Question

A beam of alpha particles is incident on a target of lead. A particular alpha particle comes in "head-on"to a particular lead nucleus and stops 7.00×10−14 m away from the center of the nucleus. (This point is well outside the nucleus). Assume that the lead nucleus, which has 82 protons, remains at rest. The mass of the alpha particle is 6.64×10−27kg.

a)Calculate the electrostatic potential energy at the instant that the alpha particle stops. Express your result in joules
b)Calculate the electrostatic potential energy at the instant that the alpha particle stops. Express your result in MeV.
c)What initial kinetic energy (in joules) did the alpha particle have?
d)What initial kinetic energy (in MeV) did the alpha particle have?
e)What was the initial speed of the alpha particle?

Answer 1

• a. The electrostatic potential energy at the instant that the alpha particle stops is 2.065 x 10⁻¹³ joules.
• b. The electrostatic potential energy at the instant that the alpha particle stops is 1.29 MeV.
• c. The initial kinetic energy of the alpha particle is 2.065 x 10⁻¹³joules.
• d. The initial kinetic energy of the alpha particle is 1.29 MeV.
• e. The initial speed of the alpha particle is approximately 7.88 x 10⁶ m/s.

Step-by-step explanation:

**a) To calculate the electrostatic potential energy at the instant that the alpha particle stops, we can use the formula:

Potential energy = (k * q₁ * q₂) / r

where k is the electrostatic constant (approximately 9.0 x 10⁹ N m²/C²), q₁ and q₂ are the charges of the alpha particle and the lead nucleus, respectively, and r is the distance between them.

The alpha particle has a charge of +2e, where e is the elementary charge (approximately 1.6 x 10⁻¹⁹ C). The lead nucleus has a charge of +82e.

Substituting these values into the formula, we get:

Potential energy = (9.0 x 10^9 N m²/C²) * (2e) * (82e) / (7.00 x 10⁻¹⁴ m)

Simplifying the expression and converting to joules, we find:

Potential energy = 2.065 x 10⁻¹³ J

Therefore, the electrostatic potential energy at the instant that the alpha particle stops is 2.065 x 10⁻¹³ joules.

**b) To express the electrostatic potential energy in MeV (mega-electron volts), we can use the conversion factor 1 MeV = 1.602 x 10⁻¹³ J.

Converting the previously calculated potential energy to MeV, we find:

Potential energy in MeV = (2.065 x 10^-13 J) / (1.602 x 10⁻¹³ J/MeV)

Simplifying the expression, we get:

Potential energy in MeV = 1.29 MeV

Therefore, the electrostatic potential energy at the instant that the alpha particle stops is 1.29 MeV.

**c) The initial kinetic energy of the alpha particle can be calculated using the conservation of energy principle. Since the alpha particle stops, its initial kinetic energy is equal to the electrostatic potential energy at that instant.

Therefore, the initial kinetic energy of the alpha particle is 2.065 x 10⁻¹³ joules.

**d) To express the initial kinetic energy in MeV, we can use the conversion factor 1 MeV = 1.602 x 10⁻¹³J.

Converting the previously calculated initial kinetic energy to MeV, we find:

Initial kinetic energy in MeV = (2.065 x 10⁻¹³ J) / (1.602 x 10⁻¹³ J/MeV)

Simplifying the expression, we get:

Initial kinetic energy in MeV = 1.29 MeV

Therefore, the initial kinetic energy of the alpha particle is 1.29 MeV.

**e) To calculate the initial speed of the alpha particle, we can use the kinetic energy formula:

Kinetic energy = (1/2) * m * v²

where m is the mass of the alpha particle and v is its speed.

Rearranging the formula, we get:

v² = (2 * Kinetic energy) / m

Substituting the values of kinetic energy (2.065 x 10⁻¹³J) and mass (6.64 x 10⁻²⁷ kg), we can solve for v:

v² = (2 * 2.065 x 10⁻¹³ J) / (6.64 x 10⁻²⁷ kg)

v² ≈ 6.21 x 10¹³ m²/s²

Taking the square root of both sides, we find:

v ≈ 7.88 x 10⁶ m/s

Therefore, the initial speed of the alpha particle is approximately 7.88 x 10⁶ m/s.