An alpha particle with a charge of +2e is fixed at the origin. A proton is aimed at the alpha particle and
shot at it from a distance of 2 m away at a speed of 3.64 × 10^5m/s
How close will the proton come to the alpha particle before coming to a stop?
To calculate how close the proton will come to the alpha particle before coming to a stop, we can use the principles of electrostatics and conservation of energy.
The electric potential energy between two particles with charges Q1 and Q2 separated by a distance r is given by the equation:
PE = (k * |Q1 * Q2|) / r
Where k is Coulomb's constant (9 × 10^9 N∙m^2/C^2), Q1 and Q2 are the charges of the particles, and r is the distance between them.
When the proton is shot at the alpha particle, it has kinetic energy (KE) due to its speed. As the proton approaches the alpha particle, the electrical potential energy between them increases, converting the proton's kinetic energy into potential energy until it eventually comes to a stop.
At this point, the sum of the initial kinetic energy (KE_initial) and the increase in potential energy (ΔPE) is equal to zero:
KE_initial + ΔPE = 0
Since ΔPE is negative when the proton comes to a stop, we can rearrange the equation to solve for the distance (r) between the particles:
r = (k * |Q1 * Q2|) / KE_initial
Given that the alpha particle has a charge of +2e and the proton has a charge of +e, we can substitute these values into the equation. Also, the charge of an electron (e) is approximately 1.6 × 10^(-19) C.
r = (9 × 10^9 N∙m^2/C^2 * |(2e)(e)|) / KE_initial
Now we need to calculate the initial kinetic energy (KE_initial) of the proton. The kinetic energy of an object with mass m moving at a speed v is given by the equation:
KE = 0.5 * m * v^2
Substituting the given values:
KE_initial = 0.5 * m_p * (3.64 × 10^5 m/s)^2
Where m_p is the mass of the proton, which is approximately 1.67 × 10^(-27) kg.
Now we can substitute this value into the equation for r and solve for the distance:
r = (9 × 10^9 N∙m^2/C^2 * |(2e)(e)|) / (0.5 * m_p * (3.64 × 10^5 m/s)^2)
To evaluate this equation, we need the value of e. Substituting e ≈ 1.6 × 10^(-19) C:
r = (9 × 10^9 N∙m^2/C^2 * |(2 * 1.6 × 10^(-19) C)(1.6 × 10^(-19) C)|) / (0.5 * 1.67 × 10^(-27) kg * (3.64 × 10^5 m/s)^2)
After substitution, simplifying, and calculation, we find:
r ≈ 4.94 × 10^(-14) m
Therefore, the proton will come closest to the alpha particle at a distance of approximately 4.94 × 10^(-14) meters before coming to a stop.