To find the intensity of the electric field at the position of the electron in a hydrogen atom, as well as the electrostatic potential energy of the system, we can use some fundamental concepts from electrostatics. Let's break this down step by step.
Electric Field Intensity Calculation
The electric field intensity (E) created by a point charge can be calculated using the formula:
E = k * |q| / r²
Where:
- E is the electric field intensity.
- k is Coulomb's constant, approximately 8.99 x 10^9 N m²/C².
- q is the charge of the nucleus (proton in hydrogen), which is approximately 1.6 x 10^-19 C.
- r is the distance from the nucleus to the electron, given as 5 x 10^-11 m.
Now, substituting the values into the formula:
E = (8.99 x 10^9 N m²/C²) * (1.6 x 10^-19 C) / (5 x 10^-11 m)²
Calculating the denominator:
(5 x 10^-11 m)² = 25 x 10^-22 m² = 2.5 x 10^-21 m²
Now substituting back into the electric field formula:
E = (8.99 x 10^9) * (1.6 x 10^-19) / (2.5 x 10^-21)
Calculating the numerator:
(8.99 x 10^9) * (1.6 x 10^-19) = 1.4384 x 10^-9 N m²/C
Now, dividing by the denominator:
E = 1.4384 x 10^-9 N m²/C / 2.5 x 10^-21 m² = 5.7536 x 10^11 N/C
Thus, the intensity of the electric field at the position of the electron is approximately 5.75 x 10^11 N/C.
Electrostatic Potential Energy Calculation
The electrostatic potential energy (U) of a system of two point charges can be calculated using the formula:
U = k * q1 * q2 / r
For a hydrogen atom, we have:
- q1 = charge of the proton = 1.6 x 10^-19 C
- q2 = charge of the electron = -1.6 x 10^-19 C
- r = distance between the proton and electron = 5 x 10^-11 m
Substituting these values into the potential energy formula:
U = (8.99 x 10^9 N m²/C²) * (1.6 x 10^-19 C) * (-1.6 x 10^-19 C) / (5 x 10^-11 m)
Calculating the numerator:
(1.6 x 10^-19) * (-1.6 x 10^-19) = -2.56 x 10^-38 C²
Now substituting back into the potential energy formula:
U = (8.99 x 10^9) * (-2.56 x 10^-38) / (5 x 10^-11)
Calculating the numerator:
(8.99 x 10^9) * (-2.56 x 10^-38) = -2.30304 x 10^-28 N m²/C²
Now dividing by the denominator:
U = -2.30304 x 10^-28 / 5 x 10^-11 = -4.60608 x 10^-18 J
Therefore, the electrostatic potential energy of the hydrogen nucleus and electron system is approximately -4.61 x 10^-18 J.
Summary of Results
In summary, we found that:
- The intensity of the electric field at the position of the electron is approximately 5.75 x 10^11 N/C.
- The electrostatic potential energy of the hydrogen nucleus and electron system is approximately -4.61 x 10^-18 J.
This analysis illustrates the fundamental principles of electrostatics and how they apply to atomic structures. If you have any further questions or need clarification on any part of this, feel free to ask!
