2024版新教材高中物理第五章原子核微點(diǎn)16核反應(yīng)類(lèi)型的判斷課時(shí)作業(yè)新人教版選擇性必修第三冊(cè)

微點(diǎn)16核反應(yīng)類(lèi)型的推斷1．“兩彈一星”可以說(shuō)長(zhǎng)了中國(guó)人的志氣，助了中國(guó)人的威歷．下列核反應(yīng)方程中，屬于探討“兩彈”的基本核反應(yīng)方程的是()①eq\o\al(\s\up1(14),\s\do1())7N＋eq\o\al(\s\up1(4),\s\do1(2))He→eq\o\al(\s\up1(17),\s\do1())8O＋eq\o\al(\s\up1(1),\s\do1(1))H②eq\o\al(\s\up1(235),\s\do1())92U＋eq\o\al(\s\up1(1),\s\do1(0))n→eq\o\al(\s\up1(90),\s\do1(38))Sr＋eq\o\al(\s\up1(136),\s\do1())54Xe＋10eq\o\al(\s\up1(1),\s\do1(0))n③eq\o\al(\s\up1(235),\s\do1())92U→eq\o\al(\s\up1(234),\s\do1())90Th＋eq\o\al(\s\up1(4),\s\do1(④eq\o\al(\s\up1(2),\s\do1(1))H＋eq\o\al(\s\up1(3),\s\do1(1))H→eq\o\al(\s\up1(4),\s\do1(2))He＋eq\o\al(\s\up1(1),\s\do1(0))nA．①②B．③④C．②④D．②③2．下列核反應(yīng)為人工核轉(zhuǎn)變的是()A．eq\o\al(\s\up1(235),\s\do1())92U＋eq\o\al(\s\up1(1),\s\do1(0))n→eq\o\al(\s\up1(140),\s\do1())54Xe＋eq\o\al(\s\up1(94),\s\do1(38))Sr＋2eq\o\al(\s\up1(1),\s\do1(0))nB．eq\o\al(\s\up1(19),\s\do1())9F＋eq\o\al(\s\up1(4),\s\do1(2))He→eq\o\al(\s\up1(22),\s\do1(10))Ne＋eq\o\al(\s\up1(1),\s\do1(1))HC．eq\o\al(\s\up1(2),\s\do1(1))H＋eq\o\al(\s\up1(3),\s\do1(1))H→eq\o\al(\s\up1(4),\s\do1(2))He＋eq\o\al(\s\up1(1),\s\do1(0))nD．eq\o\al(\s\up1(24),\s\do1(11))Na→eq\o\al(\s\up1(24),\s\do1(12))Mg＋eq\o\al(\s\up1(0),\s\do1(－1))e3．下列關(guān)于核衰變和核反應(yīng)的類(lèi)型的表述正確的是()A．eq\o\al(\s\up1(2),\s\do1(1))H＋eq\o\al(\s\up1(2),\s\do1(1))H→eq\o\al(\s\up1(3),\s\do1(2))He＋eq\o\al(\s\up1(1),\s\do1(0))n是α衰變B．eq\o\al(\s\up1(14),\s\do1())6C→eq\o\al(\s\up1(14),\s\do1(7))N＋eq\o\al(\s\up1(0),\s\do1(－1))e，是β衰變C．eq\o\al(\s\up1(238),\s\do1())92U→eq\o\al(\s\up1(234),\s\do1())90Th＋eq\o\al(\s\up1(4),\s\do1(D．eq\o\al(\s\up1(14),\s\do1())7N＋eq\o\al(\s\up1(1),\s\do1(0))n→eq\o\al(\s\up1(14),\s\do1())6C＋eq\o\al(\s\up1(1),\s\do1(1))H，是輕核聚變4．關(guān)于下列核反應(yīng)說(shuō)法正確的是()A．