Already have an account? Get multiple benefits of using own account!
Login in your account..!
Remember me
Don't have an account? Create your account in less than a minutes,
Forgot password? how can I recover my password now!
Enter right registered email to receive password!
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.
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.
The refractive index of glass is 1.54. Determine the polarizing angle when light goes from glass to air medium?
prove that equatorial line of dipole acts as equipotential surface
pedaling cycle is less friction or more
One of the most important theorems in the quantum physics in the Pauli Exclusion Principle, which is based on the experiment observations. The principle explains that “No two elect
Dc machine consists of the following parts: Field system: The object of the field system is to make a uniform magnetic field, within which the armature rotates. It having of
a) The molar mass of air is 28.97 kg/kmol and the universal gas constant is 8.314 kJ/(kmolK). Show by calculation that the particular gas constant of air is around 0.287 kJ/(kgK)
An electric motor with a load delivers 5.2 hp to its shaft (1 hp = 746 W). Under these conditions, it operates at 82.8 percent efficiency. a. How much current does the motor d
A 2.5 kg rock is located on the Earth''s surface. If the mass of the moon is 7.4 E 22 kg, and the separation distance between the center of the rock and the center of the moon is 3
Electric field lines In good metallic conductors: In good metallic conductors: (i) Static electric fields are not there (zero inside a conductor) (ii) Charges reside
Explain the Conservation of Momentum In a collision, energy is not all time conserved, if not the collision is totally elastic (all stored energy is turned into kinetic energy
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!
whatsapp: +91-977-207-8620
Phone: +91-977-207-8620
Email: [email protected]
All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd