JEE Mains · Physics · STD 12 - Atoms
- JEE Mains 2017
- Hard
- MCQ
The acceleration of an electron in the first orbit of the hydrogen atom \((z\, = 1)\) is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(\frac{{{h^2}}}{{4{\pi ^2}{m^2}{r^3}}}\)
Step-by-step Solution
Detailed explanation
Speed of electron in first orbit \((\mathrm{n}=1)\) of hydrogen atom \((z=1)\),
\(v = \frac{{{e^2}}}{{2{\varepsilon _0}h}}\)
radius of Bohr's first orbit,
\(r=\frac{h^{2} \varepsilon_{0}}{\pi m e^{2}} \Rightarrow \varepsilon_{0}=\frac{r \pi m \mathrm{e}^{2}}{h^{2}}\) ..... \((i)\)
Acceleration of electron,
\(\frac{v^{2}}{r}=\frac{e^{4}}{4 \varepsilon_{0}^{2} h^{2}} \times \frac{\pi m e^{2}}{h^{2} \varepsilon_{0}}\)
\({ = \frac{{{{\text{e}}^4} \times \pi {\text{m}}{{\text{e}}^2}}}{{4{{\text{h}}^4}\varepsilon _0^3}}}\) ..... \((ii)\)
eliminating \(\varepsilon_{0}\) from eq \((ii)\)
\(=\frac{e^{4} \pi m e^{2} h^{6}}{4 h^{4} r^{3} \pi^{3} m^{3} e^{6}}\) from \(eq^n\) \((i)\)
\(=\frac{h^{2}}{4 \pi^{2} m^{2} r^{3}}\)
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