基础物理下习题25答案.pdf

物理常数物理常数. . . . 普朗克常量 h=6.62610-34 Js，= h/2, 里德伯常量 R=1.09710-7 m-1， 真空中光速c=3108 m/s, 真空介电常数0 =8.8510-12 F/m， 真空磁导率 0 =410-7 H/m， 电子质量 me =9.1110-31 kg， 电子电量 e =1.6 10-19 C, 质子质量 mp = 1.6710-27 kg. 25 25 25 25 量子力学基础量子力学基础P255P255P255P255 25 25 25 25 量子力学基础量子力学基础P255P255P255P255 四部分四部分: : : : 波函数与概率波函数与概率6; 6; 6; 6; 一维无限深势阱一维无限深势阱4; 4; 4; 4; 氢原子定态解氢原子定态解4; 4; 4; 4; 不确定关系不确定关系5. 5. 5. 5. 25-1. 用自己的语言表述波函数的统计解释. 为什么说概率波是对微观客体 波粒二象性的统一描述. 波函数的模表达粒子在某时刻出现在某位置的概率, 设粒子在某时刻 在很大空间范围内都可能同时出现, 如同波动的运动形式分布于空间 各处, 则体现波性强; 若某时刻在某一空间位置出现的概率近似为1, 则它有具体一个占据点, 那么粒子性明显. 由此可见, 概率波是对微 观客体波粒二象性的统一描述. 25 25 25 25 量子力学基础量子力学基础P255P255P255P255 25-3. 设粒子波函数为(x,y,z) , 求在x-x+dx范围内找到粒子的概率. 25-4. 设用球坐标表示, 粒子的波函数为(r,) , (1)求粒子在球壳r-dr中被 测到的概率; (2) 粒子在(,)方向附近立体角d=sindd中出现的概率. ) ) ) ), , , , , , ,( ( ( () ) ) ), , , , , , ,( ( ( () ) ) ), , , , , , ,( ( ( () ) ) )( ( ( ( * * * * 2 2 2 2 z z z zy y y yx x x xz z z zy y y yx x x xdxdxdxdxz z z zy y y yx x x xdxdxdxdxx x x xx x x xWWWW = = = = = = =+ + + + = = = = = = = = = = = =+ + + + = = = = = = = 0 0 0 0 2 2 2 2 2 2 2 2 2 2 2 2 0 0 0 0 2 2 2 2 0 0 0 0 * * * * 2 2 2 2 0 0 0 0 2 2 2 2 0 0 0 0 2 2 2 2 2 2 2 2 sinsinsinsin ) ) ) ), , , , , , ,( ( ( ( ) ) ) )(sin(sin(sin(sin) ) ) )2 2 2 2( ( ( ( sinsinsinsin ) ) ) ), , , , , , ,( ( ( () ) ) ), , , , , , ,( ( ( ( sinsinsinsin) ) ) ) ) ) ), , , , , , ,( ( ( ( ( ( ( ) ) ) )( ( ( () ) ) )1 1 1 1( ( ( ( sinsinsinsin d d d dd d d ddrdrdrdrr r r rr r r rd d d dd d d dWWWW drdrdrdrr r r rd d d dd d d dr r r rr r r r drdrdrdrr r r rd d d dd d d dr r r rdrdrdrdrr r r rr r r rWWWW d d d ddrddrddrddrdr r r rdxdydzdxdydzdxdydzdxdydzdVdVdVdV球坐标系球坐标系球坐标系球坐标系 25-2. 设一维波函数为exp(ikx), 求粒子位置的概率分布. 此波函数是否能归一化? . . . ., , , ,. . . . 1 1 1 1) ) ) )( ( ( ( * * * * 2 2 2 2 因为分布无穷大因为分布无穷大因为分布无穷大因为分布无穷大不能归一化不能归一化不能归一化不能归一化轴上全空间轴上全空间轴上全空间轴上全空间等概率分布在等概率分布在等概率分布在等概率分布在x x x x e e e ee e e ex x x xWWWW ikxikxikxikxikxikxikxikx = = = = = = = = = = = = = 25 25 25 25 量子力学基础量子力学基础P255P255P255P255 25-5. 设粒子一维波函数为Aexp(- 2x2/2), 为正值常数, 试写出归一化的 波函数. ) ) ) )( ( ( ( 0 0 0 0 2 2 2 2 = = = = dxdxdxdxe e e e x x x x 2 2 2 2/ / / / 4 4 4 4 1 1 1 1 4 4 4 4 1 1 1 1 4 4 4 4 1 1 1 1 2 2 2 2 0 0 0 0 2 2 2 22 2 2 2 0 0 0 0 0 0 0 00 0 0 0 2 2 2 22 2 2 2/ / / /* * * *2 2 2 2/ / / / 0 0 0 0 * * * * 2 2 2 22 2 2 2 2 2 2 2 2 2 2 22 2 2 22 2 2 2 2 2 2 22 2 2 22 2 2 22 2 2 22 2 2 22 2 2 2 ) ) ) )( ( ( (: : : : . . . ., , , , , , , 