JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2024
- Medium
- Numerical
The electric potential at the surface of an atomic nucleus \((Z=50)\) of radius \(9 \times 10^{-13} cm\) is \(\alpha \times 10^6 V\). What is the value of \(\alpha\)?
(Charge of proton \(1.6 \times 10^{-19} C\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) 8
Step-by-step Solution
Detailed explanation
Potential \(=\frac{k Q}{R}=\frac{k \cdot Z e}{R}\)
\(=\frac{9 \times 10^9 \times 50 \times 1.6 \times 10^{-19}}{9 \times 10^{-13} \times 10^{-2}}\)
\(=8 \times 10^6 V\)
\(=\frac{9 \times 10^9 \times 50 \times 1.6 \times 10^{-19}}{9 \times 10^{-13} \times 10^{-2}}\)
\(=8 \times 10^6 V\)
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