The electric potential at the surface of an atomic nucleus ( z =50 . ) of radius 9 × 10 - 13 rm \;cm is rm \; × 10 6 rm \;V .

Physics · Electrostatics

JEE Main 2024 — 27 January, Shift 2 — Question 57

The electric potential at the surface of an atomic nucleus $\left(\mathrm{z}=50\right.$ ) of radius $9 \times {10^{ - 13}}{\rm{\;cm}}$ is ${\rm{\;}}$ $\times {10^{6{\rm{\;V}}}}$ .

Answer: 8

Numerical answer — enter this value.

Step-by-step solution

Potential $= \frac{{{\rm{kQ}}}}{{\rm{R}}} = \frac{{{\rm{k}} \cdot {\rm{Ze}}}}{{\rm{R}}}$

$\begin{array}{*{20}{r}}{}&{\; = \frac{{9 \times {{10}^9} \times 50 \times 1.6 \times {{10}^{ - 19}}}}{{9 \times {{10}^{ - 13}} \times {{10}^{ - 2}}}}}\\{}&{\; = 8 \times {{10}^6}{\rm{\;V}}}\end{array}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 27 January, Shift 2
Subject: Physics
Chapter: Electrostatics
Topic: Electrostatic Potential Energy and Electric Potential

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Answer: 8

Step-by-step solution

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