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实验报告19系04级姓名王承乐日期06.04.08,06.04.15评分实验题目:同位素光谱——氘原子光谱实验目的:以氘原子光谱为研究对象,研究获得同位素光谱的实验方法、分析方法及其在微观测量中的应用。实验原理:根据玻尔理论,原子的能量是量子化的,即具有分立的能级。当电子从高能级跃迁到低能级时,原子释放出能量,并以电磁波形式辐射。氢和类氢原子的巴耳末线系对应光谱线波数为)121()1()4(222320242nmmchZemZee(1)其中Zm为原子核质量,em为电子质量,e为电子电荷,h为普朗克常数,0为真空介电常数,c为光速,Z为原子序数。因此类氢原子的里德伯常数可写成)1(1)4(2320242ZeeZmmchZemR(2)若Zm,即假定原子核不动,则有chZemRe320242)4(2(3)因此)1(ZeZmmRR(4)由此可见,ZR随原子核质量Zm变化,对于不同的元素或同一元素的不同同位素ZR值不同。Zm对ZR影响很小,因此氢和它的同位素的相应波数很接近,在光谱上形成很难分辨的双线或多线。设氢和氘的里德伯常数分别为HR和DR,氢、氘光谱线的波数H、D分别为22121nRHHn=3,4,5…(5)22121nRDDn=3,4,5…(6)氢和氘光谱相应的波长差为)1()1()1(DHHDHHHDHDHRR(7)因此,通过实验测得氢和氘的巴耳末线系的前几条谱线的谱长及其波长差,可求得氢与氘的里德伯常数HR和DR。根据式(4)有)1(HeHmmRR(8))1(DeDmmRR(9)其中Hm和Dm分别为氢和氘原子核的质量。式(8)除以式(9),得DeHeHDmmmmRR11(10)从式(10)可解出HDmm11HDeHHDHDRRmmRRmm(11)式中eHmm为氢原子核质量与电子质量比,公认值为1836.1515。因此将通过实验测得的HDRR代入式(11),可求得氘与氢原子核的质量比HDmm。实验方法1.用氢氘放电管作为光源,用摄谱仪拍摄光谱,氢氘放电管是将氢气和氘气充入同一放电管中,当一定的高压加在放电管两极上时,管内的游离电子受到电场作用飞向阳极,并因此获得越来越大的动能。当它们与管中的氢、氘分子碰撞时,使氢氘分子离解为氢原子和氘原子,并进入激发状态,当它们回到低能级时产生光辐射。用碳棒与铁棒作为电极的两极,加高压击穿空气,得到铁弧光.用摄谱仪在同一张底片分别拍摄氢氘光谱和铁光谱.2.测量谱线波长采用线性插入法。其基本原理是,在光谱图片间隔很小的范围内,摄谱仪的线色散可认为是常数,即谱线间隔与谱线波长差成正比.由于铁弧光谱谱线丰富,遍布整个可见及紫外范围,其各谱线波长已被精确测定并制成铁光谱图,因此常作为测定未知谱线的标准比较光源.为此,常利用摄谱仪的哈德曼光阑,在不移动暗盒的情况下,并排拍摄未知光谱和铁光谱,并根据铁谱测定未知谱线的波长,测定方法如下:待测谱线X位于铁谱线1和2之间,1和2两条谱线相距为d,d为1和X之间的距离,则)(121ddX实验步骤:1.先在暗室中装好底片.2.调节摄谱仪,按照规定的时间分别拍下铁和氢氘光谱.在拍摄同组光谱时不能移动底片盒.3.拍摄好后,在暗室中取出底片,显影十分钟,定影十分钟.即可得到谱图.4.在映谱仪下利用标准铁谱图识别底片上氢氘光谱及其附近的铁谱线.5.用阿贝比长仪精密的测量谱线间的距离,以线性插入法计算各条光谱线的波长,并计算各谱线的里德伯常数,求HR和DR的平均值,并求出氢氘原子核质量比。实验数据及计算:已知常数:1710097373177.1mR1515.1836eHmm标准值AnmEhc429.65711429.657)4.3(51.112421AnmEhc588.48700588.487)4.3(85.012422AnmEhc739.43488739.434)4.3(544.012423AnmEhc861.41099861.410)4.3(378.01242417710096775854.11515.18361110097373177.1)1(mmmRRHeH1710097074434.1)21()1(mmmRmmRRHeDeD2HDmm1.6500A附近,即为3n时的谱线AddD370.6552)193.6546158.6569(1900.385778.431900.386391.39193.6546)(121AddH218.6554)193.6546158.6569(1900.385778.431900.380727.40193.6546)(121)1()1()1(DHHDHHHDHDHRR999718.0218.6554370.6552218.655411HDHRR1722102210098529222.1)3121(10218.65541)121(1mnRHH1722102210098839046.1)3121(10370.65521)121(1mnRDD相对误差:%26.0429.6571218.6554429.6571H%16.010096775854.110096775854.110098529222.1777H%16.010097074434.110097