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

Answer:

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

Explanation:

We have given charge on alpha particle $q=3.2\times 10^{-19}C$

Mass of the alpha particle $m=6.68\times 10^{-27}kg$

Potential difference $V=-3.45\times 10^{-3}volt$

We have to find the speed of the alpha particle

From energy conservation we know that

$\frac{1}{2}mv^2=qV$

$\frac{1}{2}\times 6.68\times 10^{-27}\times v^2=3.2\times 10^{-19}\times 3.45\times 10^{-3}$

$v=1.8180\times 10^3m/sec$