A) The electric potential energy of the alpha particle is a maximum at P.**
This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Here's an analysis of each statement:
A) The electric potential energy of the $\alpha$ particle is a maximum at P.
The electric potential energy ($U$) between two positive charges (alpha particle and gold nucleus) is given by $U = k \frac{q_1 q_2}{r}$, where $r$ is the distance between them. As the alpha particle approaches the gold nucleus, $r$ decreases, and thus $U$ increases. At point P, the alpha particle is at its closest distance to the nucleus before reversing direction, meaning $r$ is at its minimum. Therefore, the electric potential energy is at its maximum at P. This statement is correct.
B) The kinetic energy of the $\alpha$ particle is a minimum at P.
As the alpha particle approaches the positively charged gold nucleus, it experiences a repulsive electrostatic force that slows it down. At point P, the particle momentarily stops before reversing its direction. When the particle stops, its velocity is zero, and thus its kinetic energy ($KE = \frac{1}{2}mv^2$) is zero. Zero is the minimum possible kinetic energy. This statement is correct.
C) The total energy of the $\alpha$ particle is zero.
The total energy of the alpha particle is the sum of its kinetic energy and electric potential energy ($E_{total} = KE + U$). Since the electrostatic force is a conservative force, the total energy of the alpha particle is conserved throughout its motion. The alpha particle starts with some initial kinetic energy and moves towards a repulsive potential. At point P, its kinetic energy is zero, but its potential energy is at a maximum positive value (as explained in A). Therefore, the total energy at P is $E_{total} = 0 + U_{max} = U_{max}$, which is a positive value. The total energy is not zero. This statement is incorrect.
D) The total energy of the $\alpha$ particle has a constant positive value.
As mentioned in C, the total mechanical energy of the alpha particle is conserved because the electrostatic force is a conservative force. Since the alpha particle has initial kinetic energy and moves in a repulsive potential field, its total energy will be a constant positive value. This statement is correct.
The question asks for the incorrect statement.
The final answer is $\boxed{\text{C}}$
