Question

Consider the lead isotope ${}^{207}Pb$.

a. How many electrons, protons, and neutrons are in a neutral${}^{207}Pb$atom?

b. The lead nucleus has a diameter of $14.2fm$. What is the electric field strength at the surface of a lead nucleus?

Short answer

a. In neutral ${}^{207}Pb$number of neutrons are$125$, number of protons are $82$and the number of electrons are$82$.

b. The electric field strength at the surface of the nucleus is$(2.34\times {10}^{21})\frac{N}{C}$.

Step-by-step solution

Part (a) step 1: Given information 

We have to find that How many electrons, protons, and neutrons are in a neutral ${}^{207}Pb$atom.

Part (a) step 2: Simplify 

Start the solution by determining the proton number:

$Z=82$ (from Periodic Table of Elements)

$A=Z+N$

$N=A-Z$

Putting the values

$N=207-82=125$

Here, $Z$is the atomic number, A is the mass number and N is the neutron number.

Part (b) step 1: Given information 

We have to find that the electric field strength at the surface of a lead nucleus.

Part (b) step 2: Simplify 

The expression for the electric field of the point charge:

$$\begin{gathered} E=\frac{Q}{4\pi {\in }_0{r}^2} \\ E=\frac{\left(82\right)·e}{4\pi {\in }_0(7.1\times {{10}^{-15})}^2} \\ E=(2.34\times {10}^{21})\frac{N}{C} \end{gathered}$$