6ev
(a)                WHAT DOES THE NEGATIVE SIGN SIGNIFY?
(b)               HOW MUCH THE ENERGY IS REQUIRED TO TAKE AN ELECTRON IN
THIS ATOM FROM IN IHE GROUND STATE TO THE FIRST EXITED STATE ?
Q2.      TWO NUCLEI HAVE MASS NUMBER IN THE     RATIO1:2,WHATE IS THE RATIO OF THEIR NUCLEAR RADII?
Q3       COMPARE THE RADII OF TWO NUCLEI WITH MASS NUMBER 1 AND 27 RESPECTIVELY?
Q4       THE HALF LIFE OF A RADIOACTIVE SUBSTANCE IS 30 sec CALCULATE
            1.)        THE DECAY CONSTANT
            2.)        TIME TAKEN FOR THR SAMPLE TO DECAY 3/4thOF ITS INITIAL VALUE?
Q5       THE SEQUENCE OF THE STEPWISE DECAY OF RADIOACTIVE NUCLEUS IS
             D-----ά--------> D1------ β -------> D2-----ά-------->D3-----ά-------->D4
            IF NUCLEON NUMBER & ATOMIC NUMBER FOR D2 176 & 171 THEN WRITE CORRESPONDING VALUE OF  D3   &  D
Q6       Draw The Graph Showing The Varition Of Binding Energy Per Nucleous With Mass Number.Give The Reason For The Decreases Of Binding Energy Per Nucleus For Nuclei With High Mass Number?
Q7       Define Half Life And Disintegration Constant.Derive The Relation Between The
i.e. T1/2=0.693/λ
Q8       THE ACTIVITY OF A RADIOACTIVE ELEMENT DROP TO 1/6th OF ITS INITIAL VALUE IN 32 YEARS.FIND HALF LIFE AND MEAN LIFE OF THE SAMPLE .
Q9       THE ENERGY LEVEL OF AN ATOM ARE AS SHOWN BELOW.WHICH OF THEM WILL RESENT IN THE TRANSITION OF A PHOTON OF WAVE LENGTH 275mm?

Q10     WHATE IS MEANT BY BINDING ENERGY OF DEUTRON(1H2)AND α-PARTICLE (2He4) are 1.25 and 7.2 Mev/ NUCLEON RESPECTIVELY.WHICH

Saturday 10 March 2012

IMPORTANT QUESTIONS OF ELECTROSTATICS

                        IMPORTANT QUESTIONS
Q.1-  Give two properties of electric lines of force sketch them for an isolated positive charge ?
Q.2- Derive an expression for dipole field intensity at any point on – (a)- axial line of dipole
         (b)- equatorial line of dipole.
Q.3-Derive an expression for torque acting  on  an electric dipole in a uniform electric field.
Q.4-Define capacitance and write its S.I. unit .
Q.5- Define an expression for the capacitance  of a parallel plate capacitor.
Q.6- Prove that energy density stored in parallel plate capacitor is1/2 є0E2  where symbols have their usual meaning.
Q.7-  Derive an expression for energy stored in parallel plate capacitor . What is the form of this energy & where from it comes .
Q.8-   Find an expression for capacitance of parallel plate capacitor when dielectric slab is introduced between the plates of the capacitor .
Q.9-  It states gauss’s law in electrostatics. Using this law derive an expression for electric field due to charged spherical shell.
Q.10- Draw the level diagram of Van de graff generator . Explain the principle, working of  Van de graff generator.

VAN DE GRAFF GENERATOR



It was designed by Vande graff in the year 1931
PRINCIPLE-  The generator is based on –
1-       The action of sharp points  i.e,  phenomenon of CORONA DISCHARGE
2-       The property that charge is given to a hollow conductor is transferred to outer surface and distributed informally over it.
It is machine capable of building up a potential difference of few million volts & field close to the break down field of air which is about 3×108 V/m
                                   A large spherical conducting shell (few meters radius ) is supported at height several meters above the ground on the insulating column.
A long narrow endless belt insulating material like rubber or silk is wound around two pulleys . One at ground level , one at the centre of the shell. This belt is kept continuously  moving by motor driving the lower pulley . It continuously carries positive sprayed on to it by a brush at ground level to the top up. There it transfers its positive charge to another conducting brush connected to the large shell. Thus the positive charge is transferred to the shell where it spread out on the outer surface.
Application- Accelerated positive charge are used to carried out nuclear reactions.