An alpha particle having kinetic energy equal to 8.7 MeV is projected towards the nucleus of copper with Z=29.Calculate its distance of closet approach.

An alpha -particle having kinetic energy 5 MeV falls on a Cu-foil. The shortest distance from the nucleus of Cu to which alpha - particle reaches is (Atomic no. of Cu=29, K=9xx10^(9)Nm^(2)//C^(2) )

An alpha particle having 12.5 MeV kinetic energy collides with copper nucleus (Z=29) and retraces it s path .What is the size of the nucleus? Given, 1/(4 pi e_0) = 9 xx 10^9 N m^2 C^(-2) .

In Geiger-Marsden experiment,a 5 MeV alpha -particle is scattered through 180^@ from the gold foil.If the atomic number of gold is 79 ,find the distance of closet appraoch to the gold nucleus.

Direction. In the follwing questions, a statement of assertion is followed by a statement of reason. While answering a question, you are required to choose the correct are out of the given four responses and mark it as (A). If both assertion and reason are true and reason is the correct explanation of the assertion, (B) if both assertion and reason are true but reason is not correct explanation of the assertion. (C) if asssertion is true, but reason if false, (D) if both assertion and reason are false Assertion : A proton and an alpha particle having the same kinetic energy are moving in ircular paths in a uniform magnetic field. the radii of the ciruclar paths will be equal Reasons: Any two charged particles having equal kinetic energies and entering a region of uniform magnetic field vecb in a direction perpendicualr to the vecb , will describe circular trajectories of equal radii.

In a head on collision between an alpha particle and a gold nucleius (Z=79),the distnace of closet approach is 39.5 fermi.Calculate the enrgy of alpha particle.

The nucleus of an atom of ._(92)Y^(235) initially at rest decays by emitting an alpha particle. The binding energy per nucleon of parent and dougther nuclei are 7.8MeV and 7.835MeV respectively and that of alpha particles is 7.07MeV//"nucleon". Assuming the dougther nucleus to be formed in the unexcited state and neglecting its share of energy in the reaction, calculate speed of emitted alpha particle. Take mass of alpha particle to be 6.68xx10^(-27)kg .

A nucleus at rest undergoes a decay emitting an alpha -particle of de-broglie wavelength lambda = 5.76 xx 10^(-15) m . The mass of daughter nucleus is 223.40 amu and that of alpha-particle is 4.002 a.m.u.detrmine the total kinetic energy in the final state.Hence,obtain teh mass of the parent nucleus in a.m.u. Given that 1 a.m.u.=931.470 MeV .

An alpha particle (_2He^4) is moving directly towards a stationary gold nucleus (_(79)Au^(197)) .The alpha -particle and the gold nucleus may be considered to be solid spheres with the charge and mass concentrated at the centre of each sphere.When the two spheres are just toching,the sepaation of their centres is 9.6 xx 10^(-15) m . the alpha particle and the gold nucleus may be assumed to be an isolatedd system.Calculate for the alpha particle just in contact with the gold nucleus, its electric potential energy.

Two fixed, equal, positive charges, each of magnitude 5xx10^-5 coul are located at points A and B separated by a distance of 6m. An equal and opposite charge moves towards them along the line COD, the perpendicular bisector of the line AB. The moving charge, when it reaches the point C at a distance of 4m from O, has a kinetic energy of 4 joules. Calculate the distance of the farthest point D which the negative charge will reach before returning towards C.

When a deuton of mass 2.0141 u and neglible kinetic energy is absorbed by lithium ("_3^6Li) of nucleus of mass 6.0155 u, the compound nucleus disintegrates spontaneously into two alpha particles, each of mass 4.0026u. Calculate the energy in joules by each alpha particle. (1u=1.66 xx 10^-27kg)