Question

can u tell abt conservation of energy in rutherford alpha scaterring experiment

Answer 1

The total mechanical energy of the alpha particle-nucleus system is conserved throughout the Rutherford alpha scattering experiment. The initial kinetic energy of the alpha particle is converted into electrostatic potential energy as it approaches the nucleus, and this potential energy is then converted back into kinetic energy as it moves away. The alpha particle emerges with the same kinetic energy it started with, but its direction of motion is changed.

Answer 2

The conservation of energy in Rutherford's alpha scattering experiment is a fundamental principle governing the interaction between the alpha particle and the atomic nucleus. The interaction is primarily due to the electrostatic (Coulomb) force, which is a conservative force. Therefore, the total mechanical energy (kinetic energy + potential energy) of the alpha particle-nucleus system remains constant throughout the scattering process.

Here's how energy is conserved:

1. Initial State (Alpha particle far from nucleus):
   When the alpha particle is far away from the nucleus, the electrostatic interaction is negligible, so its potential energy ($U_i$) is considered zero. The alpha particle possesses only kinetic energy ($K_i$).
   Total mechanical energy $E_{total} = K_i + U_i = K_i$.

2. During Interaction (Alpha particle approaching nucleus):
   As the positively charged alpha particle approaches the positively charged nucleus, it experiences an electrostatic repulsive force. This force does negative work on the alpha particle, causing its kinetic energy ($K$) to decrease. The lost kinetic energy is simultaneously converted into electrostatic potential energy ($U$). The total mechanical energy ($K+U$) at any point remains constant and equal to the initial kinetic energy.
   The electrostatic potential energy between an alpha particle (charge $+2e$) and a nucleus (charge $+Ze$) at a distance $r$ is given by: $U = \frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{r} = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{r}$

3. At Closest Approach (for head-on collision, impact parameter $b=0$):
   If the alpha particle is directed straight towards the nucleus, it reaches a point of closest approach ($r_0$). At this point, due to the maximum repulsion, the alpha particle momentarily stops, meaning its kinetic energy ($K_{min}$) becomes zero. At this instant, all its initial kinetic energy has been completely converted into maximum electrostatic potential energy ($U_{max}$).
   Thus, by conservation of energy: $K_i = U_{max} = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{r_0}$

4. After Interaction (Alpha particle moving away from nucleus):
   After reaching the point of closest approach, the repulsive force pushes the alpha particle away from the nucleus. The electrostatic potential energy is now converted back into kinetic energy. When the alpha particle is again far away from the nucleus, its potential energy ($U_f$) becomes zero, and its kinetic energy ($K_f$) becomes equal to its initial kinetic energy ($K_i$).
   So, $K_f = K_i$.

This entire process is an example of an elastic scattering where the kinetic energy of the alpha particle is conserved before and after the collision, although its direction of motion is altered. The total mechanical energy of the system remains conserved throughout the interaction.

The interaction between the alpha particle and the nucleus is purely electrostatic and conservative. Initial kinetic energy of the alpha particle is converted into electrostatic potential energy as it approaches the nucleus, reaching maximum potential energy (zero kinetic energy) at the point of closest approach. This potential energy is then converted back into kinetic energy as the alpha particle moves away. The total mechanical energy (kinetic + potential) of the system remains constant, leading to the alpha particle having the same kinetic energy before and after scattering, making it an elastic collision.