1 1 1 1; ; ; ;0 0 0 0, , , , 1 1 1 1) ) ) )( ( ( () ) ) )( ( ( ( x x x x x x x xx x x x x x x xx x x xx x x x e e e ex x x x A A A AA A A A A A A A A A A Adxdxdxdxe e e eA A A Adxdxdxdxe e e e dxdxdxdxe e e eA A A Adxdxdxdxe e e eA A A AAeAeAeAedxdxdxdxx x x xx x x x = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = 归一化波函数归一化波函数归一化波函数归一化波函数 取最常用范围的取最常用范围的取最常用范围的取最常用范围的可以任意选择可以任意选择可以任意选择可以任意选择 Q 25 25 25 25 量子力学基础量子力学基础P255P255P255P255 25-6. 一维运动的粒子处于Axexp(-x)(x=0), 0(x0. 求(1)归一 化常量; (2)粒子坐标概率分布函数; (3)在什么位置附近, 找到粒子的概率最大. ) ) ) )/ / / / ! ! ! !( ( ( ( 0 0 0 0 1 1 1 1 + + + + = = = = n n n nx x x xn n n n n n n ndxdxdxdxe e e ex x x x . . . ., , , , , , , ,0 0 0 0, , , , , , ,/ / / /2 2 2 2, , , ,0 0 0 08 8 8 88 8 8 8) ) ) )3 3 3 3( ( ( ( , , , ,4 4 4 4) ) ) )( ( ( () ) ) )( ( ( () ) ) )( ( ( () ) ) )2 2 2 2( ( ( (; ; ; ;2 2 2 2) ) ) )1 1 1 1( ( ( ( , , , , 1 1 1 1 ) ) ) )2 2 2 2( ( ( ( 2 2 2 2 , , , ,/ / / / ! ! ! ! 1 1 1 1) ) ) )( ( ( () ) ) )( ( ( ( 2 2 2 22 2 2 24 4 4 42 2 2 23 3 3 3 2 2 2 22 2 2 23 3 3 3* * * *2 2 2 2/ / / /3 3 3 3 3 3 3 3 2 2 2 2 0 0 0 0 1 1 1 1 0 0 0 00 0 0 0 2 2 2 22 2 2 2 2 2 2 2 * * * * 0 0 0 0 * * * * 所以此点是极大点所以此点是极大点所以此点是极大点所以此点是极大点对应零概率解对应零概率解对应零概率解对应零概率解 唯一极值唯一极值唯一极值唯一极值 取最常用范围的取最常用范围的取最常用范围的取最常用范围的 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = + + + + x x x xx x x xe e e ex x x xxexexexe dxdxdxdx dWdWdWdW e e e ex x x xx x x xx x x xx x x xWWWWA A A A A A A An n n ndxdxdxdxe e e ex x x x dxdxdxdxe e e ex x x xA A A AdxdxdxdxxexexexeA A A AAxeAxeAxeAxedxdxdxdxx x x xx x x x x x x xx x x x x x x x n n n nx x x xn n n n x x x xx x x xx x x x Q 25 25 25 25 量子力学基础量子力学基础P255P255P255P255 25-7. 试求宽度为0.1nm的一维无限深势阱中, n=1,2,3,10各态电子的能量. 若势阱宽度为1cm, 结果又当如何. . . . .1010101076767676. . . .3 3 3 3101010103760376037603760100100100100 , , , ,10101010384384384384. . . . 3 3 3 3101010104 4 4 4. . . .3383383383389 9 9 9, , , ,10101010504504504504. . . . 1 1 1 1101010104 4 4 4 . . . .1501501501504 4 4 4 1010101076767676. . . .3 3 3 3 ) ) ) )10101010( ( ( ( 6 6 6 6 . . . .37373737 2 2 2 2 , , , ,1010101010101010101010101 1 1 1 . . . .3760376037603760100100100100, , , ,4 4 4 4. . . .3383383383389 9 9 9, , , ,4 4 4 4. . . .1501501501504 4 4 4, , , ,6 6 6 6. . . .37373737 6 6 6 6. . . .37373737 / / / /10101010602602602602. . . .1 1 1 1 101010106024602460246024. . . . 0 0 0 0 101010106024602460246024. . . . 