Question

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

Answer 1

Highest energy will be equal to
`9.6* 10^(-13)J`

Step-by-step explanation:

Charged on doubly ionized helium atom
`q=2e=2* 1.6* 10^(-19)C=3.2* 10^(-19)C`

It is accelerated with maximum voltage of 3 MV

So voltage
`V=3* 10^6volt`

Now energy is given by
`E=qV=3.2* 10^(-19)* 3* 10^6=9.6* 10^(-13)J`

So highest energy will be equal to
`9.6* 10^(-13)J`