Electric potential at surface of an atomic nucleus (Z = 50) of radius 9.0×{10^{ - 13}}, cm is

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2. Electric Potential and Capacitance

easy

The electric potential at the surface of an atomic nucleus $(Z = 50)$ of radius $9.0×{10^{ - 13}}\, cm$ is

A

$80\, volts$

B

$8 × {10^6}\,volts$

C

$9\, volts$

D

$9 × {10^5}\,volts$

Solution

(b) $V = 9 \times {10^9} \times \frac{{50 \times 1.6 \times {{10}^{ – 19}}}}{{9 \times {{10}^{ – 15}}}} = 8 \times {10^6}\,V$

Standard 12

Physics

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