量子物理問題 求高手幫忙!!有解答缺詳解!!~
6.8 Prior to the discovery of the neutron in 1934 it was believed that the nucleus consisted of protons and electrons. For example, 4He was envisaged as a composite of four protons and two electrons confined within the nuclear radius. What would be the kinetic energy of an electron confined within a sphere of 4 x 10^-15 m diameter, the typical diameter of a light nucleus? Is that result consistent with the energies of electrons emitted in " decay, which range from a few ke V to about 1 Me V?
KE=p^{2}/2m=\frac{(\bigtriangleup p)^{2}}{2m}=(h'/2\bigtriangleup R)^{2}/2m=\frac{(hc)^{2}}{32\pi ^{2}(\bigtriangleup R)^{2}mc^{2}}.For
\bigtriangleup R=4fm=4\times 10^{-5}nm.............we...have.....KE\geq \frac{(1240eV\cdot nm)^{2}}{32\pi ^{2}(4\times 10^{-5}nm)^{2}(5.11\times 10^{5}eV)}=600Mev.More...than...100...times...typical....\beta ....energies
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1 個解答
- 小白Lv 79 年前最佳解答
依據測不準原理，當電子被侷限在 Δx = 4 × 10^-15 m 距離內時，動量的差別為 Δp，則
圖片參考：http://upload.wikimedia.org/wikipedia/en/math/0/a/...
由此計算能量差
圖片參考：http://imgcld.yimg.com/8/n/AF03901095/o/1011111908...
被侷限在氦原子內的電子，能量差可達595 MeV。
依古典電動力學，電子動能 Ek = kQq / 2R = 9×10^9 × 2 × 1.602 × 10^-19 / ( 2 × 2 × 10^-15 ) = 7.2 × 10^5 eV = 0.72 MeV
兩者計算所得的電子動能差達 826 倍，量子理論與古典理論並不一致。
