Question

What is the electrostatic potential at the surface of A silver nucleus of A diameter 12.4 fermi? Atomic no. (Z) for silver is 47

Answer 1

Answer:

V = k Q / R     electric potential where k = 9 * 10E9

V = 9 * 10E9 * 47 * 1.6 * 10E-19 / 6.2 * 10E-15

V = 1.76 * 10E21 volts    1.1 * 10E7 volts

Answer 2

Electrostatic potential at the surface is $1.09\times 10^6\ V.$

Given :

Diameter , $D=12.4\ fermi=12.4\times 10^{-15}\ m.$

Therefore , radius $r=6.2\times 10^{-15}\ m\ .$

Also , atomic no , Z = 47 .

Now , we know potential is given by :

$V=\dfrac{kze}{r}$

Here ,  k is constant , $k=9\times 10^9\ N.m^2/C^2\ .$

Putting all these value in above equation we get :

$V=\dfrac{9\times 10^9\times 47\times 1.6\times 10^{-19}}{6.2\times 10^{-15}}\\\\V=1.09\times 10^6\ V.$

Hence , this is the required solution.

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Electrostatic potential

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