B\u00e0i 1 (trang 278 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):<\/b>\u00a0Trong m\u1ed9t ph\u1ea3n \u1ee9ng h\u1ea1t nh\u00e2n, t\u1ed5ng kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a c\u00e1c h\u1ea1t tham gia<\/span><\/p>\n

A. \u0110\u01b0\u1ee3c b\u1ea3o to\u00e0n.<\/p>\n

B. T\u0103ng.<\/p>\n

C. Gi\u1ea3m.<\/p>\n

D. T\u0103ng ho\u1eb7c gi\u1ea3m t\u00f9y theo ph\u1ea3n \u1ee9ng.<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

Ch\u1ecdn D.<\/p>\n

B\u00e0i 2 (trang 278 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):<\/b>\u00a0Trong d\u00e3y ph\u00e2n r\u00e3 ph\u00f3ng x\u1ea1\u00a0_{92<\/sub>^{235<\/sup>\u00a0X \u2192\u00a082<\/sub>207<\/sup>Y c\u00f3 bao nhi\u00eau h\u1ea1t \u03b1 v\u00e0 \u03b2 \u0111\u01b0\u1ee3c ph\u00e1t ra ?<\/span><\/p>\n}}

A. 3\u03b1 v\u00e0 4\u03b2.<\/p>\n

B. 7\u03b1 v\u00e0 4\u03b2.<\/p>\n

C. 4\u03b1 v\u00e0 7\u03b2.<\/p>\n

D. 7\u03b1 v\u00e0 2\u03b2.<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

Ch\u1ecdn B.<\/p>\n

G\u1ecdi x\u2013 s\u1ed1 h\u1ea1t \u03b1 v\u00e0 y\u2013 s\u1ed1 h\u1ea1t electron (\u03b2^{–<\/sup>). Ph\u00e2n r\u00e3 x\u1ea3y ra tu\u00e2n theo \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n s\u1ed1 kh\u1ed1i A v\u00e0 \u0111i\u1ec7n t\u00edch Z. v\u1eady ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n}

235 = 207 + x.4 + y.0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(1)<\/p>\n

92 = 82 + x.2 + y(-1). T\u1eeb (1) suy ra: x = 7. Th\u1ebf v\u00e0o (2) ta c\u00f3 y = 4.<\/p>\n

K\u1ebft qu\u1ea3 kh\u1eb3ng \u0111\u1ecbnh: h\u1ea1t nh\u00e2n\u00a0_{92<\/sub>^{235<\/sup>\u00a0X \u0111\u00e3 ph\u00f3ng ra 7 h\u1ea1t \u03b1 v\u00e0 4 electron.<\/p>\n}}

B\u00e0i 3 (trang 278 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):<\/b>\u00a0X\u00e1c \u0111\u1ecbnh h\u1ea1t X trong c\u00e1c ph\u1ea3n \u1ee9ng sau \u0111\u00e2y:<\/span><\/p>\n

F_{9<\/sub>^{19<\/sup>\u00a0+ p \u2192 O8<\/sub>16<\/sup>\u00a0+ X.<\/p>\n}}

Mg_{12<\/sub>^{25<\/sup>\u00a0+ X \u2192 Na11<\/sub>22<\/sup>\u00a0+ \u03b1<\/p>\n}}

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

\u00c1p d\u1ee5ng c\u00e1c \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n trong ph\u1ea3n \u1ee9ng h\u1ea1t nh\u00e2n ta \u0111\u01b0\u1ee3c:<\/p>\n

F_{9<\/sub>^{19<\/sup>\u00a0+ H1<\/sub>1<\/sup>\u00a0\u2192 O8<\/sub>16<\/sup>\u00a0+ He2<\/sub>4<\/sup>.<\/p>\n}}

Mg_{12<\/sub>^{25<\/sup>\u00a0+ H1<\/sub>1<\/sup>\u00a0\u2192 Na11<\/sub>22<\/sup>\u00a0+ He2<\/sub>4<\/sup><\/p>\n}}