074434.110098839046.1777D07502.2)1999718.01(1515.18361999718.0111HDeHHDHDRRmmRRmm2.4800A附近,即为4n时的谱线AddD048.4860)746.4859325.4871(8709.212523.198709.218027.21746.4859)(121AddH302.4861)746.4859325.4871(8709.212523.198709.215190.21746.4859)(121999742.0302.4861048.4860302.486111HDHRR1722102210097099776.1)4121(10302.48611)121(1mnRHH1722102210097382852.1)4121(10048.48601)121(1mnRDD相对误差%22.0588.4870048.4860588.4870H%03.010096775854.110096775854.110097099776.1777H%03.010097074434.110097074434.110097382852.1777D90109.1)1999742.01(1515.18361999742.0111HDeHHDHDRRmmRRmm3.4300A附近,即为5n时的谱线AddD206.4349)049.4337737.4352(5023.1359836.1385023.1352000.138049.4337)(121AddH450.4350)049.4337737.4352(5023.1359836.1385023.1354760.138049.4337)(121999737.0450.4350306.4349450.435011HDHRR1722102210094577518.1)5121(10450.43501)121(1mnRHH1722102210094890599.1)5121(10206.43491)121(1mnRDD相对误差%01.0739.4348206.4349739.4348H%20.010096775854.110096775854.110094577518.1777H%20.010097074434.110097074434.110094890599.1777D93281.1)1999737.01(1515.18361999737.0111HDeHHDHDRRmmRRmm4.4100A附近,即为6n时的谱线AddD612.4100)187.4098128.4104(2361.1815551.1822361.1817746.181187.4098)(121AddH714.4101)187.4098128.4104(2361.1815551.1822361.1810192.182187.4098)(121999731.0714.4101612.4100714.410111HDHRR1722102210097102333.1)6121(10714.41011)121(1mnRHH1722102210097397169.1)6121(10612.41001)121(1mnRDD相对误差%20.0861.4109714.4101861.4109H%03.010096775854.110096775854.110097102333.1777H%03.010097074434.110097074434.110097397169.1777D97704.1)1999731.01(1515.18361999731.0111HDeHHDHDRRmmRRmm综上:17777710096827212.1410097102333.110094577518.110097099776.110098529222.1mRH17777710097127417.1410097397169.110094890599.110097382852.110098839046.1mRD相对误差%005.010096775854.110096775854.110096827212.1777H%005.010097074434.110097074434.110097127417.1777D999726.010097127417.110096827212.177DHRR01361.2)1999726.01(1515.18361999726.0111HDeHHDHDRRmmRRmm%68.0201361.22m思考题1.画出氢原子巴耳末线系的能级图,并标出前四条谱线对应的能级跃迁和波长数。答:AnmEhc429.65711429.657)4.3(51.112421AnmEhc588.48700588.487)4.3(85.012422AnmEhc739.43488739.434)4.3(544.012423AnmEhc861.41099861.410)4.3(378.0124242.为什么把氢氘与铁的光谱并列拍摄在一张底片上时不用移动底片盒的方法,而是使用哈德曼光
本文标题:氢氘光谱
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