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 α = 3.20×10⁻¹⁹ C and mα = 6.68×10⁻²⁷ kg , respectively.

Respuesta :

Answer:

Speed of the alpha particle is [tex]v=1.8180\times 10^3m/sec[/tex]      

Explanation:

We have given charge on alpha particle [tex]q=3.2\times 10^{-19}C[/tex]

Mass of the alpha particle [tex]m=6.68\times 10^{-27}kg[/tex]

Potential difference [tex]V=-3.45\times 10^{-3}volt[/tex]

We have to find the speed of the alpha particle

From energy conservation we know that

[tex]\frac{1}{2}mv^2=qV[/tex]

[tex]\frac{1}{2}\times 6.68\times 10^{-27}\times v^2=3.2\times 10^{-19}\times 3.45\times 10^{-3}[/tex]

[tex]v=1.8180\times 10^3m/sec[/tex]