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Question

In the Rutherford's scattering experiment, deflection of alpha (α) particles is due to

The correct answer is

Force of repulsion

Rutherford's Scattering Experiment and Alpha Particle Deflection

Rutherford's alpha ($\alpha$) particle scattering experiment was a landmark study that led to the discovery of the atomic nucleus. In this experiment, a beam of positively charged alpha particles was directed at a thin gold foil. The observation of how these alpha particles behaved after interacting with the gold atoms provided crucial insights into the structure of the atom.

Alpha Particle Interaction with the Nucleus

Alpha particles are composed of two protons and two neutrons, giving them a net positive charge ($+2e$). According to Rutherford's model, an atom consists of a tiny, dense, positively charged nucleus at its center, with electrons orbiting around it. When an alpha particle approaches an atom in the gold foil, its interaction with the atom's components determines its path.

  • Interaction with Electrons: Electrons are very light and negatively charged. The electrostatic force between the heavy alpha particle and the light electrons is negligible. Therefore, electrons do not cause significant deflection of alpha particles.
  • Interaction with the Nucleus: The atomic nucleus is positively charged, just like the alpha particles. When two positively charged particles come close to each other, they experience a strong electrostatic force of repulsion. This force pushes them away from each other.

Understanding the Force of Repulsion

The deflection or scattering of the alpha particles in Rutherford's experiment is primarily due to the electrostatic force of repulsion between the positively charged alpha particle and the positively charged atomic nucleus. As an alpha particle approaches the nucleus, this repulsive force causes its trajectory to bend, leading to deflection. The closer the alpha particle gets to the nucleus, the stronger the repulsive force, and thus, the greater the angle of deflection. Some alpha particles even experienced large deflections, some almost 180 degrees, indicating a direct collision with the highly dense, positively charged nucleus.

Why Other Options are Not the Primary Cause of Deflection

  • Increase in Kinetic energy: The deflection is not caused by an increase in kinetic energy. In fact, as a positively charged alpha particle approaches the positively charged nucleus, its kinetic energy is momentarily converted into potential energy due to the repulsive force. The deflection happens because of the change in direction due to the force, not because of a kinetic energy increase.
  • Force of attraction: A force of attraction would imply that the alpha particle and the nucleus have opposite charges. Since both are positively charged, they repel each other, rather than attract. If there were a significant force of attraction, the alpha particles would be drawn towards the nucleus, leading to a different scattering pattern or even capture, which was not observed.
  • Decrease in Kinetic energy: While the kinetic energy of an alpha particle does decrease as it approaches the nucleus (being converted to potential energy), this decrease is a *consequence* of the repulsive force acting on the particle, not the *cause* of the deflection. The deflection itself is a direct result of the force changing the particle's momentum and direction.

Therefore, the fundamental reason for the observed deflection of alpha particles in Rutherford's scattering experiment is the strong electrostatic force of repulsion between the positively charged alpha particles and the positively charged atomic nuclei.

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Important Questions from Atoms

  1. If $M$ is the mass of water that rises in a capillary tube of radius $r$, then what would be the total mass of water that rises if a capillary tube of radius $r$ and another capillary tube of radius $2r$ are simultaneously placed in water, assuming identical liquid and material properties?

  2. The diameter of an atom is

  3. The ratio of specific charge of a proton and a α-particle is

  4. The ratio of radii of two nuclei having atomic mass numbers 27 and 8 respectively, will be:

  5. A $Be^{3+}$ ion, initially in its second excited state, absorbs a photon of wavelength $601.6\text{ A}$. The radius of the ion in the resulting excited state in terms of Bohr radius $a_0$ will be (Take $hc = 12500\text{ eV-A}$)

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