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Câu hỏi: Biết bán kính Bo $\mathrm{r}_{0}=5,3.10^{-11} \mathrm{~m}$. Trong nguyên tử hyđro, bán kính quỹ đạo dừng L là
A. $2,12.10^{-10} \mathrm{~m}$.
B. $10,6.10^{-11} \mathrm{~m}$.
C. $2,12.10^{-11} \mathrm{~m}$.
D. $10,6.10^{-10} \mathrm{~m}$.
A. $2,12.10^{-10} \mathrm{~m}$.
B. $10,6.10^{-11} \mathrm{~m}$.
C. $2,12.10^{-11} \mathrm{~m}$.
D. $10,6.10^{-10} \mathrm{~m}$.
$r={{n}^{2}}{{r}_{0}}={{2}^{2}}.5,{{3.10}^{-11}}=2,{{12.10}^{-10}}m$.
Đáp án A.