Potassium-40 can decay in three modes .It can decay by `beta^(-)` - emission,`beta^(+)` -emission or electron capature. (a) Write the equation showing the end products. (b) Find the Q-value in each of the three cases. Atomic masses of `Ar_18^40`,`K_19^40`and `Ca_20^40` are 39.9624 u,39.9640 u,and 39.9626 u respectively.
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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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Show that the solution of the equation `[(x, y),(z, t)]^(2)=O` is `[(x,y),(z,t)]=[(pm sqrt(alpha beta),-beta),(alpha,pm sqrt(alpha beta))]` where `alp
Given equation is `[(x,y),(z,t)]^(2)=[(0,0),(0,0)]` `implies [(x,y),(z,t)][(x,y),(z,t)]=[(x^(2)+yz,xy+yt),(zx+tz,zy+t^(2))]=[(0,0),(0,0)]` `implies x^(2)+yz=0` (1) `y(x+t)=0` (2) `z(x+t)=0` (3) `yz+t^(2)=0` (4) From (1) and (4), we have `x^(2)=t^(2)` or `x= pm t` Case I : If...
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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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An element `_(z)M^(A)` undergoes an `alpha -` emission followed by two successive `beta- emission.` The element formed is ……
Correct Answer - `zM^(A-4) `
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Nuclei of `.^(64)Cu` can decay be electron capture (probability 61%) or by `beta^(+)` decay (39%). Half life of `.^(64)Cu` is 12.7 hour. Find the part
Correct Answer - 20.8hr.
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The ratio isotope `._(29)^(64)Cu` can decay through positron emission and electron capture. In both the processes neutrino (v) is emitted. Calculate t
Correct Answer - `(k_("max"))_(Ec)-(k_("max"))_(beta^(+))=2m_(e)c^(2)=1.02MeV`
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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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