0 0 0 0 101010101 1 1 1. . . .0 0 0 01010101011111111. . . .9 9 9 98 8 8 8 10101010626626626626. . . . 6 6 6 6 2 2 2 2 , , , , 1 1 1 1 , , , ,1 1 1 1 . . . .0 0 0 0, , , , 2 2 2 2 10101010626626626626. . . .6 6 6 6 , , , ,1010101011111111. . . .9 9 9 9 , , , , 2 2 2 2 , , , , 2 2 2 2 , , , , 2 2 2 2 , , , ,0 0 0 0sinsinsinsin 0 0 0 0sinsinsinsin2 2 2 2, , , ,0 0 0 0, , , ,; ; ; ;0 0 0 0, , , ,0 0 0 0, , , , 0 0 0 0 2 2 2 2 , , , , , , ,) ) ) ) 2 2 2 2 ( ( ( (), ), ), ),( ( ( ( 2 2 2 2 ) ) ) )( ( ( (: : : :0 0 0 0 ) ) ) )( ( ( () ) ) )( ( ( () ) ) )( ( ( ( 2 2 2 2 ( ( ( () ) ) )( ( ( (: : : :, , , , ) ) ) )0 0 0 0, , , ,( ( ( ( ) ) ) )0 0 0 0( ( ( (0 0 0 0 ) ) ) )( ( ( (: : : : 1313131316161616 1 1 1 110101010 1414141416161616 1 1 1 13 3 3 3 1414141416161616 1 1 1 12 2 2 2 15151515 2 2 2 28 8 8 82 2 2 2 2 2 2 22 2 2 2 1 1 1 1 101010108 8 8 82 2 2 2 1 1 1 1101010101 1 1 13 3 3 31 1 1 12 2 2 21 1 1 1 19191919 17171717 17171717 2 2 2 2* * * *9 9 9 92 2 2 231313131 686868682 2 2 2 2 2 2 2 2 2 2 22 2 2 2 1 1 1 1 34343434 31313131 2 2 2 2 2 2 2 22 2 2 2 1 1 1 11 1 1 1 2 2 2 2 2 2 2 2 2 2 2 22 2 2 22 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 22 2 2 2 2 2 2 22 2 2 2 2 2 2 2 eVeVeVeVeVeVeVeVE E E EE E E E eVeVeVeVeVeVeVeVE E E EE E E EeVeVeVeVeVeVeVeVE E E EE E E E eVeVeVeV eVeVeVeV mamamama E E E Em m m mm m m mcmcmcmcma a a a eVeVeVeVE E E EE E E EeVeVeVeVE E E EE E E EeVeVeVeVE E E EE E E EeVeVeVeVE E E E eVeVeVeV eVeVeVeVJ J J J J J J J J J J J mamamama E E E En n n n nmnmnmnma a a a s s s sJ J J J kgkgkgkgm m m m mamamama E E E EE E E En n n n mamamama n n n n E E E Ea a a a mEmEmEmE n n n nwawawawawawawawa wawawawaA A A AAeAeAeAeAeAeAeAea a a ax x x xB B B BA A A Ax x x x mEmEmEmE w w w wBeBeBeBeAeAeAeAe mEmEmEmE x x x x x x x xm m m m x x x xE E E Ea a a ax x x x x x x xE E E Ex x x xx x x xV V V V m m m m p p p p x x x x t t t t i i i i x x x xa a a ax x x x a a a ax x x x x x x xV V V V n n n n iwaiwaiwaiwaiwaiwaiwaiwa x x x xx x x x iwxiwxiwxiwxiwxiwxiwxiwx x x x xx x x xx x x x = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =+ + + += = = = = = = = = = =+ + + += = = = = = = = = = = = = = = =+ + + += = = = = = = = h h h hh h Q QQ hh ); ); );0 0 0 0( ( ( (0 0 0 0) ) ) )( ( ( (: : : : 2 2 2 2 maxmaxmaxmax1 1 1 1 2 2 2 22 2 2 2 * * * * 1 1 1 11 1 1 11 1 1 1 2 2 2 2 2 2 2 2 2 2 2 22 2 2 2 1 1 1 11 1 1 1 2 2 2 2 2 2 2 2 2 2 2 22 2 2 