Electron quay quanh hạt nhân nguyên tử Hidro theo quỹ đạo tròn bán kính R= 5.10^-11m a) Tính độ lớn lực hướng tâm đặt lên electron b) Tính vận tốc và tần số chuyển động của electorn

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Đáp án:

Lời giải:

Đáp án:

\(a)\,\,9,{216.10^{ - 8}}\,\,N;\,\,b)\,\,2,{24.10^6}\,\,m/s;\,\,7,{13.10^{15}}\,\,Hz\)

Giải thích các bước giải:

 a) Lực hướng tâm chính là lực tương tác giữa electron và hạt nhân:

\({F_{ht}} = k.\frac{{\left| {{q_1}{q_2}} \right|}}{{{r^2}}} = {9.10^9}.\frac{{\left| {1,{{6.10}^{ - 19}}.\left( { - 1,{{6.10}^{ - 19}}} \right)} \right|}}{{{{\left( {{{5.10}^{ - 11}}} \right)}^2}}} = 9,{216.10^{ - 8}}\,\,\left( N \right)\)

b) Gia tốc hướng tâm của electron là:

\({a_{ht}} = \frac{{{F_{ht}}}}{{{m_e}}} = \frac{{9,{{216.10}^{ - 8}}}}{{9,{{1.10}^{ - 31}}}} \approx {1.10^{23}}\,\,\left( {m/{s^2}} \right)\)

Vận tốc của electron là:

\(v = \sqrt {{a_{ht}}.r}  = \sqrt {{{1.10}^{23}}{{.5.10}^{ - 11}}}  \approx 2,{24.10^6}\,\,\left( {m/s} \right)\)

Tần số chuyển động của electron là:

\(f = \frac{v}{{2\pi r}} = \frac{{2,{{24.10}^6}}}{{2\pi {{.5.10}^{ - 11}}}} = 7,{13.10^{15}}\,\,\left( {Hz} \right)\)

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