In the reaction , if the binding energies of are respectively, a, b and c (in MeV), then the energy (in MeV) released in this reaction is [2005]
requires a and b amount of energies for their nucleons to be separated.
releases c amount of energy in its formation i.e., in assembling the nucleons as nucleus.
Hence, Energy released =c – (a + b) = c – a – b
In any fission process, the ratio [2005]
Binding energy per nucleon for fission products is higher relative to Binding energy per nucleon for parent nucleus, i.e., more masses are lost and are obtained as kinetic energy of fission products. So, the given ratio < 1.
Fission of nuclei is possible because thebinding energy per nucleon in them [2005]
Binding energy per nucleons decreases with increase in mass numbers.
The decrease of binding energy per nucleon for nuclei with a high mass number is due to increased Coulombic repulsion between protons inside the nucleus.
The binding energy of deuteron is 2.2 MeV and that of is 28 MeV. If two deuterons are fused to form one then the energy released is [2006]
Energy released = 28 – 2 × 2.2 = 23.6 MeV (Binding energy is energy released on formation of Nucleus)
In a radioactive material the activity at time t_{1} is R_{1} and at a later time t_{2}, it is R_{2}. If the decay constant of the material is λ, then [2006]
Let at time t_{1} & t_{2}, number of particles be N_{1} & N_{2}. So,
Two radioactive substances A and B have decay constants 5λ and λ respectively. At t = 0 they have the same number of nuclei. The ratio of number of nuclei of A to those of B will be (1/e)^{2} after a time interval [2007]
λ_{ A} = 5λ and λ_{B} = λ
At t = 0, (N_{0})_{A} = (N_{0})_{B}
Given,
According to radioactive decay,
From (1) and (2),
In a radioactive decay process, the negativelycharged emitted β particles are [2007]
In beta min us decay (β–), a neutron is transformed into a proton, and an electron is emitted from the nucleus along with antineutrino.
A nucleus ^{A}_{Z}X has mass represented by M (A, Z). If M_{p} and M_{n} denote the mass of proton and neutron respectively and B.E. the binding energy in MeV, then [2007]
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 = ΔMc^{2}
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.
ΔM= [ZM_{p} + (AZ)M_{n} M(A,Z)
Where M (A, Z) is the mass of the atom of mass number A and atomic number Z. Hence, the binding energy of the nucleus is
BE = [ZM_{p} + (AZ)M_{n} M(A,Z)]c^{2}
Where N = A Z = number of neutrons.
If the nucleus has nuclear radius of about 3.6 fm, then would have its radius approximately as [2007]
It has been known that a nucleus of mass number A has radius R = R_{0}A^{1/3},
where R_{0 }= 1.2 × 10^{–15}m and A = mass number
In case of let nuclear radius be R_{1}
and for nuclear radius be R_{2}
For
For
Two radioactive materials X_{1} and X_{2} have decay constants 5λ and λ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of X_{1} to that of X_{2} will be 1/e after a time
Let the required time be t. Then
Where N_{1} = number of nuclei of X_{1 }after time t
N_{2 }= number of nuclei of X_{2} after time t
N_{0 }= initial number of nuclei of X_{1} and X_{2} each.
Now,
λ_{1 }= 5λ; λ_{2} = λ
If M (A; Z), M_{p} and M_{n} denote the masses of the nucleus proton and neutron respectively in units of u ( 1u = 931.5 MeV/c^{2}) and BE represents its bonding energy in MeV, then [2008]
Mass defect = ZM_{p }+ (A –Z)M_{n}–M(A,Z)
ZM_{p} + (A–Z) M_{n}–M(A,Z)
∴ M (A, Z) = ZM_{p} + (A–Z)M_{n}–
Two nuclei have their mass numbers in the ratioof 1 : 3. The ratio of their nuclear densities wouldbe [2008]
Requird ratio of nuclear densities =
In the nuclear decay given below: [2009]
the particles emitted in the sequence are
mass number and charge number of a nucleus remains unchanged during γ decay)
The mass of a nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of nucleus is nearly
The activity of a radioactive sample is measured as N0 counts per minute at t = 0 and N0/e counts per minute at t = 5 minutes. The time (in minutes) at which the activity reduces to half its value is [2010]
The binding energy per nucleon in deuteriumand helium nuclei are 1.1 MeV and 7.0 MeV,respectively. When two deuterium nuclei fuseto form a helium nucleus the energy released inthe fusion is :
Binding energy of two_{ 1}H^{2} nuclei = 2(1.1 × 2) = 4.4 MeV
Binding energy of one _{2}He^{4 }nucelus = 4 × 7.0 =28 MeV
∴ Energy released = 28 – 4.4 = 23.6 MeV
The decay constant of a radio isotope is λ. If A_{1} and A_{2} are its activities at times t_{1} and t_{2} respectively, the number of nuclei which have decayed during the time (t_{1} – t_{2}) :
Activity is given by
Activity at time t_{1} is A_{1} = λN_{1}
and activity at time t_{2 }is A_{2} = – l N_{2} As t_{1} > t_{2}, therefore, number of atoms remained after time t_{1 }is less than that remained after time t_{2}. That is, N_{1} < N_{2}.
