Multiple Choice QuestionsThe activity of a radioactive sample is measured as No counts per minute at t = 0 and No/e counts per minute at t = 5 min. The time (in minute) at which the activity reduces to half its value is
loge 2 /5
5/ loge 2
5 log10 2
5 log10 2
The decay constant of a ratio isotope is λ. If A1 and A2 are its activities at times t1 and t2 respectively, te number of nuclei which have decayed during the time (t1-t2)
A1t1 - A2t2
A1 - A2
(A1 - A2)/λ
(A1 - A2)/λ
The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV, respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is
23.6 MeV
2.2 MeV
28.0 MeV
28.0 MeV
In the nuclear decay given below,
the particle emitted in the sequence are
β, α, γ
γ, β, α,
β,γ, α
β,γ, α
The number of beta particles emitted by radioactive substance is twice the number of alpha particle emitted by it. The resulting daughter is an
isobar of parent
isomer of parent
isotone of parent
isotone of parent
Two radioactive materials X1 and X2 have decay constant 5 λ respectively. If initially, they have the same number of nuclei, then the ratio of the number of nuclei of X1 to that of X2 will be 1/e after a time
λ
λ/2
1/4λ
1/4λ
If M (A, Z), MP and Mn denote the masses of the nucleus 
proton and neutron respectively in units of u (1 u = 931.5 MeV/c2) and BE represents its binding energy in MeV, then
M (A,Z) = ZMp + (A-Z)Mn - BE/c2
M (A,Z) = ZMp + (A-Z)Mn + BE
M (A,Z) = ZMp + (A-Z)Mn - BF
M (A,Z) = ZMp + (A-Z)Mn - BF
Two nuclei have their mass number in the ratio of 1:3. The ratio of their nuclear densities would be
1:3
3:1
(3)1/3 : 1
(3)1/3 : 1
In radioactive decay process, the negatively charged emitted beta particles are:
the electrons present inside the nucleus
the electrons produced as a result of the decay of neutrons inside the nucleus
the electrons produced as a result of collisions between atoms
the electrons produced as a result of collisions between atoms
A nucleus has 
mass represented by M (A, Z). If Mp and Mn denote the mass of proton and neutron respectively and BE the binding energy (in MeV), then:
BE = [M(A,Z)-ZMp - (A-Z)Mn]c2
BE = [ZMp + (A-Z)Mn -M(A,Z)]c2
BE = [ZMp + AMn - M (A,Z)]c2
BE = [ZMp + AMn - M (A,Z)]c2
B.
BE = [ZMp + (A-Z)Mn -M(A,Z)]c2
In the case of formation of a nucleus, the evolution of energy equal to the binding energy of the nucleus takes place due to the disappearance of a fraction of the total mass. If the quantity of mass disappearing is ΔM, then the binding energy is
BE = ΔMc2
From the above discussion, it is clear that the mass of the nucleus must be less than the sum of the masses of the constituent neutrons and protons. We can then write.
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