Rutherford’s gold - foil experiment , Physics

Assignment Help:

An α particle is a positively charged particle.
Rutherford observed the behaviour of these particles when they approach the interior of the atoms as shown in fig. 

Rutherford observed that most of the α particles like a, a1 , pass through the atom un deflected.

2343_Rutherfords gold foil experiment with.png

Some of the particles like b, b1  get scattered by the atom at smaller angles of deflections. The particles like c, c1  undergo large deflection at an angle less than 180o  but more than 90o . The particles like ‘d' get deflected such that they are sent back as d1  with an angle of 180o.
Such large - angle deflections require strong forces to be acting on the α particles. Rutherford argued that this would be possible if all the positive charges and mass of the atom was concentratedin a very small central region which he called the nucleus of the atom. Then the large angle deflections of α particles are c, c1, due to electric repulsive forces caused by the nucleus. From the data obtained in this experiment Rutherford calculated the radius of the nucleus and found it to be shorter than 2.4×10-15m. Because the mass of the electron is only about 1/7000th mass of an alpha particle, the effect of the presence of electrons inside the atom on the deflection of the alpha particles can safely be ignored.

 


Related Discussions:- Rutherford’s gold - foil experiment

Find the voltage gain after feedback, Q. When there is no feedback, the gai...

Q. When there is no feedback, the gain of the amplifier is 100. If 5% of the output voltage is fed back into the input through a negative network, find the voltage gain after feedb

Mass of a sphere of diameter 1cm whose density is 900 kg/m^3, mass=volume *...

mass=volume * density.By using this formula we can find the mass of sphere. Volume of sphere=(4/3)pi*r^3 V=(4/3)*3.14*(1*10^-2)^3=4.187*10^-6 Mass=(4.187*10^-6)*900=3768*10^-6kg

Semi conductor , how potential barrier is formed at the pn junction diode

how potential barrier is formed at the pn junction diode

What is alpha decay and beta decay, The graph below shows a sequence of alp...

The graph below shows a sequence of alpha and beta decays, labeled 1, 2, 3, and 4. Consult Table 30-1 as required.   a) Which shows alpha decays, and which represen

Superconductivity, Superconductivity The phenomena from which, at suff...

Superconductivity The phenomena from which, at sufficiently low temperatures, a conductor can conduct charge along with zero resistance. The current theory for describing supe

Determine the diameter of ring in newtons rings experiment, In a Newton's r...

In a Newton's rings arrangement the diameter of a bright ring is 0.5cm. Determine the diameter of ring if the lens placed on the plane glass plate is replaced by another one having

Half-life simulation, Aim : To obtain the half-life of two radioisotopes by...

Aim : To obtain the half-life of two radioisotopes by graphical means, using data from a simulated experiment. Theory :  Half-life( t ½ ) is the time it takes any particular ma

Kelvin effect, Thomson experiment; Kelvin effect (Sir W. Thomson [later Lor...

Thomson experiment; Kelvin effect (Sir W. Thomson [later Lord Kelvin]): While an electric current pass through a conductor whose ends are maintained at distinct temperatures,

Explain point charge - electric field, What are the magnitude and the sign ...

What are the magnitude and the sign of a point charge that experiences a force of   0.48 N east when placed in an electric   field of 1.6 x 10 5 N/C west? Normal 0

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd