contestada

What is the highest energy to which doubly ionized helium atoms (alpha particles) can be accelerated in a DC accelerator with 3 MV maximum voltage

Respuesta :

Answer:

Highest energy will be equal to [tex]9.6\times 10^{-13}J[/tex]

Explanation:

Charged on doubly ionized helium atom [tex]q=2e=2\times 1.6\times 10^{-19}C=3.2\times 10^{-19}C[/tex]

It is accelerated with maximum voltage of 3 MV

So voltage [tex]V=3\times 10^6volt[/tex]

Now energy is given by [tex]E=qV=3.2\times 10^{-19}\times 3\times 10^6=9.6\times 10^{-13}J[/tex]

So highest energy will be equal to [tex]9.6\times 10^{-13}J[/tex]