22 2 2 2 2 2 2 2 处概率最大处概率最大处概率最大处概率最大 定态解定态解定态解定态解 一维无限深势阱一维无限深势阱一维无限深势阱一维无限深势阱 a a a a x x x xA A A AWWWW a a a a x x x x A A A AWWWW a a a a mamamama m m m m w w w wE E E En n n n mamamama n n n n E E E E mEmEmEmE w w w wa a a ax x x xwxwxwxwxA A A A x x x xa a a ax x x xa a a ax x x xx x x xV V V V n n n n x x x x = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = h h h h 25 25 25 25 量子力学基础量子力学基础P255P255P255P255 25-9. 设粒子在 一维无限 深势阱中 运动, 能 量量子数 为 n. 试求: (1) 粒子在 0-a/4区间 出现的概 率; (2) n为何 值时, 该概 率最大; (3) n 时该概率 的极限为 多少. . . . . 4 4 4 4 1 1 1 1 0 0 0 0 4 4 4 4 1 1 1 1 , , , ,0 0 0 0 2 2 2 2 2 2 2 2 sinsinsinsin 2 2 2 2 2 2 2 2 sinsinsinsin , , , ,) ) ) )3 3 3 3( ( ( ( ) ) ) )6 6 6 68 8 8 8( ( ( ( 1 1 1 1 4 4 4 4 1 1 1 1 , , , , , , ,3 3 3 34 4 4 4 , , , , 2 2 2 2 3 3 3 3 2 2 2 2 2 2 2 2 , , , , 1 1 1 1 2 2 2 2 sinsinsinsin, , , , 2 2 2 2 1 1 1 1 4 4 4 4 1 1 1 1 ) ) ) )2 2 2 2( ( ( ( ), ), ), ), 2 2 2 2 2 2 2 2 sinsinsinsin 4 4 4 4 1 1 1 1 ( ( ( () ) ) )1 1 1 1( ( ( ( ) ) ) ) 2 2 2 2 2 2 2 2 sinsinsinsin 4 4 4 4 1 1 1 1 ( ( ( () ) ) ) 4 4 4 4 2 2 2 2 sinsinsinsin 2 2 2 24 4 4 4 ( ( ( (2 2 2 2) ) ) ) 2 2 2 2 sinsinsinsin 2 2 2 24 4 4 4 ( ( ( (2 2 2 2 ) ) ) ) 2 2 2 2 2 2 2 2 coscoscoscos 2 2 2 2 1 1 1 1 ( ( ( (4 4 4 4sinsinsinsin4 4 4 4 2 2 2 2 1 1 1 1 : : : :, , , ,1 1 1 12 2 2 2) ) ) ) 2 2 2 2 coscoscoscos 2 2 2 2 1 1 1 1 ( ( ( (4 4 4 4sinsinsinsin4 4 4 4 , , , , , , ,sinsinsinsin2 2 2 2, , , , , , , 2 2 2 2 , , , , 2 2 2 2 ), ), ), ),0 0 0 0( ( ( (, , , ,sinsinsinsin2 2 2 2: : : : ); ); ); );0 0 0 0, , , ,( ( ( (); ); ); );0 0 0 0( ( ( (0 0 0 0) ) ) )( ( ( (: : : : ) ) ) )( )( )( )(4 4 4 4/ / / /, , , ,0 0 0 0( ( ( ( maxmaxmaxmax) ) ) )4 4 4 4/ / / /, , , ,0 0 0 0( ( ( ( maxmaxmaxmax) ) ) )4 4 4 4/ / / /, , , ,0 0 0 0( ( ( ( ) ) ) )4 4 4 4/ / / /, , , ,0 0 0 0( ( ( ( 2 2 2 24 4 4 4/ / / / 0 0 0 0 2 2 2 2 4 4 4 4/ / / / 0 0 0 0 2 2 2 2 4 4 4 4/ / / / 0 0 0 0 2 2 2 22 2 2 2 4 4 4 4/ / / / 0 0 0 0 * * * * ) ) ) )4 4 4 4/ / / /, , , ,0 0 0 0( ( ( ( 2 2 2 2 0 0 0 0 2 2 2 2 0 0 0 0 2 2 2 22 2 2 2 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 22 2 2 22 2 2 2 2 2 2 2 = = = = = = = = = = = = = = = + + + + + + + += = = =+ + + += = = = + + + += = = = = = = =+ + + += = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = n n n na a a a a a a a a a a a a a a a a a a a a a a aa a a aa a a a n n n nn n n na a a a a a a aa a a a n n n nn n n nn n