A radioactive element decays by `beta^(-)` emission. If mass of parent and daughter atom are `m_(1)` and `m_(2)` respectively, calculate energy liberated during the emission. Mass of an electron = m.
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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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A radioactive nucleus decays by emitting one alpha and two beta particles, the daughter nucleus is ….. Of the parent.
Correct Answer - isotope
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The number of neutrons in the parent nucleus which gives `N^(14)` on beta emission is `…………………………`.
Correct Answer - eight
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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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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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In a reactor, an element X decays to a radioactive element Y, at a constant rate r atoms per second. Each decay reaction releases energy `E_(1)`. Half
Correct Answer - `[(i)" "eta r[E_(1)+E_(2)(1-e^((t log 2)/T))], (ii)" "eta r(E_(1)+E_(2))]`
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`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 d
Correct Answer - [1.86 MeV]
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Potassium-40 can decay in three modes .It can decay by `beta^(-)` - emission,`beta^(+)` -emission or electron capature. (a) Write the equation showing
Correct Answer - [(a) `._(19)^(40)K rarr ._(20)^(40)Ca + e^(-) + bar(v), ._(19)^(40)K rarr ._(18)^(40)Ar + e^(+) +v, ._(19)^(40)K + e^(-1) rarr ._(18)^(40)Ar+v` (b) 1.3034 MeV, 0.4676 MeV, 1.490 MeV]
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