`P^32` beta-decays to `S^32`.Find the sum of the energy of the antineutrino and the kinetic energy of the `beta`-particle. Neglect the recoil of the daughter nucleus. Atomic mass of `P^32=31.974 u` and that of `S^32=31.972 u`.
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If `"sin"(alpha + beta) "sin" (alpha-beta) = "sin" gamma(2"sin" beta + "sin"gamma) " where " 0 lt alpha, beta, lt pi,` then the straight line whose eq
Correct Answer - C `"sin"(alpha+beta)"sin"(alpha-beta)="sin" gamma(2"sin"beta + "sin"gamma)` `"or ""sin"^(2)alpha-("sin"beta + "sin" gamma)^(2)=0` `"or "("sin"alpha+"sin" beta+"sin"gamma)("sin"alpha-"sin"beta-"sin"gamma)=0` `"Since " 0 lt alpha, beta, gamma lt pi," we have"` `"sin"alpha+"sin"beta+"sin"gamma ne 0` `therefore...
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If `alpha, beta, gamma` are three real numbers and `A=[(1,cos (alpha-beta),cos(alpha-gamma)),(cos (beta-alpha),1,cos (beta-gamma)),(cos (gamma-alpha),
Correct Answer - A::B::D `A=[(cos alpha,sin alpha,0),(cos beta,sin beta,0),(cos gamma,sin gamma,0)][(cos alpha,cos beta,cos gamma),(sin alpha,sin beta,sin gamma),(0,0,0)]` Clearly, A is symmetric and `|A|=0`, hence, singular and not invertiable. Also, `A A^(T)...
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The value of `(alpha^3)/2cos e c^2(1/2tan^(-1)alpha/beta)+(beta^3)/2sec^2(1/2tan^(-1)(beta/alpha))i se q u a lto` `(alpha+beta)(alpha^2+beta^2)` (b) `
Correct Answer - C `(alpha^(3))/(2) cosec^(2) ((1)/(2) tan^(-1). (alpha)/(beta)) + (beta^(3))/(2) sec^(2) ((1)/(2) tan^(-1).(beta)/(alpha))` `= alpha^(3) (1)/(1 - cos(tan^(-1) ((alpha)/(beta)))) + beta^(3) (1)/(1 + cos (tan^(-1).(beta)/(alpha)))` `= alpha^(3) (1)/(1 -cos (cos^(-1)...
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In the thorium decay series that begins with `.^(232)Th,.^(228)Th` alpha-decays into `.^(224)Ra` with a half life of 1.9yr. `.^(224)Ra` alpha-decays i
Correct Answer - 5.9kg,1149kg
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A radioactive sample consists of two isotopes one of then decays by `alpha-` emission with a half life of `tau_(1)=405s` and the other one decays by `
Correct Answer - (a) `(1)/(9)` (b) 1215s
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A radioactive nucleus A decays into B with a decay constnat `lambda_(1)`. B is alos radioactive and it decays into C with a decay constant of `lambda_
Correct Answer - (a)`(dN_(2))/(dt)=-lambda_(2)N_(2)+lambda_(1)N_(1)^(0)e^(-lambda_(1)t)` (b) `(A_(2))/(A_(1))=(lambda_(2))/(lambda_(2)-lambda_(1))` (c) `((1)/(lambda_(2)-lambda_(1)))ln((lambda_(2))/(lambda_(1)))`
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Let `alpha,beta` be two real numbers satisfying the following relations `alpha^2+beta^2=5, 3(alpha^5+beta^5)=11(alpha^3+beta^3)1.` Possible value of `
Correct Answer - A `alpha^(2)+beta^(2)=5` `3(alpha^(5)+beta^(5))=11(alpha^(3)+beta^(3))` `(alpha^(5)+beta^(5))/(alpha^(3)+beta^(3))=(11)/(3)` ` :. ((alpha^(3)+beta^(3))(alpha^(2)+beta^(2))-(alpha^(2)beta^(2)(alpha+beta)))/(alpha^(3)+beta^(3))=(11)/(3)` `:. alpha^(2)+beta^(2)-(alpha^(2)beta^(2)(alpha+beta))/((alpha+beta)(alpha^(2)+beta^(2)-alphabeta))=(11)/(3)` ` :. 5-(alpha^(2)beta^(2))/(5-alphabeta)=(11)/(3)` ltbrlt `:. (25-5alphabeta-alpha^(2)beta^(2))/(5-alphabeta)=(11)/(3)` Let `alphabeta=t` `(25-5t-t^(2))/(5-t)=(11)/(3)` `75-15t-3t^(2)=55-11 t` `75-15t-3t^(2)-55+11t=0` `-3t^(2)-4t+20=0` `(t-2)(3t+10)=0` ` :. t=2` or `(-10)/(3)` So `alpha...
