物质对β射线的吸收Pb06210263赵媛实验目的：学习和掌握与物质对射线吸收的有关知识和实验方法，测量吸收曲线并求出β射线的射程和最大能量。同时进一步加深对计数管、定标器等仪器的了解。实验原理：(见预习报告)数据处理：1.相应参数薄片质量：557.78mg，厚片质量：3550mg薄片：长cm6.726.716.696.656.60宽cm5.205.105.155.205.15面积cm234.94434.21134.453534.5833.99平均面积＝34.4357cm2质量厚度d1＝557.78mg÷34.4357cm2＝16.20mg/cm2厚片：长cm6.006.166.006.175.92宽cm5.305.205.315.365.20面积cm231.832.03231.8633.071230.784平均面积＝31.90944cm2质量厚度d2＝3550mg÷31.90944cm2＝111.25mg/cm22.测量对应于不同厚度铝箔的β射线强度。N10I/I0=(N1-N0)/(485.5-N0)本底记数：N=95（t=300s）平均每秒N0=0.32s-11.由知用以上数据作图，从线性拟合后的数据可得质量厚度类型质量厚dmg/cm2间隔时间s记数NN1/sI/I0ln(I/I0)00104855485.510d116.20104466446.60.919824-0.083572d132.40103952395.20.813884-0.205943d148.60103551355.10.731234-0.313024d164.80103275327.50.674348-0.394015d181.00102986298.60.614782-0.486496d197.20102843284.30.585309-0.53562d2111.25205675283.750.584175-0.53755d2＋d1127.45205217260.850.536976-0.6218d2＋2d1143.65204899244.950.504205-0.68477d2＋3d1159.85204319215.950.444433-0.81096d2＋4d1176.05204109205.450.422792-0.86088d2＋5d1192.25203883194.150.399501-0.91754d2＋6d1208.45203389169.450.348592-1.053852d2222.50203507175.350.360753-1.019563d2333.7520195397.650.200606-1.606414d2445.0020151375.650.155262-1.862645d2556.2550117223.440.047652-3.043826d2667.501009479.470.018859-3.970776d2+d1683.701008358.350.016551-4.101346d2+2d1699.901006566.560.012861-4.353546d2+3d1716.1020011355.6750.011037-4.506496d2+4d1732.302009084.540.008698-4.744686d2+5d1748.502008384.190.007976-4.831277d2778.752006453.2250.005987-5.118097d2+d1794.953008532.8433330.005201-5.258947d2+2d1811.153006532.1766670.003827-5.56574从线性拟合后的数据可得质量吸收系数≈0.00654b.上图中纵坐标为-4即I/I0=10-4已标出,根据图线的弯曲走向,去掉最后九个点和前七个点,剩下的点重新做图,线性拟合,如下:对照图4.3.2-3吸收曲线,由以上两图知h≈10-0.5ln(I/I0)=-0.006d+0.23726取纵坐...
