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 Since almost, whole the mass of an atom is in the nucleus, Mass of F nucleus = Mass of F atom `=(19)/(6.02xx10^(23))g` `=3.15xx10^(-23)g` (At. Mass of F...

Correct Answer - C `U^(238) rarr U^(234)+._(2)He^(4)` Let recoil speed be V then by COLM `234 V=4uimplies V=(4u)/234`

Correct Answer - D Number of nuclei decreases experientially `N=N_(0)e^(-lambdat)` and Rate of decay `(-(dN)/(dt))=lambdaN` Therefore, decay process lasts upto `t=oo`. Therefore a given nucleus may decay at any time after...

Correct Answer - A `4 ._(2)He^(4) rarr _(8)O^(16)` Mass defect `=4 xx 4.0026 -15.834 a.m.u.=0.1764 a.m.u.` Total energy released `=0.1764 xx 931 `MeV Energy released /Nucleon`=(0.1764 xx 931)/(16)=10.24 `MeV

Correct Answer - D When an `alpha` particle of the mass m moving with velocity v bombards on a heavy nucleus of charge Ze, then there will be no less of...

Correct Answer - B Momentum remains constant, `K=(p^(2))/(2m)or Kprop(1)/(m)implies(K_(n))/(K_(alpha))=(m_(alpha))/(m_(n))=(4)/(206)=(2)/(103)` `therefore` Kinetic energy of the residual nucleus `K_(n)=((2)/(103))K_(alpha)=((2)/(103))E`

Correct Answer - A We know that `alpha` particle carries two units of positve charge and four units of mass In `alpha` decays charge number of parent nucleus decreses by 2...

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.

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)`

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...