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In the Rutherford's experiment, a partic...

In the Rutherford's experiment, a particles were bombarded towards the copper atoms so as to arrives a distance of `10^(-13)` metre from the nucleus of copper and then getting either deflected or traversing back. The a-particles did not move further closer
It can also be concluded that the electrostatic potential energy is equal to

A

`(1)/(4pi epsilon_(0))(q_(1)q_(2))/(r)`

B

`(1)/(4pi epsilon_(0)) (5Ze^(2))/(mv^(2))`

C

`(Ze^(2))/(r)`

D

`mv^(2)`

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The correct Answer is:
A
`(1)/(4pi epsilon_(0))(q_(1)q_(2))/(r)`
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In the Rutherford's experiment, a particles were bombarded towards the copper atoms so as to arrives a distance of 10^(-13) metre from the nucleus of copper and then getting either deflected or traversing back. The a-particles did not move further closer The velocity of the a-particles must be

With what velocity should as alpha - particle travel toward the nucless of a copper atom at a distance of 10^(13)m from the nucless of the copper atom ?

Rutherford model: The approximate size of the nucleus can be calculated by using energy conservation theorem in Rutherford's alpha -scattering experiment. If an alpha -particle is projected from infinity with speed v towards the nucleus having Z protons, then the alpha -particle which is reflected back or which is deflected by 180^@ must have approached closest to the nucleus .It can be approximated that alpha particle collides with the nucleus and gets back. Now if we apply the energy conservation equation at initial point and collision point then: (P.E.)_i= 0 , since P.E. of two charge system separated by infinite distance is zero. Finally the particle stops and then starts coming back. 1/2m_alpha v_alpha^2+0=0+(Kq_1q_2)/Rimplies 1/2m_alphav_alpha^2=K(2exxZe)/R implies R=(4KZe^2)/(m_alphav_alpha^2) Thus the radius of nucleus can be calculated using above equation. The nucleus is so small a particle that we can't define a sharp boundary for it An alpha -particle with initial speed v_0 is projected from infinity and it approaches up to r_0 distance from a nuclie. Then, the initial speed of alpha -particle, which approaches upto 2r_0 distance from the nucleus is :

Read the following passage carefully and answer the questions The approximate size of the nucleus can be calculated by using energy conservation theorem in Rutherford's alpha- scattering experiment. If an alpha- perticle is projected from infinity with speed v, towards the nucleus having z protons then the apha- perticle which is reflected back or which is deflected by 180^(@) must have approach closest to the nucleus. It can be approximated that alpha- particles collides with the nucleus and gets back. Now if we apply the energy conservation at initial and collision point then: ("Total Energy")_("initial") = ("Total Enregy")_("final") (KE)_(i) +(PE)_(i)=(KE)_(f)+(PE)_(f) (PE)_(i) =0, "since"PE of two charge system separated by infinite distance is zero, finally the particle stops and then starts coming back. 1/2m_(alpha)v_(alpha)^(2) + 0=0 + (kq_(1)q_(2))/(R) rArr1/2m_(alpha)v_(alpha)^(2)=k(2exxze)/(R)rArrR=(4kze^(2))/(m_(alpha)v_(alpha)^(2) Thus the radius of nucleus can be calculated using above equation. The nucleus is so small a particle that we can't define a sharp boundary for it . Experiments show that the average radius R fo a nucleus may be written as R=R_(0)(A)^(1//3) where R_(0) = 1.2xx10^(-15)m A= atomic mass number R=radius of nucleus If alpha- particle with speed v_(0) is projected from infinity and it approaches upto r_(0) distance from the nuclei. Then the speed of alpha- particle which approaches 2r_(0) distance from the nucleus is

Read the following passage carefully and answer the questions The approximate size of the nucleus can be calculated by using energy conservation theorem in Rutherford's alpha- scattering experiment. If an alpha- perticle is projected from infinity with speed v, towards the nucleus having z protons then the apha- perticle which is reflected back or which is deflected by 180^(@) must have approach closest to the nucleus. It can be approximated that alpha- particles collides with the nucleus and gets back. Now if we apply the energy conservation at initial and collision point then: ("Total Energy")_("initial") = ("Total Enregy")_("final") (KE)_(i) +(PE)_(i)=(KE)_(f)+(PE)_(f) (PE)_(i) =0, "since"PE of two charge system separated by infinite distance is zero, finally the particle stops and then starts coming back. 1/2m_(alpha)v_(alpha)^(2) + 0=0 + (kq_(1)q_(2))/(R) rArr1/2m_(alpha)v_(alpha)^(2)=k(2exxze)/(R)rArrR=(4kze^(2))/(m_(alpha)v_(alpha)^(2) Thus the radius of nucleus can be calculated using above equation. The nucleus is so small a particle that we can't define a sharp boundary for it . Experiments show that the average radius R fo a nucleus may be written as R=R_(0)(A)^(1//3) where R_(0) = 1.2xx10^(-15)m A= atomic mass number R=radius of nucleus Radius of a particular nucleus is calculated by the projection of alpha- particle from infinity at a particular speed. Let this radius be the true radius. If the radius calculation for the same nucleus is made by , alpha- particle with half of the earlier speed then the percentage error involved in the radius calculation is :

The potential energy at a point, relative to the reference point is always defined as the negative of work done by the force as the object moves from the reference point to the point considered. The value of potential energy at the reference point itself can be set equal to zero because we are always concerned only with differences of potential energy between two points and the associated change of kinetic energy. A particles A is fixed at origin of a fixed coordinate system. A particle B which is free to move experiences an force vec(F) = (-(2a)/(r^(3)) + (beta)/(r^(2))) hat(r) due to particle A where hat(r) is the position vector of particle B relative to A. It is given that the force is conservative in nature and _potential energy at infinity is zero. If B has to be removed from the influence of A, energy has to be supplied for such a process. The ionization energy E 0 is work that has to be done by an external agent to move the particle from a distance r_(0) to infinity slowly. Here r_(0) is the equilibrium position of the particle What is potential energy function of particle as function of r.

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