n n x x x x WWWW n n n n n n n n n n n n n n n n k k k k WWWWk k k kk k k kn n n n k k k k n n n nn n n n n n n n WWWW n n n n n n n n WWWW n n n n n n n n n n n n n n n n a a a aa a a a A A A A a a a a x x x xn n n n n n n n a a a aa a a a A A A A dxdxdxdx a a a a x x x xn n n n A A A Adxdxdxdx a a a a x x x xn n n n A A A AdxdxdxdxWWWW a a a a A A A Aa a a aA A A Adxdxdxdx a a a a x x x xn n n n A A A Adxdxdxdx a a a a x x x xn n n n A A A A A A A A a a a a x x x xn n n n A A A A a a a a n n n n w w w wE E E En n n n mamamama n n n n E E E E mEmEmEmE w w w wa a a ax x x xwxwxwxwxA A A A x x x xa a a ax x x xa a a ax x x xx x x xV V V V 为任意整数为任意整数为任意整数为任意整数 选正的数值选正的数值选正的数值选正的数值 归一化求归一化求归一化求归一化求 定态解定态解定态解定态解 一维无限深势阱一维无限深势阱一维无限深势阱一维无限深势阱 h h 25 25 25 25 量子力学基础量子力学基础P255P255P255P255 ). ). ). ). 2 2 2 2 ( ( ( (0 0 0 0) ) ) )( ( ( (); ); ); ); 2 2 2 2 ,.,.,.,.,6 6 6 6 , , , ,4 4 4 4 , , , ,2 2 2 2( ( ( ( sinsinsinsin 2 2 2 2 ) ) ) )( ( ( (); ); ); ); 2 2 2 2 ,.,.,.,.,5 5 5 5 , , , , 3 3 3 3, , , , 1 1 1 1( ( ( (coscoscoscos 2 2 2 2 ) ) ) )( ( ( ( ). ). ). ). 2 2 2 2 ( ( ( ( , , , ,) ) ) )( ( ( (), ), ), ), 2 2 2 2 ( ( ( ( , , , ,0 0 0 0) ) ) )( ( ( (, , , ,. . . .1010101025252525 a a a a x x x xx x x x a a a a x x x xn n n n a a a a x x x xn n n n a a a a x x x x a a a a x x x xn n n n a a a a x x x xn n n n a a a a x x x x a a a a x x x xx x x xV V V V a a a a x x x xx x x xV V V V n n n n n n n nn n n n = = = = + + + + = = = = = = = = = = = = = = + + + + = = = = = = =+ + + + + + + += = = = = = = = = = =+ + + += = = = = = = = = = + + + + = = = = = = = + + + + = = = = + + + + = = = = = = =+ + + + + + + = = = =+ + + += = = = = = = = = =+ + + += = = = = = = = = = = =+ + + += = = = = = = = = = = = = = = =+ + + += = = = = = = = 为为为为粒子的波函数可以表示粒子的波函数可以表示粒子的波函数可以表示粒子的波函数可以表示 归一化条件归一化条件归一化条件归一化条件 又又又又 定态解定态解定态解定态解一维无限深势阱一维无限深势阱一维无限深势阱一维无限深势阱 Q h hh QQ hh ; ; ;2 2 2 2) ) ) )1 1 1 1( ( ( (; ; ; ;4 4 4 4 . . . . 3 3 3 3 4 4 4 4 6 6 6 6 . . . .13131313 , , , ,6 6 6 6 . . . .13131313 2 2 2 2) ) ) )4 4 4 4( ( ( ( 1 1 1 1 , , , , 1 1 1 1, , , , 1 1 1 1, , , ,2 2 2 2: : : : 2 2 2 2 1 1 1 1 2 2 2 2 2 2 2 2 4 4 4 4 2 2 2 2 0 0 0 0 1 1 1 11 1 1 1, , , , 1 1 1 1, , , ,2 2 2 2 25-11. 在氢原子中, 若主量子数 n=5, 角量子数可取哪些数值? 对于l=3, 轨道磁量子数可取哪些数值? 在氢原子中, 若主量子数 n, l=0,1,2,n-1; 今 n=5, l=0,1,2,3,4; 角量子数l, 轨道磁量子数ml=-l, ,+l; 今l=3, ml=-3,-2,-1,0,1,2,3 25-12. 试写出氢原子2,1,-1态的能量, 角动量, 和角动量的z分量的表达式 25-13. 