Question

A proton, an alpha particle (a bare helium nucleus), and a singly ionized helium atom are accelerated through a potential difference of 100 V. Find the energy that proton gains..

Answer 1

Answer:

The correct question is:

"Find the energy each gains"

The energy gained by a charged particle accelerated through a potential difference is given by

$\Delta U = q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

For a proton,

$q=+e=1.6\cdot 10^{-19}C$

And since $\Delta V=100 V$

The energy gained by the proton is

$\Delta U=(1.6\cdot 10^{-19})(100)=1.6\cdot 10^{-17}J$

For an alpha particle,

$q=+2e=3.2\cdot 10^{-19}C$

Therefore, the energy gained is

$\Delta U=(3.2\cdot 10^{-19})(100)=3.2\cdot 10^{-17}J$

Finally, for a singly ionized helium nucleus (a helium nucleus that has lost one electron)

$q=+e=1.6\cdot 10^{-19}C$

So the energy gained is the same as the proton:

$\Delta U=(1.6\cdot 10^{-19})(100)=1.6\cdot 10^{-17}J$