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Let `alpha`, `beta` be two real numbers satisfying the following relations `alpha^(2)+beta^(2)=5`, `3(alpha^(5)+beta^(5))=11(alpha^(3)+beta^(3))` Poss
Correct Answer - B `alpha^(2)+beta^(2)=5` `3(alpha^(5)+beta^(5))=11(alpha^(3)+beta^(3))` `(alpha^(5)+beta^(5))/(alpha^(3)+beta^(3))=(11)/(3)` ` :. ((alpha^(3)+beta^(3))(alpha^(2)+beta^(2))-(alpha^(2)beta^(2)(alpha+beta)))/(alpha^(3)+beta^(3))=(11)/(3)` `:. alpha^(2)+beta^(2)-(alpha^(2)beta^(2)(alpha+beta))/((alpha+beta)(alpha^(2)+beta^(2)-alphabeta))=(11)/(3)` ` :. 5-(alpha^(2)beta^(2))/(5-alphabeta)=(11)/(3)` ltbrlt `:. (25-5alphabeta-alpha^(2)beta^(2))/(5-alphabeta)=(11)/(3)` Let `alphabeta=t` `(25-5t-t^(2))/(5-t)=(11)/(3)` `75-15t-3t^(2)=55-11 t` `75-15t-3t^(2)-55+11t=0` `-3t^(2)-4t+20=0` `(t-2)(3t+10)=0` ` :. t=2` or `(-10)/(3)` So `alpha...
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Let `alpha`, `beta` be two real numbers satisfying the following relations `alpha^(2)+beta^(2)=5`, `3(alpha^(5)+beta^(5))=11(alpha^(3)+beta^(3))` Quad
Correct Answer - D `alpha^(2)+beta^(2)=5` `3(alpha^(5)+beta^(5))=11(alpha^(3)+beta^(3))` `(alpha^(5)+beta^(5))/(alpha^(3)+beta^(3))=(11)/(3)` ` :. ((alpha^(3)+beta^(3))(alpha^(2)+beta^(2))-(alpha^(2)beta^(2)(alpha+beta)))/(alpha^(3)+beta^(3))=(11)/(3)` `:. alpha^(2)+beta^(2)-(alpha^(2)beta^(2)(alpha+beta))/((alpha+beta)(alpha^(2)+beta^(2)-alphabeta))=(11)/(3)` ` :. 5-(alpha^(2)beta^(2))/(5-alphabeta)=(11)/(3)` ltbrlt `:. (25-5alphabeta-alpha^(2)beta^(2))/(5-alphabeta)=(11)/(3)` Let `alphabeta=t` `(25-5t-t^(2))/(5-t)=(11)/(3)` `75-15t-3t^(2)=55-11 t` `75-15t-3t^(2)-55+11t=0` `-3t^(2)-4t+20=0` `(t-2)(3t+10)=0` ` :. t=2` or `(-10)/(3)` So `alpha...
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A stationary `._(82)^(200)Pb` nucleus emits an `alpha`-particle with kinetic energy `T_(alpha) = 5.77 MeV`. Find the recoil velocity of a daughter nuc
Correct Answer - `[3.4 xx 10^(5) m//s, 0.020]`
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