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

Answer 1

Speed of the alpha particle is
`v=1.8180* 10^3m/sec`

Step-by-step explanation:

We have given charge on alpha particle
`q=3.2* 10^(-19)C`

Mass of the alpha particle
`m=6.68* 10^(-27)kg`

Potential difference
`V=-3.45* 10^(-3)volt`

We have to find the speed of the alpha particle

From energy conservation we know that

`(1)/(2)mv^2=qV`

`(1)/(2)* 6.68* 10^(-27)* v^2=3.2* 10^(-19)* 3.45* 10^(-3)`

`v=1.8180* 10^3m/sec`