∴ number of nuclei decayed in (t_{1} – t_{2})
The half life of a radioactive isotope 'X' is 50 years. It decays to another element 'Y' which isstable. The two elements 'X' and 'Y' were foundto be in the ratio of 1 : 15 in a sample of agiven rock. The age of the rock was estimatedto be [2011]
The power obtained in a reactor using U^{235} disintegration is 1000 kW. The mass decay of U^{235} per hour is [2011]
E = mc^{2}
So, mass decay per second
(Power in watt)
and mass decay per hour =
= 4 × 10^{–8} kg
= 40 microgram
A radioactive nucleus of mass M emits a photonof frequency μ and the nucleus recoils. The recoilenergy will be [2011]
Momentum
Recoil energy
A nucleus emits one αparticle and two βparticles. The resulting nucleus is [2011]
When emits one particle then its atomic mass decreases by 4 units and atomic number by 2. Therefore, the new nucleus becomes mn . But as it emits two β^{–} particles, its atomic number increases by 2. Thus the resulting nucleus is
Fusion reaction takes place at high temperaturebecause [2011]
When the coulomb repulsion between the nuclei is overcome then nuclear fusion reaction takes place. This is possible when temperature is too high.
Two radioactive nuclei P and Q, in a given sample decay into a stable nucleolus R. At time t = 0, number of P species are 4 N_{0} and that of Q are N_{0}. Halflife of P (for conversion to R) is 1 minute where as that of Q is 2 minutes. Initially there are no nuclei of R present in the sample. When number of nuclei of P and Q are equal, the number of nuclei of R present in the sample would be [2011M]
Initially P → 4N_{0}
Q → N_{0}
Half life T_{P} = 1 min.
T_{Q} = 2 min.
Let after time t number of nuclei of P and Q are equal, that is
⇒ t = 4 min
at t = 4 min.
or population of R
If the nuclear radius of ^{27}Al is 3.6 Fermi, the approximate nuclear radius of ^{64}Cu in Fermi is : [2012]
The radius of the nuclears is directly proportional to cube root of atomic number i.e. R ∝ A^{1/3} ⇒ R = R_{0} A^{1/3}, where R_{0} is a constant of proportionality
where R_{1} = the radius of ^{27}Al, and A_{1} = Atomic mass number of Al R_{2} = the radius of^{ 64}Cu and A_{2} = Atomic mass number of C_{4}
A mixture consists of two radioactive materials A_{1} and A_{2} with half lives of 20 s and 10 srespectively. Initially the mixture has 40 g of A_{1 }and 160 g of A_{2}. The amount of the two in themixture will become equal after : [2012]
Let, the amount of the two in the mixture will become equal after t years.
The amount of A_{1}, which remains after t years
The amount of A_{2}, which remains, after t years
According to the problem N_{1 }= N_{2}
t = 40 s
The half life of a radioactive nucleus is 50 days. The time interval (t_{2} – t_{1}) between the time t_{2 }when 2/3 of its has decayed and the time t_{1} when 1/3 of it had decayed is : [2012]
...(i)
...(ii)
Dividing equation (i) by equation (ii)
λ (t_{2}  t_{1}) = ln2
= 50 days
A certain mass of Hydrogen is changed to Heliumby the process of fusion. The mass defect infusion reaction is 0.02866 a.m.u. The energyliberated per a.m.u. is(Given : 1 a.m.u = 931 MeV) [NEET 2013]
Mass defect Δm = 0.02866 a.m.u.
Energy = 0.02866 × 931 = 26.7 MeV
Energy liberated per a.m.u = 13.35/2 MeV
= 6.675 MeV
The half life of a radioactive isotope ‘X’ is20 years. It decays to another element ‘Y’ whichis stable. The two elements ‘X’ and ‘Y’ were foundto be in the ratio of 1 : 7 in a sample of a thegiven rock. The age of the rock is estimated tobe [NEET 2013]
The value of x is
t = 3T = 3 × 20 = 60
years Hence the estimated age of the rock is 60 years
t = 3T = 3 × 20 = 60 years
αparticles, βparticles and γrays are all havingsame energy. Their penetrating power in a givenmedium in increasing order will be [NEET Kar. 2013]
Increasing order of penetrating power : α < β < γ.
For same energy, lighter particle has higher penentrating power.
How does the binding energy per nucleon varywith the increase in the number of nucleons ? [NEET Kar. 2013]
From the graph of BE/A versus mass number A it is clear that, BE/A first increases and then decreases with increase in mass number.
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