Question

The electric potential at the surface of an atomic nucleus z=50 of radius 9×10^-5

Answer 1

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Answer 2

Explanation:

The electric potential is given by :

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

q is total electric charge and q = number of proton × e

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

In this question, Z = 50

Radius of the nucleus, r = 9 × 10⁻⁵ m

So, $V=\dfrac{9\times 10^9\times 50\times 1.6\times 10^{-19}}{9\times 10^{-5}}$

V = 0.0008 Volts

or V = 8 × 10⁻⁴ Volts

Hence, this is the required solution.