You cause a particle to move from point A, where the electric potential is 14.1 V, to point B, where the electric potential is −28.5 V. Calculate the change that occurs in the particle's electrostatic potential energy, when the particle is an electron, a proton, a neutral hydrogen atom, and a singly ionized helium atom (i.e., lacking one electron from its neutral state).

Respuesta :

Explanation:

It is given that,

The electric potential at point A, [tex]V_A=14.1\ V[/tex]

The electric potential at point B, [tex]V_B=-28.5\ V[/tex]

We know that the relationship between the particle's electrostatic potential energy and the electric potential is given by :

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

q is the charge

U is the electrostatic potential energy

For an electron, [tex]q_e=-1.6\times 10^{-19}\ C[/tex]

For a proton, [tex]q_p=1.6\times 10^{-19}\ C[/tex]

Change in electric potential,

[tex]\Delta V=V_B-V_A[/tex]

[tex]\Delta V=-28.5-14.1=-42.6\ V[/tex]

Electrostatic potential energy for an electron is :

[tex]U=-42.6\times (-1.6\times 10^{-19})[/tex]

[tex]U=6.81\times 10^{-18}\ J[/tex]

Electrostatic potential energy for a proton is :

[tex]U=-42.6\times (1.6\times 10^{-19})[/tex]

[tex]U=-6.81\times 10^{-18}\ J[/tex]

Electrostatic potential energy for a neutral hydrogen atom (q = 0) is :

U = 0

Electrostatic potential energy for a singly ionized helium atom is same as the proton.

The electrostatic potential energy for:

electron = [tex]6.81\times 10^{-18}J[/tex]

proton = [tex]-6.81\times 10^{-18}J[/tex]

Hydrogen atom = 0

Singly ionized Helium atom = [tex]-6.81\times 10^{-18}J[/tex]

According to the question:

let the electric potential at point A be [tex]V_A=14.1V[/tex]

and at point B be [tex]V_B=-28.5V[/tex]

The potential difference between the points B and A is(since we are bringing the charge from point A to B):

[tex]\triangle V=V_B-V_A\\\\\triangle V=28.5-14.1V\\\\\triangle V=-42.6V[/tex]

Now the potential energy U is given by:

U = ΔV×q

where q is the charge of the particle.

(i) for electron [tex]q=-1.6\times 10^{-19}C[/tex]

so,

[tex]U=-42.6\times (-1.6\times 10^{-19})J\\\\U=6.81\times 10^{-18}J[/tex]

(ii) for proton [tex]q=1.6\times 10^{-19}C[/tex]

[tex]U=42.6\times (-1.6\times 10^{-19})J\\\\U=-6.81\times 10^{-18}J[/tex]

(iii)for neutral Hydrogen atom q=0

so, U = 0

(iv) Singly ionized Helium atom has charge same as a proton [tex]q=1.6\times 10^{-19}C[/tex]

therefore, [tex]U=-6.81\times 10^{-18}J[/tex]

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