The electric potential at the surface of an atomic nucleus $(Z=50)$ of radius $9 \times 10^{-15}…
The electric potential at the surface of an atomic nucleus $(Z=50)$ of radius $9 \times 10^{-15} \mathrm{~m}$ is
- $4 \times 10^6 \mathrm{~V}$
- $8 \times 10^6 \mathrm{~V}$
- $4 \times 10^{-6} \mathrm{~V}$
- $8 \times 10^{-6} \mathrm{~V}$
Solution
Atomic number, $Z=50$
Radius, $r=9 \times 10^{-15} \mathrm{~m}$
Electric potential at the surface of atomic nucleus,
$
\begin{aligned}
V & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Z e}{r} \quad \quad[\because q=Z e] \\
& =9 \times 10^9 \times \frac{50 \times 1.6 \times 10^{-19}}{9 \times 10^{-15}}=8 \times 10^6 \mathrm{~V}
\end{aligned}
$
Asked in: AP EAMCET 2020 (22 Sep Shift 1)
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