Rutherford scattering show

1. SCATTERING OF

Khemendra
Shukla
MSc(Applied Physics) sem. I
B.B.A. University, Lucknow
1
PARTICLES

2. 2

3. Sir J.J. Thomson discovered electron in 1897. After
one year he suggested an atomic model.
“Atoms are just positively charged lumps of matter in
which electrons are embedded in them.”
3
Thomson’s atomic model
Fig. 1.1

4. Thismodelcouldnotexplainallfeaturesofvisible
spectrumofhydrogenatomandotherelements.
Rutherfordperformedanumberofexperimentwith
Geiger&Marsdenonthescatteringofalphaparticles
byaverythingoldfoil.Thomson’smodelcouldn’t
explaintheexperimentalresults.
4

5. Born:30August1871,aNewZealand-British
chemist&physicist.
Knownasfatherofnuclearphysics.
5
(1871 –1937)
@McGillUniversity,discoveredtheconceptof
radioactivehalf-life,provedthatradioactivityinvolved
thetransmutationofonechemicalelementtoanother,
andalsodifferentiatedandnamedalphaandbeta
radiation,provingthattheformerwasessentiallyhelium
ions.
Continued…

6. AwardedwithNobelPrizeinChemistryin1908"for
hisinvestigationsintothedisintegrationofthe
elements,andthechemistryofradioactive
substances.“
HeremainstheonlyNobelPrizewinnertoperformed
hismostfamousworkafterreceivingtheprize.
“That was Rutherford Atomic Model and obviously
discovery of NUCLEUS.”
6

7. Rutherfordmodelwasacceptedbecauseheproved
scatteringofparticleswithmathematicalformula.
Atomconsistsofcentralmassivenucleusinwhichall
thepositivechargeandmostofthemassare
concentrated.
Acloudofnegativelychargedelectronssurroundsthis
nucleus.Theyaremovingaroundthenucleus.
Mostofthespaceinatomisempty.
7

8. 8
Experiment
Theory
Mathematicalanalysis
Experimental verification

9. 9
Small aperture
Fig. 1.2

10. Mostofthe–particlewerescatteredbysmall
deviationswhilepassingthroughgoldfoil.
Therewereafewparticlesthatwerescatteredthrough
largeangle.
Oneofabout8000particlessufferedanglesof
scattering>90
Afewofthemgoheadoncollision.
10

11. Assuming-particlesandnucleusarepointcharges.
Impactparameterbistheminimumdistancetowhich
particlewouldapproachthenucleusifthereisno
forcesbetweenthem.
b=0forheadoncollision
DistanceofclosestapproachDistheminimum
distancetowhichparticleapproachesnucleushead
on.
11
Continued…

12. Scatteringangleistheanglebetweentheasymptotic
directionofapproachof-particleandtheasymptotic
directioninwhichitrecedes.
Head on collision
12
b=impact parameter
b
-particle1P 2P P
Fig. 1.32 2

13. At the instant of closest approach KE of -particle
is the no. of particles
per unit area that reach the
screen at scattering angle of
is this no. of backward
scattering.
13D
Ze
PEKE
initial
2
0
2
4
1 initial
KE
Ze
D
0
2
4
2 0
60 0
80 0
100 0
120 0
140 o
160 o
180
N(
Fig. 1.4o
N180 o
N180

14. As a result of impulse F dtgiven it by the nucleus, the
momentum of the -particle changes by from
initial value to the final value .
Hence the magnitude of its momentum is also same
before and after, and
We have magnitude for momentum change
Impulse is in same direction as
14 P 1P 2P FdtPPP
12 mvPP
21 2
sin2mvP P Fdt dtFFdt cos

15. b=impact parameter
b
-particle1P 2P P F P 2 1P 2P 2
sin
sin
mvP mvPP
21 2

16. 162
2
cos
2
sin2
cos
2
sin2
d
d
dt
Fmv
dtFmv 2
2
2
2
0
2
2
22
2
22
2
cos2cos
2
sin
4
cos
2
sin2
d
Ze
bmv
dFrbmv
vb
r
d
dt
mvb
dt
d
mrrm

17. Theelectroncannotbestationaryinthismodel
becausethereisnothingtobalancetheforceof
nucleus.
Iftheelectroninmotion,however,dynamicallystable
orbitsarepossiblelikethoseoftheplanetsaroundthe
sun.
Rutherfordassumedcircularorbitforconvenience,
thoughitmightbereasonablybeassumedtobe
ellipticalinshape.
18
Continued…

18. 192
2
mv
F
C 2
2
0
4
1
r
e
F
e ecFF 2
2
0
2
1
2 r
emv
The centripetal force holding the electron in
an orbit r from the nucleus is provided
by the electric force
The condition for dynamical stable orbit
ise
F
F
Electron
r mr
e
v
0
4
Continued…
Fig. 1.2

19. 20
Total energy of atom (hydrogen) E0
22
42
emv
E
PEKEE r
e
E
r
e
r
e
E
0
2
0
2
0
2
8
48
(Putting value of ‘v’ )
The total energy is –vethis holds for every atomic electron.
This reflects the fact that it is bound to the nucleus. 0
22
42
emv
E
PEKEE r
e
E
r
e
r
e
E
0
2
0
2
0
2
8
48
(Putting value of ‘v’ )

20. Bytheapplicationofnewton’slawofmotionsand
Coulomb’slawexperimentalobservationthatatoms
arestable.But,
“Accelerated Electric Charges Radiates Energy In The
Form Of EM Waves.”
Anelectronacceleratingincurvedpathshould
continuouslyloseenergy,spiralingintothenucleusin
afractionofasecond.
Butatomsdonotcollapse.
21

21. Thank You

22. 12
v
o
m
Impact
parameter
Fig. 1.3a
Fig. 1.3b
P = P = mv
1 2