A protom of energy 1 MeV is incident head-on on a gold nucleus (Z = 79), and is scattered through an angle of `180^(@)`. Calculate the distance of nearest approach.
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Gold number of haemoglobin is `0.03`. Hence, `100 mL` of gold sol will require haemoglobin so that gold is not coagulated by `1 mL` of `10%` NaCl solu
Correct Answer - C `0.03`= weight to Hb in mg xx `10//100` weight of Hb in mg `=0.30`
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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`
2 Answers 1 views
A deuterium nucleus at rest absorbs a Gamma ray photon of energy 2.21 MeV and splits into a proton and a neutron. The neutron was found to be at rest
Correct Answer - 1.008968u
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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`.
2 Answers 1 views
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`
2 Answers 1 views
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`
2 Answers 1 views
Assuming the splitting of `U^235` nucleus liberates `200 MeV` energy, find (a) the energy liberated in the fission of 1 kg of `U^235` and (b) the mass
Correct Answer - [(a) `8.2 xx 10^(10)kJ, 2.7 xx 10^(6) kg` (b) `1.5 g`]
2 Answers 1 views
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]`
2 Answers 1 views