A n `alpha`-particle of velocity `1.6xx10^(7) m s^(-1)` approaches a gold nucleii `(Z=79)`. Calculate the distance of closest approach. Mas of an `alpha`-particle `=6.6xx10^(-27) kg`.
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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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If `A(alpha, beta)=[("cos" alpha,sin alpha,0),(-sin alpha,cos alpha,0),(0,0,e^(beta))]`, then `A(alpha, beta)^(-1)` is equal to
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)`
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If `alpha, beta, gamma` are three real numbers and `A=[(1,cos (alpha-beta),cos(alpha-gamma)),(cos (beta-alpha),1,cos (beta-gamma)),(cos (gamma-alpha),
Correct Answer - A::B::D `A=[(cos alpha,sin alpha,0),(cos beta,sin beta,0),(cos gamma,sin gamma,0)][(cos alpha,cos beta,cos gamma),(sin alpha,sin beta,sin gamma),(0,0,0)]` Clearly, A is symmetric and `|A|=0`, hence, singular and not invertiable. Also, `A A^(T)...
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If `f(x)=x^(11)+x^9-x^7+x^3+1` and `f(sin^(-1)(sin8)=alpha,alpha` is constant, then `f(tan^(-1)(t a n8)` is equal to `alpha` (b) `alpha-2` (c) `alpha+
Correct Answer - D `f(x) + f(-x) = 2` Now `(sin^(-1) (sin 8)) = 3pi - 8 = y` And `(tan^(-1) (tan 8)) = (8 - 3pi) = -y` Hence, `f(y)...
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The value of `(alpha^3)/2cos e c^2(1/2tan^(-1)alpha/beta)+(beta^3)/2sec^2(1/2tan^(-1)(beta/alpha))i se q u a lto` `(alpha+beta)(alpha^2+beta^2)` (b) `
Correct Answer - C `(alpha^(3))/(2) cosec^(2) ((1)/(2) tan^(-1). (alpha)/(beta)) + (beta^(3))/(2) sec^(2) ((1)/(2) tan^(-1).(beta)/(alpha))` `= alpha^(3) (1)/(1 - cos(tan^(-1) ((alpha)/(beta)))) + beta^(3) (1)/(1 + cos (tan^(-1).(beta)/(alpha)))` `= alpha^(3) (1)/(1 -cos (cos^(-1)...
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If `alpha` is a root of `x^4 = 1` with negative principal argument then the principal argument of `Delta(alpha) = |(1,1,1), (alpha^n, alpha^(n+1), alp
Correct Answer - B `(b)` Clearly `alpha=-i` where `i^(2)=-1` So `Delta(alpha)=0+alpha^(2)+1-1-0-alpha^(3)` `=(-i)^(2)+1-1-(-i)^(3)` =-1+1-1-i=-1-i` So, principal argument of `Delta(alpha)` is `-(3pi)/(4)`.
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An alpha particle of velocity `1.6xx10^(7) ms^(-1)` approaches a gold nucleus (Z=79). Calculate the distance of the closest approach. Mass of an alpha
Correct Answer - `4.3 xx 10^(-14)m` , the size of the nucleus cannot be greater than `4.3 xx 10^(-14)m`.
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In a head-on collision between and `alpha`-particle and a gold nucleus, the distance of closest approach is 4.13 fermi. Calculate the energy of the pa
Correct Answer - 5.51 MeV
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In a head-on collision between an `alpha`-particle and a gold nucleus, the minimum distance of approach is `3.95 xx 10^(4)m`. Calculate the energy of
Correct Answer - 6 MeV
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What is the distance of closest approach to the nucleus for an `alpha`-particle of energy 5J MeV which undergoes scattering in the Gieger-Marsden expe
Correct Answer - `4.13 xx 10^(-14) m`
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