An alpha particle (Z = 2, mass 6.64 × 10-27 kg) approaches to within 1.00 × 10-14 m of a carbon nucleus (Z = 6). What are (a) the maximum Coulomb force on the alpha particle, (b) the acceleration of the alpha particle at this time, and (c) the potential energy of the alpha particle at the same time?
The Correct Answer and Explanation is :
Let’s go step by step to solve each part of the problem:
Given:
- Alpha particle: ( Z_1 = 2 ), mass ( m_1 = 6.64 \times 10^{-27} \, \text{kg} )
- Carbon nucleus: ( Z_2 = 6 )
- Distance between particles: ( r = 1.00 \times 10^{-14} \, \text{m} )
- Coulomb’s constant: ( k_e = 8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2 )
(a) Maximum Coulomb Force on the Alpha Particle
The Coulomb force between two charged particles is given by Coulomb’s Law:
[
F = k_e \frac{Z_1 Z_2 e^2}{r^2}
]
Where:
- ( e ) is the elementary charge ( e = 1.60 \times 10^{-19} \, \text{C} )
- ( Z_1 ) and ( Z_2 ) are the atomic numbers of the particles
- ( r ) is the distance between the particles
Substituting the values:
[
F = \left(8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2 \right) \times \frac{(2)(6) \left(1.60 \times 10^{-19} \right)^2}{\left(1.00 \times 10^{-14} \right)^2}
]
[
F = 2.30 \times 10^{23} \, \text{N}
]
(b) Acceleration of the Alpha Particle
The acceleration of the alpha particle can be calculated using Newton’s second law:
[
a = \frac{F}{m}
]
Where:
- ( F ) is the Coulomb force
- ( m ) is the mass of the alpha particle
Substituting the values:
[
a = \frac{2.30 \times 10^{23} \, \text{N}}{6.64 \times 10^{-27} \, \text{kg}} = 3.47 \times 10^{53} \, \text{m/s}^2
]
(c) Potential Energy of the Alpha Particle
The potential energy between two charged particles is given by:
[
U = k_e \frac{Z_1 Z_2 e^2}{r}
]
Substituting the values:
[
U = \left(8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2 \right) \times \frac{(2)(6) \left(1.60 \times 10^{-19} \right)^2}{1.00 \times 10^{-14}}
]
[
U = 1.76 \times 10^{-12} \, \text{J}
]
Summary of Results:
- (a) The maximum Coulomb force on the alpha particle: ( F = 2.30 \times 10^{23} \, \text{N} )
- (b) The acceleration of the alpha particle: ( a = 3.47 \times 10^{53} \, \text{m/s}^2 )
- (c) The potential energy of the alpha particle: ( U = 1.76 \times 10^{-12} \, \text{J} )
Explanation:
In this scenario, the alpha particle, with charge ( +2e ), is attracted to the carbon nucleus with charge ( +6e ) due to their electrostatic interaction. The Coulomb force is strongest when the distance between the particles is at its minimum, which is given as ( 1.00 \times 10^{-14} \, \text{m} ). The force calculated from Coulomb’s law shows a very strong repulsive force at this short distance.
From Newton’s second law, the alpha particle experiences an enormous acceleration, which is also a result of this large Coulomb force acting on its relatively small mass.
The potential energy represents the energy stored in the system due to the electrostatic interaction between the two charged particles. This energy is positive because both particles have like charges, meaning they repel each other.
These calculations demonstrate the significance of Coulomb forces in interactions at the atomic level, where such immense forces and accelerations can occur despite the relatively small masses involved.