4. What is the electric potential at the surface of a gold atom’s nucleus, which is assumed to be spherical? The radius of a gold nucleus is 6.6 x 10-15 meters and the atomic number of gold is 79.
V= kq/r ---------1.
k= 9*(10^9)
r= 6.6*(10^-15)m
q= ne = 79* 1.6*(10^-19)C
put these values in eq. 1.
V= 172.36 * 10^5
V= 1.7236 * 10^7 = (1.724 * 10^7)V
V=kq/r q= 79*(-e)/r where e is the charge on an electron.
1.72×10^-7
To find the electric potential at the surface of a gold atom's nucleus, we will use the formula for electric potential due to a point charge. The electric potential at a point due to a point charge is given by:
V = k * (q / r)
Where:
V is the electric potential
k is the Coulomb's constant (k = 8.99 x 10^9 N*m^2/C^2)
q is the charge of the nucleus
r is the distance from the nucleus to the point where we want to find the potential
In this case, we want to find the electric potential at the surface of the gold nucleus, so the distance (r) will be equal to the radius of the nucleus. The charge (q) of the nucleus is equal to the product of the atomic number (Z) and the elementary charge (e), where e = 1.6 x 10^-19 C.
Let's calculate the electric potential:
First, we find the charge:
q = Z * e = 79 * 1.6 x 10^-19 C
Now, we calculate the electric potential:
V = k * (q / r) = 8.99 x 10^9 N*m^2/C^2 * (79 * 1.6 x 10^-19 C) / (6.6 x 10^-15 m)