Fill in the blanks with appropriate items :
When `._(c)^(a) X` changes to `._(d)^(b) Y` number if ` alpha` and `beta` particle emitted respectively are ______
1 Answers
Talk Doctor Online in Bissoy App
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^(+)`...
2 Answers 2 views
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)`
2 Answers 1 views
Fill in the blanks with appropriate items : The number of ` alpha` and `beta ` - particles emitted, when the following nuclear transformation takes pl
Correct Answer - 8,6 `._(92)^(238)X rarr. _(82)^(206)Y + x alpha + y alpha` `._(92)^(238)X rarr ._(82)^(206) Y + x(._(2)He^(4))+y(._(-1)^(0)e)` Comparing mass no. `238 rArr 206 + 4x +0y` `x=8 i.e. 8...
2 Answers 1 views
Fill in the blanks with appropriate items : The `alpha` -activity of `1 gm .^(226) Ra. (t_(1//2)=1620 ` year) is _______ dpm.
Correct Answer - `2.16 xx 10^(12) dpm` `A=N lambda =((1)/(226)xxN_(A)) xx(0.693)/(1620 xx12 xx24xx30xx60)` `2.16 xx 10^(-12)` d.p.m
2 Answers 1 views
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...
2 Answers 1 views
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.
2 Answers 1 views
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)`
2 Answers 1 views
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...
2 Answers 1 views
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...
2 Answers 1 views
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...
2 Answers 1 views