试证量子数为 n, l 时, 电子分布的平均半径可由下式求出 并证明电子在氢原子基态中的平均半径为3a0/2(a0为玻尔半径). 0 0 0 0, , , , , , , ! ! ! ! , , , ,) ) ) )( ( ( (: : : : 1 1 1 1 0 0 0 00 0 0 0 2 2 2 2 3 3 3 3 0 0 0 0 1 1 1 1, , , ,1 1 1 1 , , , ,2 2 2 2 = = = = = = = = = = + + + + a a a an n n n a a a a n n n n dxdxdxdxe e e ex x x xdrdrdrdrR R R Rr r r rdrdrdrdrr r r rrwrwrwrwr r r r n n n n axaxaxaxn n n n nlnlnlnlnlnlnlnl 为正整数为正整数为正整数为正整数 . . . .* * * * 0 0 0 0 = = = = = = = drdrdrdrr r r rr r r r 是电子的平均半径是电子的平均半径是电子的平均半径是电子的平均半径分量的统计平均数值就分量的统计平均数值就分量的统计平均数值就分量的统计平均数值就电子出现位置对应径向电子出现位置对应径向电子出现位置对应径向电子出现位置对应径向 现的概率现的概率现的概率现的概率在空间任意位置电子出在空间任意位置电子出在空间任意位置电子出在空间任意位置电子出电子的波函数的模表示电子的波函数的模表示电子的波函数的模表示电子的波函数的模表示 25 25 25 25 量子力学基础量子力学基础P255P255P255P255 25-14.在氢原子中, 若主量子数 n=3, 它可能具有的状态数为多少? 18181818) ) ) )5 5 5 53 3 3 31 1 1 1( ( ( (2 2 2 2, , , , 5 5 5 52 2 2 2, , , , 1 1 1 1, , , , 0 0 0 0, , , , 1 1 1 1, , , , 2 2 2 2, , , , 2 2 2 2 3 3 3 3, , , , 1 1 1 1, , , , 0 0 0 0, , , , 1 1 1 1, , , , 1 1 1 1 1 1 1 1, , , ,0 0 0 0, , , ,0 0 0 0 2 2 2 2, , , , 1 1 1 1, , , ,0 0 0 0, , , ,3 3 3 3 = = = =+ + + + + + + = = = = = = = = = = = = = = = = = = = = = = = = = = = = 状态数总计状态数总计状态数总计状态数总计旋方向旋方向旋方向旋方向每种状态又可取两种自每种状态又可取两种自每种状态又可取两种自每种状态又可取两种自 状态数为状态数为状态数为状态数为 状态数为状态数为状态数为状态数为 状态数为状态数为状态数为状态数为 l l l l l l l l l l l l m m m ml l l l m m m ml l l l m m m ml l l l l l l ln n n n 25-15.设原子线度10-10m, 求原子中电子速度的不确定量. s s s sm m m mv v v vm m m mx x x x s s s sm m m m x x x x v v v vv v v vx x x x s s s sJ J J Jkgkgkgkgm m m mmvmvmvmvp p p pp p p px x x x / / / /1010101079797979. . . .5 5 5 5, , , ,10101010 ) ) ) )/ / / /( ( ( ( 1010101079797979. . . .5 5 5 5 , , , ,101010105275527552755275. . . .0 0 0 01010101011111111. . . .9 9 9 9 10101010055055055055. . . .1 1 1 1, , , ,1010101011111111. . . .9 9 9 9, , , , , , , 2 2 2 2 5 5 5 510101010 5 5 5 5 3434343431313131 3434343431313131 = = = = = = = = = = = = = = = Q h h 电子电子电子电子 25-16.小球质量为210-6kg, 确定其质心位置可精确测量到2m=2 10-6m , 按不确定关系求小球速度的不确定量. s s s sm m m ms s s sm m m mv v v v m m m mx x x xv v v vm m m mp p p ps s s sJ J J Jkgkgkgkgm m m mp p p px x x x / / / /1010101032323232. . . . 1 1 1 1) ) ) )/ / / /( ( ( ( 101010102 2 2 2101010102 2 2 22 2 2 2 10101010055055055055. . . .1 1 1 1 101010102 2 2 2, , , , , , ,10101010055055055055. . . .1 1 1 1, , , ,101010102 2 2 2, , , , 2 2 2 2 23232323 6 6 6 66 6 6 6 34343434 6 6 6 6343434346 6 6 6 = = = = = =