Question

The nuclei of large atoms, such as uranium, with 9292 protons, can be modeled as spherically symmetric spheres of charge. The radius of the uranium nucleus is approximately 7.4×10−15m7.4×10−15m.

Answer 1

Answer:

Step-by-step explanation:

Here is the complete question:

The nuclei of large atoms, such as uranium, with 92 protons, can be modeled as spherically symmetric spheres of charge. The radius of the uranium nucleus is approximately 7.4 x 10^-15 meters

What is the electric field this nucleus produces just outside its surface?

Express your answer using two significant figures.

_________?????____N/C

solution

The concepts that are to be used to solve the given problem are quantization of charge and the concept of electric field.

Use the formula for total charge and the electric field to solve this problem. Use the formula for the quantization of charge to find the total charge of the nucleus. Then use the concept of electric field to calculate the electric field.

From the quantization of charge, the total charge is the number of protons times the charge of each proton.

The total charge enclosed in the nucleus is given by,

$q = ne$

Here, n is the number of electrons and e is the electric charge.

The magnitude of electric field due to a point charge at a distance from the point charge is,

$E = \frac{{kq}}{{{r^2}}}$

Here, k is the electrostatic constant, q is the charge, and r is the distance.

 

The total charge of the nucleus is,

$q = ne$

Substitute 92 for n and $1.6 \times {10^{ - 19}}{\rm{ C}}$ for e.

[tex[\begin{array}{c}\\q = 92\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)\\\\ = 1.47 \times {10^{ - 17}}{\rm{ C}}\\\end{array}[/tex]  

The electric field due to the nucleus just outside the surface is,

$E = \frac{{kq}}{{{r^2}}}$

Substitute $9.0 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}$ for k, $1.47 \times {10^{ - 17}}{\rm{ C}}$ for q , and $7.4 \times {10^{ - 15}}{\rm{ m}}$ for r .

$\begin{array}{c}\\E = \frac{{\left( {9.0 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right)\left( {1.47 \times {{10}^{ - 17}}{\rm{ C}}} \right)}}{{{{\left( {7.4 \times {{10}^{ - 15}}{\rm{ m}}} \right)}^2}}}\\\\ = 2.415 \times {10^{21}}{\rm{ N/C}}\\\end{array}$

Round off to two significant figures. So, the magnitude of electric field is $2.4 \times {10^{21}}{\rm{ N/C}}$

The magnitude of electric field is $2.4 \times {10^{21}}{\rm{ N/C}}$