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

Answer:

Highest energy will be equal to $9.6\times 10^{-13}J$

Explanation:

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

It is accelerated with maximum voltage of 3 MV

So voltage $V=3\times 10^6volt$

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

So highest energy will be equal to $9.6\times 10^{-13}J$