Fill in the blanks with appropriate items :
The number of ` alpha` and `beta ` - particles emitted, when the following nuclear transformation takes place are _______ and _______ respectively.
`._(92)^(238) Xrarr ._(82)^(206)Y`
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^(+)`...
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)`
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
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...
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)`
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...