B\u00e0i 4 (trang 278 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):<\/b>\u00a0Cho ph\u1ea3n \u1ee9ng h\u1ea1t nh\u00e2n:<\/span><\/p>\n

<\/p>\n

a) X\u00e1c \u0111\u1ecbnh s\u1ed1 kh\u1ed1i, nguy\u00ean t\u1eed s\u1ed1 v\u00e0 t\u00ean g\u1ecdi h\u1ea1t nh\u00e2n X.<\/p>\n

b) Ph\u1ea3n \u1ee9ng \u0111\u00f3 t\u1ecfa ra hay thu n\u0103ng l\u01b0\u1ee3ng. t\u00ednh \u0111\u1ed9 l\u1edbn c\u1ee7a n\u0103ng l\u01b0\u1ee3ng t\u1ecfa ta hay thu \u0111\u00f3 theo \u0111\u01a1n v\u1ecb jun.<\/p>\n

Cho bi\u1ebft : m_{Ar<\/sub>\u00a0= 36,956889u; mCl<\/sub>\u00a0= 36,956563u; mn<\/sub>\u00a0= 1,008665u; mp<\/sub>\u00a0= 1,007276u.<\/p>\n}

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

a) X\u00e1c \u0111\u1ecbnh s\u1ed1 kh\u1ed1i, nguy\u00ean t\u1eed s\u1ed1 v\u00e0 t\u00ean g\u1ecdi h\u1ea1t nh\u00e2n X.<\/p>\n

T\u1eeb ph\u01b0\u01a1ng tr\u00ecnh ph\u1ea3n \u1ee9ng ta c\u00f3 h\u1ea1t X c\u00f3 Z = 1.A – 1. \u0110\u00f3 l\u00e0 h\u1ea1t proton<\/p>\n

V\u1eady ph\u01b0\u01a1ng tr\u00ecnh ph\u1ea3n \u1ee9ng \u0111\u1ea7y \u0111\u1ee7:<\/p>\n

<\/p>\n

b) Ph\u1ea3n \u1ee9ng \u0111\u00f3 t\u1ecfa ta hay thu n\u0103ng l\u01b0\u1ee3ng. t\u00ednh \u0111\u1ed9 l\u1edbn c\u1ee7a n\u0103ng l\u01b0\u1ee3ng t\u1ecfa ra hay thu v\u00e0o:<\/p>\n

Ta c\u00f3 :<\/p>\n

M_{0<\/sub>\u00a0= m(Cl) + m(P) = 37,963839u ;<\/p>\n}

M = m(Ar) + m(n) = 37,965559u<\/p>\n

Ta th\u1ea5y M > M_{0<\/sub>: ph\u1ea3n \u1ee9ng thu n\u0103ng l\u01b0\u1ee3ng. \u0111\u1ed9 l\u1edbn c\u1ee7a n\u0103ng l\u01b0\u1ee3ng thu l\u00e0:<\/p>\n}

Q = (M – M_{0<\/sub>)c^{2<\/sup>\u00a0– 0,001720u.c2<\/sup>Q = 1,6022 MeV = 2,56.10(-13)<\/sup>\u00a0J.<\/p>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"}}

C\u00c2U H\u1eceI C\u00e2u 1 (trang 278 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):\u00a0Th\u1ebf n\u00e0o l\u00e0 ph\u1ea3n \u1ee9ng h\u1ea1t nh\u00e2n ? L\u1eddi gi\u1ea3i: Xem m\u1ee5c 1a ph\u1ea7n KTCB. C\u00e2u 2 (trang 278 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):\u00a0N\u00eau v\u00e0 gi\u1ea3i th\u00edch c\u00e1c \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n trong ph\u1ea3n \u1ee9ng h\u1ea1t nh\u00e2n. L\u1eddi gi\u1ea3i: Xem m\u1ee5c 1b ph\u1ea7n […]<\/p>\n","protected":false},"author":3,"featured_media":24957,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1304],"tags":[1432,1431],"yoast_head":"\n