The electric potential at the surface of an atomic nucleus (Z=50) of radius 9.0 x 10-15 m is :
Answer & explanation
Correct answer: option 2
The correct answer is Option 2: 8 x 106 V
-
Atomic Number ($Z$): 50
-
Charge of a proton ($e$): $1.6 \times 10^{-19}\text{ C}$
-
Total Charge ($Q$): $Z \times e = 50 \times 1.6 \times 10^{-19}\text{ C} = 80 \times 10^{-19}\text{ C}$
-
Radius ($r$): $9.0 \times 10^{-15}\text{ m}$
-
Coulomb's Constant ($k$): $9 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2$
$V = \frac{(9 \times 10^9) \times (80 \times 10^{-19})}{9.0 \times 10^{-15}}$
$V = 80 \times 10^5\text{ V}$
$V = 8 \times 10^6\text{ V}$