When an `alpha` particle of mass m moving with velocity v bombards on a heavy nucleus of charge Ze, its distance of closest approach form the nucleus depends on m as
A. `(1)/(sqrt(m))`
B. `(1)/(m^(2))`
C. m
D. `(1)/(m)`
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 energy , so intial kinetic energy of `alpha` particle at closet apporach.
`rArr" " (1)/(2)mv^(2)=(2Ze^(2))/(4piepsi_(0)r_(0))rArrr_(0)prop(1)/(m)`
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 - B
According to the question,
For the distance of closest approach kinetic energy will totally be converted to potential energy
Hence, `" " (1)/(2) mv^(2) =(1)/(4piepsi_(0)) (Qq)/(r_(0)) rArr...
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 - C
Let d be the distance of closest approach then by the conservation of energy. Initial kinetic energy of incoming `alpha`-particle K.
= Final electric potential energy U...
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)`
Correct Answer - A
We have,
`A(alpha, beta)^(-1)=1/e^(beta) [(e^(beta) cos alpha,-e^(beta) sin alpha,0),(e^(beta) sin alpha,e^(beta) cos alpha,0),(0,0,1)]`
`=A(-alpha, -beta)`