eq\o\al(\s\up1(2),\s\do1(1))H＋eq\o\al(\s\up1(3),\s\do1(1))H→eq\o\al(\s\up1(4),\s\do1(2))He＋eq\o\al(\s\up1(1),\s\do1(0))n，此反應(yīng)放出能量，所以結(jié)合能降低B．eq\o\al(\s\up1(14),\s\do1())7N＋eq\o\al(\s\up1(4),\s\do1(2))He→eq\o\al(\s\up1(17C．eq\o\al(\s\up1(235),\s\do1())92U＋eq\o\al(\s\up1(1),\s\do1(0))n→eq\o\al(\s\up1(144),\s\do1())56Ba＋eq\o\al(\s\up1(89),\s\do1(36))Kr＋3eq\o\al(\s\up1(1),\s\do1(0))n，此反應(yīng)是核裂變，質(zhì)量虧損，放出能量D．eq\o\al(\s\up1(235),\s\do1())92U→eq\o\al(\s\up1(234),\s\do1())90Th＋eq\o\al(\s\up1(4),\s\do1(2))He，此反應(yīng)是α衰變，4個(gè)eq\o\al(\s\up1(5．[2024·江蘇泰州中學(xué)高二階段練習(xí)]有四個(gè)核反應(yīng)方程如下①eq\o\al(\s\up1(235),\s\do1())92U＋eq\o\al(\s\up1(1),\s\do1(0))n→eq\o\al(\s\up1(95),\s\do1(38))Sr＋eq\o\al(\s\up1(138),\s\do1())54Xe＋3X1②eq\o\al(\s\up1(2),\s\do1(1))H＋X2→eq\o\al(\s\up1(3),\s\do1(2))He＋eq\o\al(\s\up1(1),\s\do1(0))n③eq\o\al(\s\up1(238),\s\do1())92U→④eq\o\al(\s\up1(24),\s\do1(12))Mg＋eq\o\al(\s\up1(4),\s\do1(2))He→eq\o\al(\s\up1(27),\s\do1(13))Al＋X4下列說(shuō)法正確的是()A．①是核裂變B．X2為eq\o\al(\s\up1(3),\s\do1(1))HC．③是人工核反應(yīng)D．X4為中子6．[2024·遼寧沈陽(yáng)東北育才學(xué)校高二期中]下列關(guān)于核反應(yīng)方程及其表述均正確的是()A．eq\o\al(\s\up1(235),\s\do1())92U→eq\o\al(\s\up1(89),\s\do1(36))Kr＋eq\o\al(\s\up1(144),\s\do1())56Ba＋2eq\o\al(\s\up1(1),\s\do1(0))n為重核裂變反應(yīng)B．eq\o\al(\s\up1(238),\s\do1())92U→eq\o\al(\s\up1(234),\s\do1())90Th＋eq\o\al(\s\up1(4),\s\do1(C．eq\o\al(\s\up1(2),\s\do1(1))H＋eq\o\al(\s\up1(3),\s\do1(1))H→eq\o\al(\s\up1(4),\s\do1(2))He＋eq\o\al(\s\up1(1),\s\do1(0))n為輕核聚變反應(yīng)D．eq\o\al(\s\up1(3),\s\do1(2))He＋eq\o\al(\s\up1(27),\s\do1(13))Al→eq\o\al(\s\up1(30),\s\do1(15))P＋eq\o\al(\s\up1(1),\s\do1(1))H為原子核的人工轉(zhuǎn)變7．[2024·四川德陽(yáng)五中高二階段練習(xí)]下列核反應(yīng)方程正確的是()A．鈾核發(fā)生α衰變的核反應(yīng)是eq\o\al(\s\up1(238),\s\do1())92U→eq\o\al(\s\up1(234),\s\do1())93Th＋eq\o\al(\s\up1(0),\s\do1(B．