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  

                                            L_n  =mv_nr_n

                                 Mv_nr_n=nh/2π              WHERE    r_n= RADIUS OF n^th ORBIT.

                                                                                                V_n=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=E_i-E_f

E_i=ENERGY OF INITIAL STATE.

                                                                         E_f=ENERGY OF FINAL STATE.

  TOTAL ENERGY OF THE ELECTRON IN STATIONARY STATE OF HYDROGEN ATOM IS GIVEN BY

                

   E_n=-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/n_f²-1/n_i²]     WHERE n_i=QUANTUM NUMBER OF HIGHER ENERGY STATE

                                          n_f=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/n_i²],

WHERE n_i=2,3,4,…..

                                       THIS SERIES LIES IN ULTRA VIOLET REGION.

2.BALMER SERIES:

                                                             ὺ=R[1/2²-1/n_i²],

WHERE n_i=3,4,5…

                                      THIS SERIES LIES IN THE VISIBLE REGION.

3.PASCHEN SERIES:

                                                           ὺ=R[1/3²-1/n_i²],

WHERE n_i=4,5,6…

                                    THIS SERIES LIES IN THE INFRA RED REGION.

4.BRACKTT SERIES:

                                                             ὺ=R[1/4²-1/n_i²],

WHERE n_i=5,6,7…

                                     THIS SERIES LIES IN THE INFRA RED REGION

5.PFUND SERIES:

                                                            ὺ=R[1/5²-1/n_i²],

   WHERE n_i=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 αA^1/3                             

                                          =>R =R₀A^1/3

(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 1/12^th 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 E_b WILL REALSE IN THIS PROCESS.THIS ENERGY E_b IS CALLED THE BINDING ENERGY.

                              B.E. = ∆mc²

                              ∆M =Z_mp+(A-Z)m_n-M_N(ZX^A)

                              Z_mp =MASS OF PROTONS.

                              (A-Z)m_n=MASS OF NEUTRON.

                              M_N(ZX^A)=MASS OF NUCLEUS.

B.E.=[ Z_mp+(A-Z)m_n+ Z_me-M_N(ZX^A)-m_e]C²          ( 931.5Mev)

B.E.=[Z(m_p+m_e)+ (A-Z)m_n-{m_n(ZX^A)+ m_e}] C²     (   931.5Mev)

B.E.=[Zm_H+(A-Z)m_n -m(ZX^A)] C²                                      (  931.5Mev)

m_H=MASS OF HYDROGEN ATOM.

B.E. PER  NUCLEOUS=     TOTAL B.E./A

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

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

1.LIGHT NUCLI LIKE ₁H¹, ₁H², ₁H³  HAVE 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 E_bn 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.  E_bn 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

                              _No∫^N dN/N=-λ ∫₀^tdt

=> _No∫^Nlog_e N= -λ[t]_o^t = -λt

=> log_eN- log_eN₀ = - λt

log_eN/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 T_1/2

   t= T_{1/2  ;} N= N₀/2

                 ·.·  N=N₀e^-λt

               => N₀/2=N₀e^-λ*1/λ= N₀ e^{- λT½}                          ·.·t=T_½

               =>1/2= e^{- λT½} =1/ e^λT½

                2= e^λT½

               TAKING log ON BOTH SIDE

                  log_e2  =  log_ee^λT½

                          =λT^½ log_ee

     3.303* log_e10²=λT^½

                   λT^½  =3.303*0.3010=0.693

                                    T^½ = 0.693/λ

  IMPORTANT

    t=    T^½  =T

    N= N₀/2

               IN ANOTHER HALF LIFE       (i.e.AFTER 2 HALF LIFE t=2T)

                N=1/2*N₀/2  = N₀/4  = N₀  (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.

  ZX^A→(Z-2)Y^(A-4)+₂He⁴+Q

₉₂U²³⁸→₉₀Th²³⁴+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 .

              

                                          ZX^A→(Z+1)Y^A  +  _-1e⁰+   Q

  IN -DECAY:-   A NUCLEUS SPONTANOUSLY EMITS AN ELECTRON (β^-DECAY)OR A POSISOTRON (β⁺DECAY)

   EX.OF β^- DECAY -

                                             ₁₅P³²→₁₆S³²+_-1e⁰+ύ       _-1e⁰=ELECTRON , ύ=ANTI NUTRINO

        EX.OF β⁺   DECAY                           

                                             ₁₁Na²²→₁₀Ne²²+₊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 3/4^thOF 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  D_{3   &}  D₄ 

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. T^1/2=0.693/λ

Q8       THE ACTIVITY OF A RADIOACTIVE ELEMENT DROP TO 1/6^th 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