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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..

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skyluke89

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

[tex]\Delta U = q\Delta V[/tex]

where

q is the charge of the particle

[tex]\Delta V[/tex] is the potential difference

For a proton,

[tex]q=+e=1.6\cdot 10^{-19}C[/tex]

And since [tex]\Delta V=100 V[/tex]

The energy gained by the proton is

[tex]\Delta U=(1.6\cdot 10^{-19})(100)=1.6\cdot 10^{-17}J[/tex]

For an alpha particle,

[tex]q=+2e=3.2\cdot 10^{-19}C[/tex]

Therefore, the energy gained is

[tex]\Delta U=(3.2\cdot 10^{-19})(100)=3.2\cdot 10^{-17}J[/tex]

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

[tex]q=+e=1.6\cdot 10^{-19}C[/tex]

So the energy gained is the same as the proton:

[tex]\Delta U=(1.6\cdot 10^{-19})(100)=1.6\cdot 10^{-17}J[/tex]

The energy that proton and alpha particle is [tex]1.6\times 10^{-17}{\rm J[/tex] and [tex]3.2\times 10^{-17}{\rm J[/tex] respectively.

The energy gained by proton can be determined by,

[tex]\Delta U = q \times \Delta V[/tex]

Where,

[tex]\Delta U[/tex] -  energy gained

[tex]q[/tex] - charge = [tex]1.6\times 10^{-19}\rm C[/tex]

[tex]\Delta V[/tex] - change in velocity  = 100 V

Put the values,

[tex]\Delta U =1.6\times 10^{-19}{\rm C }\times 100 \rm \ V\\\\\Delta U = 1.6\times 10^{-17}{\rm J }[/tex]

Since, an alpha particle has  2 protons, hence they have two times higher energy ([tex]3.2\times 10^{-17}{\rm J }[/tex]).

Therefore, the energy that proton and alpha particle is [tex]1.6\times 10^{-17}{\rm J[/tex] and [tex]3.2\times 10^{-17}{\rm J[/tex] respectively.

To know more about charge of particles,

https://brainly.com/question/6903736

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