輕核聚變反應(yīng)方程eq\o\al(\s\up1(2),\s\do1(1))H＋eq\o\al(\s\up1(3),\s\do1(1))H→eq\o\al(\s\up1(4),\s\do1(2))He＋X中，X表示中子C．鈾核裂變的一種核反應(yīng)方程eq\o\al(\s\up1(235),\s\do1())92U→eq\o\al(\s\up1(144),\s\do1())56Ba＋eq\o\al(\s\up1(89),\s\do1(36))Kr＋2eq\o\al(\s\up1(1),\s\do1(0))nD．14C發(fā)生β衰變的核反應(yīng)方程式為eq\o\al(\s\up1(14),\s\do1())6C→eq\o\al(\s\up1(14),\s\do1())5B＋eq\o\al(\s\up1(0),\s\do1(微點(diǎn)16核反應(yīng)類(lèi)型的推斷1．答案：C解析：“兩彈”是指原子彈和氫彈，原子彈是鈾核裂變eq\o\al(\s\up1(235),\s\do1(92))U＋eq\o\al(\s\up1(1),\s\do1(0))neq\o\al(\s\up1(90),\s\do1(38))Sr＋eq\o\al(\s\up1(136),\s\do1(54))Xe＋10eq\o\al(\s\up1(1),\s\do1(0))n氫彈是輕核聚變eq\o\al(\s\up1(2),\s\do1(1))H＋eq\o\al(\s\up1(3),\s\do1(1))Heq\o\al(\s\up1(4),\s\do1(2))He＋eq\o\al(\s\up1(1),\s\do1(0))n，故選C.2．答案：B解析：依據(jù)核裂變定義可知，eq\o\al(\s\up1(235),\s\do1(92))U＋eq\o\al(\s\up1(1),\s\do1(0))neq\o\al(\s\up1(140),\s\do1(54))Xe＋eq\o\al(\s\up1(94),\s\do1(38))Sr＋2eq\o\al(\s\up1(1),\s\do1(0))n為核裂變，故A錯(cuò)誤；eq\o\al(\s\up1(19),\s\do1(9))F＋eq\o\al(\s\up1(4),\s\do1(2))Heeq\o\al(\s\up1(22),\s\do1(10))Ne＋eq\o\al(\s\up1(1),\s\do1(1))H為人工轉(zhuǎn)變?cè)雍说倪^(guò)程，用快速粒子穿透核內(nèi)部，將核轉(zhuǎn)變?yōu)榱硪粋€(gè)核的過(guò)程，故B正確；依據(jù)核聚變的定義可知，eq\o\al(\s\up1(2),\s\do1(1))H＋eq\o\al(\s\up1(3),\s\do1(1))Heq\o\al(\s\up1(4),\s\do1(2))He＋eq\o\al(\s\up1(1),\s\do1(0))n為核聚變，故C錯(cuò)誤；依據(jù)變更過(guò)程可知，eq\o\al(\s\up1(24),\s\do1(11))Naeq\o\al(\s\up1(24),\s\do1(12))Mg＋eq\o\al(\s\up1(0),\s\do1(－1))e為原子核發(fā)生β衰變的過(guò)程，故D錯(cuò)誤．故選B.3．答案：B解析：eq\o\al(\s\up1(2),\s\do1(1))H＋eq\o\al(\s\up1(2),\s\do1(1))Heq\o\al(\s\up1(3),\s\do1(2))He＋eq\o\al(\s\up1(1),\s\do1(0))n是輕核聚變，故A錯(cuò)誤；eq\o\al(\s\up1(14),\s\do1(6))Ceq\o\al(\s\up1(14),\s\do1(7))N＋eq\o\al(\s\up1(0),\s\do1(－1))e是β衰變，故B正確；eq\o\al(\s\up1(238),\s\do1(92))Ueq\o\al(\s\up1(234),\s\do1(90))Th＋eq\o\al(\s\up1(4),\s\do1(2))He是α衰變，故C錯(cuò)誤；eq\o\al(\s\up1(14),\s\do1(7))N＋eq\o\al(\s\up1(1),\s\do1(0))neq\o\al(\s\up1(14),\s\do1(6))C＋eq\o\al(\s\up1(1),\s\do1(1))H是原子核的人工轉(zhuǎn)變，故D錯(cuò)誤．