RUTHERFORD ATOM MODEL

(ATOMIC STRUCTURE)

by

M.KANNAN

INTRODUCTION• knowing atomic structure is very important to understand• elements, their chemical reactions and nuclear reactions too. We human beings are very inquisitive to know correct atomic structure. Rutherford

conducted a fantastic experiment to find out atomic structure. Rutherford atom model gives more or less correct picture of an atom.

Alpha

particles

start

Lead screen

Thin gold foil Fluorescent screen

EXPERIMENT

DESCRIPTION

Alpha particles from the polonium sample passed through lead slit for a narrow beam. Then they passed through thin gold foil.A t the centre of the fluorescent screen more scintillations were observed .some were observed little above and few further above as it is shown in the figure.

Rutherford told positive charge of atom should be contained at the centre of the atom;only then repulsive force 2Ze*e/r Newtons (Z is atomic number, e is electron charge and r is radius of nucleus) could have deflected alpha particles.Deflection occurred because alpha particles too are positively charged.

THEORY

The fact that more alpha particles went straight shows that there is more space between electrons and the positively charged nucleus.

The angle of scattering depends upon the factor called ‘impact parameter.’ closer the approach of alpha particle more the de4flection was observed.

The impact parameter is given by the factor,b= Ze*e* cot0.5

0 .5mv*v 4

where is angle of scattering,m is mass of alpha particle&v is its velocity.

DEFINITION FOR IMPACT PARAMETER

Impact parameter(b) is the perpendicular distance between the velocity vector and centre of the nucleus.

b

Conclusion 1

Positive charge of atom is contained at the core called nucleus.

CONCLUSION 2

There is more space between positively charged nucleus andnegatively charged electrons.

CONCLUSION 3

The negatively charged electrons revolve around the positively charged nucleus.

SUMMARY

Conclusions 1,2 &3 constitute Rutherford Atom Model.

EXERCISE

Calculate angle of scattering if alpha particle retraces its path.

SOLUTION FOR EXERCISE

b=Ze*ecot0.5 0.5mv*v

Here b= 0so cot0.5hence 0.5and

ALL THE BEST! GOD BLESS YOU,DEAR

STUDENTS,AS YOU LEARN

MORE OF HIS CREATION.