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Câu hỏi: Biết bán kính Bo là ${{r}_{0}}={{5,3.10}^{-11}}m$. Bán kính quỹ đạo dừng M trong nguyên tử hiđrô là
A. ${{132,5.10}^{-11}}m.$
B. ${{84,8.10}^{-11}}m.$
C. ${{21,2.10}^{-11}}m.$
D. ${{47,7.10}^{-11}}m.$
A. ${{132,5.10}^{-11}}m.$
B. ${{84,8.10}^{-11}}m.$
C. ${{21,2.10}^{-11}}m.$
D. ${{47,7.10}^{-11}}m.$
Quỹ đạo M có n = 3
$\Rightarrow {{r}_{n}}={{n}^{2}}{{r}_{0}}={{3}^{3}}{{.5,3.10}^{-11}}={{47,7.10}^{-11}}\left( m \right)$.
$\Rightarrow {{r}_{n}}={{n}^{2}}{{r}_{0}}={{3}^{3}}{{.5,3.10}^{-11}}={{47,7.10}^{-11}}\left( m \right)$.
Đáp án D.