close
網址:http://dreamstore.info/r/661e8d07ba5a67bccaa7b0590cc07164
平時我自己上網搜尋資料就還蠻喜歡看 HP 4芯 RA04 日系電芯 電池 HSTTN-UB4L HSTTN-IB4L HSTTN-W01C 430 G0 430 G1 430 G2 430 G3
的
因為可以一網打盡真的是太方便!!!!!
就算沒買過肯定逛過聽過看過 HP 4芯 RA04 日系電芯 電池 HSTTN-UB4L HSTTN-IB4L HSTTN-W01C 430 G0 430 G1 430 G2 430 G3
吧!!!
HP 4芯 RA04 日系電芯 電池 HSTTN-UB4L HSTTN-IB4L HSTTN-W01C 430 G0 430 G1 430 G2 430 G3
功能:
HP 4芯 RA04 日系電芯 電池 HSTTN-UB4L HSTTN-IB4L HSTTN-W01C 430 G0 430 G1 430 G2 430 G3
描述:
HP 4芯 RA04 日系電芯 電池 HSTTN-UB4L HSTTN-IB4L HSTTN-W01C 430 G0 430 G1 430 G2 430 G3
相關 HP 4芯 RA04 日系電芯 電池 HSTTN-UB4L HSTTN-IB4L HSTTN-W01C 430 G0 430 G1 430 G2 430 G3
商品推薦
標題:
量子物理問題 求高手幫忙!!有解答缺詳解!!~
發問:
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... 顯示更多 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 這段文字打在以下的網站的 Launch CodeCogs Equation Editor 上 http://www.homeschoolmath.net/worksheets/equation_editor.php 感謝大大了
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/1/0a1c02498125a255a2f5b0e58908a8ae.png 由此計算能量差 圖片參考:http://imgcld.yimg.com/8/n/AF03901095/o/101111190870313869464170.jpg 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。
其他解答:
HP 4芯 RA04 日系電芯 電池 HSTTN-UB4L HSTTN-IB4L HSTTN-W01C 430 G0 430 G1 430 G2 430 G3
34F8E429B5D3BD34
平時我自己上網搜尋資料就還蠻喜歡看 HP 4芯 RA04 日系電芯 電池 HSTTN-UB4L HSTTN-IB4L HSTTN-W01C 430 G0 430 G1 430 G2 430 G3
的
因為可以一網打盡真的是太方便!!!!!
就算沒買過肯定逛過聽過看過 HP 4芯 RA04 日系電芯 電池 HSTTN-UB4L HSTTN-IB4L HSTTN-W01C 430 G0 430 G1 430 G2 430 G3
吧!!!
HP 4芯 RA04 日系電芯 電池 HSTTN-UB4L HSTTN-IB4L HSTTN-W01C 430 G0 430 G1 430 G2 430 G3
功能:
HP 4芯 RA04 日系電芯 電池 HSTTN-UB4L HSTTN-IB4L HSTTN-W01C 430 G0 430 G1 430 G2 430 G3
描述:
電池適用型號
RA04 E5H00PA H6L28AA?
707618-121 768549-001?
HSTTN-UB4L?
HSTTN-IB4L?
HSTTN-W01C
?
?
筆電適用型號
HP ProBook?
430 G1
?
? 產品訊息
HP 4芯 RA04 日系電芯 電池 HSTTN-UB4L HSTTN-IB4L HSTTN-W01C 430 G0 430 G1 430 G2 430 G3
相關 HP 4芯 RA04 日系電芯 電池 HSTTN-UB4L HSTTN-IB4L HSTTN-W01C 430 G0 430 G1 430 G2 430 G3
商品推薦
標題:
量子物理問題 求高手幫忙!!有解答缺詳解!!~
發問:
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... 顯示更多 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 這段文字打在以下的網站的 Launch CodeCogs Equation Editor 上 http://www.homeschoolmath.net/worksheets/equation_editor.php 感謝大大了
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/1/0a1c02498125a255a2f5b0e58908a8ae.png 由此計算能量差 圖片參考:http://imgcld.yimg.com/8/n/AF03901095/o/101111190870313869464170.jpg 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。
其他解答:
HP 4芯 RA04 日系電芯 電池 HSTTN-UB4L HSTTN-IB4L HSTTN-W01C 430 G0 430 G1 430 G2 430 G3
34F8E429B5D3BD34
文章標籤
全站熱搜
留言列表