To solve this problem we will apply the concepts related to the calculation of work through the equations of the electric variables that define the energy stored on a body, such as the product of the load and the voltage. Later we will find the change in potential energy, such as the product between the electric field and the charge of the electron.
[tex]W = -q \Delta V[/tex]
Here,
q = Charge of nucleus
[tex]\Delta V =[/tex] Electric Potential
But at the same time, we have that Work is equal to the electric field for the charge of a electron, then the work made is,
[tex]W = Eq[/tex]
Where,
[tex]q = 9.6110^{-19}C[/tex]
[tex]E = 2.75*10^{3}eV[/tex]
Then replacing we have that,
[tex]W = (2.75*10^3eV)(1.6*10^{-19}C)[/tex]
[tex]W = 4.4*10^{-16}J[/tex]
Then from the first equation we have,
[tex]W = q \Delta V[/tex]
[tex]4.4*10^{-16}= 9.61*10^{-19} \Delta V[/tex]
[tex]\Delta V = 457.85V[/tex]
PART B)
Change in [tex]PE = \Delta W[/tex]
[tex]PE = 4.4*10^{-16}J[/tex]
The change in the nucleus' electric potential is "457.85 V" and electric potential energy is "4.4×10⁻¹⁶ J".
According to the question,
As we know the formula,
→ [tex]W = Eq[/tex]
By substituting the values, we get
→ [tex]= (2.75\times 10^3)(1.6\times 10^{-19})[/tex]
→ [tex]= 4.4\times 10^{-16} \ J[/tex]
then,
→ [tex]W = q \Delta V[/tex]
→ [tex]4.4\times 10^{-16} = 9.61\times 10^{-19} \ \Delta V[/tex]
→ [tex]\Delta V = 457.85 \ V[/tex]
and,
→ Change in P.E will be:
= [tex]\Delta W[/tex]
= [tex]4.4\times 10^{-16} \ J[/tex]
Thus the above response is right.
Learn more:
https://brainly.com/question/13488748