故選B.4．答案：C解析：eq\o\al(\s\up1(2),\s\do1(1))H＋eq\o\al(\s\up1(3),\s\do1(1))Heq\o\al(\s\up1(4),\s\do1(2))He＋eq\o\al(\s\up1(1),\s\do1(0))n是核聚變，此反應(yīng)放出能量，所以結(jié)合能變大，故A錯(cuò)誤；eq\o\al(\s\up1(14),\s\do1())7N＋eq\o\al(\s\up1(4),\s\do1(2))Heeq\o\al(\s\up1(17),\s\do1())8O＋X此反應(yīng)是原子核的人工轉(zhuǎn)化，X是質(zhì)子，故B錯(cuò)誤；eq\o\al(\s\up1(235),\s\do1(92))U＋eq\o\al(\s\up1(1),\s\do1(0))neq\o\al(\s\up1(144),\s\do1(56))Ba＋eq\o\al(\s\up1(89),\s\do1(36))Kr＋3eq\o\al(\s\up1(1),\s\do1(0))n此反應(yīng)是核裂變，質(zhì)量虧損，放出能量，故C正確；eq\o\al(\s\up1(235),\s\do1(92))Ueq\o\al(\s\up1(234),\s\do1(90))Th＋eq\o\al(\s\up1(4),\s\do1(2))He此反應(yīng)是α衰變，半衰期是大量放射性原子衰變的統(tǒng)計(jì)規(guī)律，對(duì)個(gè)別的放射性原子沒(méi)有意義，故D錯(cuò)誤．5．答案：A解析：重核分裂成質(zhì)量較小的核的反應(yīng)為重核的裂變，①是核裂變，A正確；依據(jù)質(zhì)量數(shù)守恒和核電荷數(shù)守恒，知eq\o\al(\s\up1(2),\s\do1(1))H＋eq\o\al(\s\up1(2),\s\do1(1))Heq\o\al(\s\up1(3),\s\do1(2))He＋eq\o\al(\s\up1(1),\s\do1(0))n，所以X2為eq\o\al(\s\up1(2),\s\do1(1))H，B錯(cuò)誤；原子核放出α粒子或β粒子，變成另一種原子核的變更稱(chēng)為原子核的衰變，依據(jù)質(zhì)量數(shù)守恒和核電荷數(shù)守恒，知eq\o\al(\s\up1(238),\s\do1(92))Ueq\o\al(\s\up1(234),\s\do1(90))Th＋eq\o\al(\s\up1(4),\s\do1(2))He，③是α衰變，C錯(cuò)誤；依據(jù)質(zhì)量數(shù)守恒和核電荷數(shù)守恒，知eq\o\al(\s\up1(24),\s\do1(12))Mg＋eq\o\al(\s\up1(4),\s\do1(2))Heeq\o\al(\s\up1(27),\s\do1(13))Al＋eq\o\al(\s\up1(1),\s\do1(1))H，X4為質(zhì)子，D錯(cuò)誤．6．答案：C解析：重核裂變反應(yīng)方程為eq\o\al(\s\up1(235),\s\do1(92))U＋eq\o\al(\s\up1(1),\s\do1(0))neq\o\al(\s\up1(89),\s\do1(36))Kr＋eq\o\al(\s\up1(144),\s\do1(56))Ba＋3eq\o\al(\s\up1(1),\s\do1(0))n，A錯(cuò)誤；eq\o\al(\s\up1(238),\s\do1(92))Ueq\o\al(\s\up1(234),\s\do1(90))Th＋eq\o\al(\s\up1(4),\s\do1(2))He為α衰變，B錯(cuò)誤；eq\o\al(\s\up1(2),\s\do1(1))H＋eq\