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Câu hỏi: Trong nguyên tử Hiđrô, bán kính Bo là ${{r}_{0}}=5,{{3.10}^{-11}} m$. Bán kính quỹ đạo dùng M là
A. $47,{{7.10}^{-11}} m$
B. $21,{{2.10}^{-11}} m$
C. $84,{{8.10}^{-11}} m$
D. $132,{{5.10}^{-11}} m$
A. $47,{{7.10}^{-11}} m$
B. $21,{{2.10}^{-11}} m$
C. $84,{{8.10}^{-11}} m$
D. $132,{{5.10}^{-11}} m$
Ta có:
${{r}_{n}}={{n}^{2}}{{r}_{0}}={{\left( 3 \right)}^{2}}.\left( 5,{{3.10}^{-11}} \right)=4,{{77.10}^{-10}}$ m.
${{r}_{n}}={{n}^{2}}{{r}_{0}}={{\left( 3 \right)}^{2}}.\left( 5,{{3.10}^{-11}} \right)=4,{{77.10}^{-10}}$ m.
Đáp án A.