he line spectrum of hydrogen, Bohr started with this idea, assuming that electrons move in circular orbits around the nucleus. According to classical physics, however, an electrically charged particle(such as an electron) that moves in a circular path should continuously lose energy by emitting electromagnetic radiation. As the electron loses energy, it should spiral into the nucleus. Bohr approached this problem in much the same way that Planck had pproached the problem of the nature of the radiation emitted by hot objects: He assumed that the prevailing laws of physics were inadequate to describe atoms. Furthermore, he adopted Planck's idea that energies are quantized 1.Bohr' s hypothesis:1913年玻尔在普朗克量子论,爱因斯坦光子学说和卢瑟福有核原子 模型的基础上提出了如下假设: 假设一:氢原子是由一个质子的原子核与一个沿着原子核以圆形轨道半径为r运动的 电子所构成 由 Coulomb’slaw可知f= 4 (负号表示吸引),F,=m 假设二:并非所有的圆形轨道均为电子所容许的,只有电子的轨道角动量(mr)等于h /2x的正整数倍,才是电子运动所容许的轨道 n nh.即v 两边平方得 4r2mr2 nh 由①,4m4nm, 解得 ame 把h=663×10-34Js,m=9.1×10-31kg,ao=8.85×10-12C.m-1入上式得 r=0.53m2(A)=0.053n2(mm) 假设三:由于轨道角动量的限制,在一定的允许圆形轨道上的电子所具有的能量是固 定的,这些值显然是不连续的,这些固定轨道上,电子既不吸收能量也不放 出能量: E=E+E 由①y2 代入得 Eg 4To/ 2 4Er rsr nh5代入得 82h2n2 假设四:电子从一个允许圆形轨道向另一个允许圆形轨道跃迁时,放出(或吸收)的能 量必须等于两个轨道之间的能量差 AE=E-E 8c2h2101 the line spectrum of hydrogen, Bohr started with this idea, assuming that electrons move in circular orbits around the nucleus. According to classical physics, however, an electrically charged particle (such as an electron) that moves in a circular path should continuously lose energy by emitting electromagnetic radiation. As the electron loses energy, it should spiral into the nucleus. Bohr approached this problem in much the same way that Planck had approached the problem of the nature of the radiation emitted by hot objects: He assumed that the prevailing laws of physics were inadequate to describe atoms. Furthermore, he adopted Planck’s idea that energies are quantized. 1．Bohr’s hypothesis：1913 年玻尔在普朗克量子论,爱因斯坦光子学说和卢瑟福有核原子 模型的基础上,提出了如下假设： 假设一：氢原子是由一个质子的原子核与一个沿着原子核以圆形轨道半径为 r 运动的 电子所构成； 由 Coulomb’s law 可知 2 2 0 4 e f  r = − （负号表示吸引）， 2 mv F r = 离心 ∴ 2 2 2 0 4 mv e r r  = ， 即 2 2 0 4 e v  mr = ① 假设二：并非所有的圆形轨道均为电子所容许的，只有电子的轨道角动量 (mvr )等于 h / 2π 的正整数倍，才是电子运动所容许的轨道； 2 nh mvr  = ，即 2 nh v  mr = ，两边平方得 2 2 2 2 2 2 4 n h v  m r = ② 由①，②得： 2 2 2 2 2 2 0 4 4 e n h   mr m r = ， 解得 2 2 0 2 n h r me   = 把 h = 6.6310−34 J·s，m = 9.110−31 kg，ε0 = 8.8510−12 C 2·m−1 入上式得 r = 0.53·n 2 (Å) = 0.053·n 2 (nm) 假设三：由于轨道角动量的限制，在一定的允许圆形轨道上的电子所具有的能量是固 定的，这些值显然是不连续的，这些固定轨道上，电子既不吸收能量也不放 出能量； 2 2 0 1 4 2 e E E E mv  r = + = − + 总 势 动 由① 2 2 0 4 e v  mr = 代入 得 2 2 2 0 0 0 1 4 2 4 8 e e e E    r r r = − +  = − 总 以 2 2 0 2 n h r me   = 代入 得 4 18 2 2 2 2 2 0 1 1 1 13.6 (eV) 2.18 10 (J) 8 me E  h n n n − = −  = −  = −   总 （ ） 假设四：电子从一个允许圆形轨道向另一个允许圆形轨道跃迁时，放出(或吸收)的能 量必须等于两个轨道之间的能量差。 4 4 2 1 2 2 2 2 2 2 2 2 0 2 1 0 1 2 1 1 1 1 ( ) ( ) 8 8 me me E E E   h n n h n n  = − = − − = −