Question

An alpha particle (α), which is the same as a helium-4 nucleus, is momentarily at rest in a region of space occupied by an electric field. The particle then begins to move. Find the speed of the alpha particle after it has moved through a potential difference of −3.45×10^−3 V. The charge and the mass of an alpha particle are qα = 3.20×10^−19 C and mα = 6.68×10^−27 kg , respectively.

Mechanical energy is conserved in the presence of which of the following types of forces?

Select all that apply.
A- electrostatic
B- frictional
C- magnetic
D- gravitational

Part B- Which of the following quantities are unknown?
A- the initial speed of the alpha particle
B- the value of the electric potential at the initial position of the alpha particle
C- the value of the electric potential at the final position of the alpha particle
D- the final speed of the alpha particle
E- the charge of the alpha particle
F- the difference in potential between the initial and final positions of the alpha particle
G- the mass of the alpha particle

Part C- Use conservation of energy (Ki+qVi=Kf+qVf) to solve for the final kinetic energy. Then use this value to solve for the final velocity of the alpha particle.

Answer 1

Answer:

Explanation:

Kinetic energy gained by alpha particle

= charge x potential difference

1/2 mv² = 3.2 x 10⁻¹⁹ x 3.45 x 10⁻³

.5 x 6.68 x 10⁻²⁷ V² = 11.04 x 10⁻²²

V² = 3.3  x 10⁵

V = 5.74 x 100

= 574 m / s

Mechanical energy is conserved in respect of A , C and D .

Part B

B , C, are unknown .