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 experiment.
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An `alpha`-particle having kinetic energy x MeV falls on a Cu-foil. The shortest distance from the nucleus of Cu to which `alpha`-particle reaches is
Correct Answer - 5 `K.E.=(KZe.2e)/(r ), " "K.E.=(9xx10^(9)xx29xx2(1.6 xx 10^(-19))^(2))/(1.67xx10^(-14))=5Mev`
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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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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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An `alpha` particle of `K.E. 10^(-12)J` exhibits back scattering from a gold nucleus Z=79. What can be the maximum possible radius of the gold nucleus
Correct Answer - `3.64 xx 10^(-14)m`
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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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A stationary `._(82)^(200)Pb` nucleus emits an `alpha`-particle with kinetic energy `T_(alpha) = 5.77 MeV`. Find the recoil velocity of a daughter nuc
Correct Answer - `[3.4 xx 10^(5) m//s, 0.020]`
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And `alpha`-particle with kinetic enregy `T_(alpha) = 7.0 MEV` is scattered elastically by an initially stationary `.^(6)Li` nucleus. Find the kinetic
Correct Answer - `[(T)/([1+(M-m)^(2)//4mM cos^(2) theta]) = 6.0 MeV]`
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Find the energy of the reaction `.^(14)N(alpha, p) .^(17)O`, if the kinetic energy of the incoming `alpha`-particle is `T_(alpha) = 4.0 MeV` & the pro
Correct Answer - `[Q = (1-eta_(p))T_(p) - (1-eta_(alpha))T_(alpha)-2 sqrt(eta_(p)eta_(alpha)T_(p)T_(alpha)) cos theta =-12 MeV, "where" n_(p) = m_(p)//m_(o), n_(alpha) = (m_(alpha))/(m_(0))]`
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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 `alp
Correct Answer - `4.3xx10^(-14) m`
2 Answers 1 views