TS EAMCET · Physics · Atomic Physics
The radius of the first orbit of hydrogen is \(\mathrm{r}_{\mathrm{H}}\), and the energy in the ground state is - \(13.6 \mathrm{eV}\). Considering a \(\mu^{-}\)-particle with a mass \(207 m_e\) revolving round a proton as in hydrogen atom, the energy and radius of proton and \(\mu^{-}\)- combination respectively in the first orbit are (assume nucleus to be stationary)
- A \(-13.6 \times 207 \mathrm{eV}, \frac{\mathrm{C}^{\mathrm{H}}}{207}\)
- B \(-207 \times 13.6 \mathrm{eV}, 207 r_{\mathrm{H}}\)
- C \(-\frac{13.6}{207} \mathrm{eV}, \frac{\mathrm{F}}{207}\)
- D \(-\frac{13.6}{207} \mathrm{eV}, 207 \mathrm{r}_{\mathrm{H}}\)
Answer & Solution
Correct Answer
(A) \(-13.6 \times 207 \mathrm{eV}, \frac{\mathrm{C}^{\mathrm{H}}}{207}\)
Step-by-step Solution
Detailed explanation
The total energy of \(n\)th orbit \(E_n=-\frac{m e^4}{8 \varepsilon_0^2 h^2} \cdot \frac{1}{n^2}\) Obviously \(E_n \propto m\) \(\begin{array}{llll} & \therefore & \frac{E_\mu}{E_e} & =\frac{m_\mu}{m_e} \\ \Rightarrow & E_\mu & =\frac{m_\mu}{m_e} \times E_e \end{array}\) Ground…
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