Question

3．天然放射元素${\;}_{94}^{239}Pu$變成鉛的同位素${\;}_{82}^{207}Pb$要經(jīng)過n₁次α衰變，n₂次β衰變，中子數(shù)減少的個數(shù)為n₃，則（　�。�

• A． n₁=8，n₂=6，n₃=32
• B． n₁=5，n₂=6，n₃=20
• C． n₁=8，n₂=4，n₃=20
• D． n₁=5，n₂=6，n₃=32

Answer 1

分析 α衰變電荷數(shù)少2，質(zhì)量數(shù)少4，β衰變電荷數(shù)多1，質(zhì)量數(shù)不變．根據(jù)該規(guī)律判斷衰變的次數(shù)．抓住質(zhì)量數(shù)等于中子數(shù)與質(zhì)子數(shù)之和求出中子數(shù)減少的個數(shù)．

解答 解：${\;}_{94}^{239}{P}_{U}^{\;}{→}_{82}^{207}{P}_^{\;}+{{n}_{1}^{\;}}_{2}^{4}{H}_{e}^{\;}$${{+n}_{2}^{\;}}_{-1}^{0}e$
根據(jù)質(zhì)量數(shù)守恒和電荷數(shù)守恒，
$239=207+4{n}_{1}^{\;}$…①
$94=82+2{n}_{1}^{\;}+{n}_{2}^{\;}（-1）$…②
解得：${n}_{1}^{\;}=8$
${n}_{2}^{\;}=4$
${\;}_{94}^{239}{P}_{U}^{\;}$的中子數(shù)239-94=145
${\;}_{82}^{207}{P}_^{\;}$的中子數(shù)207-82=125
中子數(shù)減少的個數(shù)為：${n}_{3}^{\;}=145-125=20$
故選：C

點評 解決本題的關(guān)鍵知道衰變的過程中電荷數(shù)守恒、質(zhì)量數(shù)守恒．