JEE Mains · Physics · STD 12 - Nuclei
- JEE Mains 2023
- Medium
- Numerical
Experimentally it is found that \(12.8\,eV\) energy is required to separate a hydrogen atom into a proton and an electron. So the orbital radius of the electron in a hydrogen atom is \(\frac{9}{ x } \times 10^{-10}\,m\). The value of the \(x\) is \(\left(1\,eV =1.6 \times 10^{-19}\,J , \frac{1}{4 \pi \epsilon_0}=9 \times 10^9\,Nm ^2 / C ^2\right.\) and electronic charge \(=1.6 \times 10^{-19}\,J\,C\) )
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Answer & Solution
Correct Answer
(C) \(16\)
Step-by-step Solution
Detailed explanation
Binding energy of system \(=\frac{k e^2}{2 r}\) joule and \(\frac{k e^2}{2 r}=12.8\,ev\)
\(\frac{9 \times 10^9 \times\left(1.6 \times 10^{-19}\right)^2}{2 r}=12.8 \times 1.6 \times 10^{-19}\)
\(\Rightarrow r=\frac{9 \times 10^9 \times 1.6 \times 10^{-19}}{12.8 \times 2}\)
\(\Rightarrow r=\frac{9 \times 10^{-10}}{16}\)
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