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To practice Problem-Solving Strategy 21.1 Conservation of energy in charge interactions. 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.

Respuesta :

Answer:

Explanation:

kinetic energy of alpha particle

= Q X V ( Q is charge on the particle and V is potential difference )

= 3.2 x 10⁻¹⁹ x 3.45 x 10⁻³

= 11.04 x 10⁻²² J

1/2 m v² = 11.04 x 10⁻²²

1/2 x 6.68 x 10⁻²⁷ x v² = 11.04 x 10⁻²²

v² = 3.305 x 10⁵

v = 5.75 x 10² m /s

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