True or False Statements :
`5 alpha` and `4 beta^(-)` are emitted during the radioactive decay chain starting from `._(88)^(226) Ra` and ending at `._(82)^(206) Pb`.
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A nucleus with Z =92 emits the following in a sequence: `alpha,beta^(-),beta^(-),alpha,alpha,alpha,alpha,alpha,beta^(-),beta^(-),alpha,beta^(+),beta^(
Correct Answer - B `8 alpha` have been emitted `4 beta^(-)` have been emitted `2 beta^(+)` have emitted `alpha` reduces atomic number by 2 `beta^(-)` increases atomic number by 1 `beta^(+)`...
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Let `N_(beta)` be the number of `beta` particles emitted by `1` gram of `Na^(24)` radioactive nuclei (half life `= 15` hrs) in `7.5` hours, `N_(beta)`
Correct Answer - A No. of nuclei remains at any instant `N= N_(0)/(2^(t//T_(1//2)))=N_(0)/sqrt(2)` `N_("decayed")=N_(0)-N=N_(0) [1-1/sqrt(2)]` `=1/24xx6.023xx10^(23)xx0.3` `~~7.5xx10^(21)`
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Assuming that only particles emitted from atoms during natural radioactive decay are alpha and beta particles, which of the following atoms could not
Correct Answer - C `._(92)U^(235) overset (-alpha)rarr _(90)U^(231) overset(-alpha)rarr U_(88)^(227) overset(-beta)rarr _(89)U^(231)` So C and D are not possible
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The `Th_(90)^(232)` atom has successive alpha and beta decays to the end product `Pb_(82)^(208)` . The numbers of alpha and beta. Particles. Emitted i
Correct Answer - D Mass number get charge by `rArr4xxn` because atomic number charge occur only in `alpha` decay 4n=232-208 4n=24 4n=6 `alpharArr4` Mass number get charge by `rArr24` New...
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When the radioactive `isotope _(88)Ra^(226)` decays in a series by emission of three aplha `(alpha)` and a beta `(beta)` particle, the isotope X which
Correct Answer - A Due to emission os `3alpha` particles, the mass number is reduced by 12 while atomic number is decresed by 6. Due to emission of `beta` particle the...
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Using peoperties of determinants in questions 11 to 15, prove that : `|{:(alpha,alpha^(2),beta+gamma),(beta,beta^(2),gamma+alpha),(gamma,gamma^(2),alp
L.H.S.=`|{:(alpha,alpha^(2),beta+gamma),(beta,beta^(2),gamma+alpha),(gamma,gamma^(2),alpha+beta):}|=|{:(alpha,alpha^(2),alpha+beta+gamma),(beta,beta^(2),alpha+beta+gamma),(gamma,gamma^(2),alpha+beta+gamma):}|` `(C_(3)toC_(3)+C_(1))` `=(alpha+beta+gamma)|{:(alpha,alpha^(2),1),(beta,beta^(2),1),(gamma,gamma^(2),1):}|` `(alpha+beta+gamma)|{:(alpha,alpha^(2),1),(beta-alpha,beta^(2)-alpha^(2),0),(gamma-alpha,gamma^(2)-alpha^(2),0):}|` `(R_(2)toR_(2)-R_(1),R_(3)toR_(3)-R_(1))` `(alpha+beta+gamma)(beta-alpha)(gamma-alpha)|{:(alpha,alpha^(2),1),(1,beta-alpha,0),(1,gamma-alpha,0):}|` `=-(alpha-beta)(gamma-alpha)(alpha+beta+gamma)1.|{:(1,beta+alpha),(1,gamma+alpha)` `=-(alpha-beta)(gamma-alpha)(alpha+beta+gamma)(gamma+alpha-beta-alpha)` `=(alpha-beta)(beta-gamma)(gamma-alpha)(alpha+beta+gamma)` =R.H.S.
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If a variable takes the discrete values `alpha-4`, `alpha -(7)/(2), alpha-(5)/(2), alpha-2,alpha+(1)/(2), alpha-(1)/(2), alpha+5(alpha gt 0)`, then th
Correct Answer - A Arrange the data as follows : `alpha-(7)/(2), alpha-3, alpha-(5)/(2),alpha-2, alpha-(1)/(2),alpha+(1)/(2),alpha+4,alpha+5` Median `=(1)/(2)` [value of 4th item+value of 5th item] `therefore " Median"=(alpha-2+alpha-(1)/(2))/(2)=(2alpha-(5)/(2))/(2)=alpha-(5)/(4)`
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If alpha + beta is equal to minus 3 and alpha beta is equal to minus 5 by 2 then find the cube roots are alpha and beta
α + β = - 3 .......(1) and αβ = \(\frac{-5}{2}\) .........(2) (α - β)2 = (α + β)2 - 4αβ = (-3)2 - 4 x \(\frac{-5}{2}\) = 9 + 10 = 19 α - β = ±√19 ........(3) By adding equations (1) and...
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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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