If a polonium-210 (Po-210) atom has just decayed, how close can the ejected
alpha particle get to a neighboring Po-210 nucleus?
Hint: W = F*r
F=kq1q2/r
2
The particle has initial KE, lets say K.
That energy can be converted to PE as it approaches the nucleus.
When equal, the particle can go no further.
K= INTEGRAL F*dr= -kq1q2/r evaluated from infinity to x
K= kq1q2/x
so the minimum distance x= kq1q2/K
So look up the energy lost when the Po nucleus decays, the charge on the decayed nucleus, and the charge on the alpha particle. Then, it is just calculator work.
I don't quite understand
According to my physics text, Po-210 decays by alpha decay, leaving an alpha particle (+2e charge) , a Pb-206(-2) ion that will eventually neautalize, and 45.17 - 36.15 = 9.02 MeV of energy.
Most, but not all, of the energy will be kinetic energy of the alpha particle. The adjacent Po atom has nuclear charge of +84 e.
They probably expect you to ignore the energy loss as the alpha particle penetrates the electron cloud around the Po atom.
Where does the 45.17-36.15 come from and the +84 e?
To determine how close the ejected alpha particle can get to a neighboring Po-210 nucleus, we need to consider the electrostatic force between the two particles.
We can calculate the force using Coulomb's law formula:
F = (k * q1 * q2) / r^2
Where:
- F is the force between the two particles
- k is the electrostatic constant (approximately equal to 9 x 10^9 N*m^2/C^2)
- q1 and q2 are the charges of the particles
- r is the distance between the two particles
In the case of the alpha particle (He2+), it has a charge of +2e (where e represents the elementary charge, approximately 1.6 x 10^-19 C).
The Polonium-210 nucleus (Po-210) also has a charge of +2e.
Substituting the values into the formula, we get:
F = (k * (2e) * (2e)) / r^2
F = (4ke^2) / r^2
Since the question asks for how close the particles can get, we need to find the minimum distance at which the force between them becomes significant.
The closest distance occurs when the force of attraction between the two particles is equal to the repulsive force due to kinetic energy.
The kinetic energy of the alpha particle (K.E.) can be calculated using the following formula:
K.E. = (1/2) * m * v^2
Where:
- m is the mass of the alpha particle (approximately 6.64 x 10^-27 kg)
- v is the velocity of the alpha particle
The force due to kinetic energy (F_ke) is equal to K.E. divided by the distance between the two particles (r):
F_ke = K.E. / r
Now, we can set the electrostatic force equal to the force due to kinetic energy and solve for r:
(4ke^2) / r^2 = K.E. / r
Multiplying both sides by r^2:
4ke^2 = K.E.
Substituting the value of the kinetic energy:
4ke^2 = (1/2) * m * v^2
Solving for r:
r^2 = (4ke^2) / [(1/2) * m * v^2]
r^2 = (8ke^2) / (m * v^2)
r = sqrt((8ke^2) / (m * v^2))
Now, we can plug in the known values:
- k ≈ 9 x 10^9 N*m^2/C^2
- e ≈ 1.6 x 10^-19 C
- m ≈ 6.64 x 10^-27 kg
- v (velocity of the alpha particle) ≈ a significant fraction of the speed of light (approx. 3 x 10^8 m/s)
After plugging in these values, we can calculate the minimum distance at which the ejected alpha particle can get to a neighboring Polonium-210 nucleus.