问题标题:
钾(42K)的半衰期=12.36h,衰变2h,30h,60h,计算残留百分比?
问题描述:
钾(42K)的半衰期=12.36h,衰变2h,30h,60h,计算残留百分比?
陈毛狗回答:
一级反应:Lnc0/c=K*△tLn2=k*t1/2k=Ln2/t1/2=Ln2/12.36h=0.056(1/h)Lnc0/c=K*2h=0.056(1/h)*2hc0/c=1.12即c/c0=0.98398.3%Lnc0/c=K*2h=0.056(1/h)*30hc0/c=5.38即c/c0=0.18618.6%Lnc0/c=K*2h=...
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