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What conclusion is drawn from Rutherford...

What conclusion is drawn from Rutherford.s scattering experiment of `alpha`-particles?

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**Step-by-Step Solution:** 1. **Understanding the Experiment**: Rutherford conducted an experiment using a radioactive substance that emitted alpha particles. These particles were directed towards a very thin gold foil. 2. **Observation of Alpha Particles**: During the experiment, it was observed that: - Most of the alpha particles passed through the gold foil without any deflection. - A small number of alpha particles were deflected at small angles. - A very few alpha particles were deflected back along their original path. ...
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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 :

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 Radius of a particular nucleus is calculated by the projection of alpha -particle from infinity at a particular speed. Let this radius is the true radius . If the radius calculation for the same nucleus is made by another alpha -particle with half of the earlier speed, then the percentage error involved in the radius calculation is :

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Enlist the conclusions drawn by Rutherford from his alpha -ray scattering experiment.