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CBSE Questions for Class 12 Medical Physics Nuclei Quiz 4 - MCQExams.com
CBSE
Class 12 Medical Physics
Nuclei
Quiz 4
The binding energy of an $$\alpha $$ -particle is
(Given that mass of proton$$=$$$$1.0073$$ amu, mass of neutron$$=$$ $$1.0087$$ amu, and mass of $$\alpha $$-particle$$=$$$$4.0015$$ amu)
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$$28.4 MeV$$
0%
$$35.6 MeV$$
0%
$$40.3 MeV$$
0%
$$44.7 MeV$$
Explanation
$$\Delta m = (2\times 1.0073 + 2\times 1.0087 - 4.0015)$$
$$= 0.00305 \mu$$
$$E = 0.0305\times 931.478 MeV$$
$$= 28.41\ MeV$$
The binding energy per nucleon of Uranium in $$MeV$$ is
(Atomic mass of uranium $$m_{a}=\ 238.0508 amu;$$
Mass of hydrogen atom is$$M_{H}=\ 1.0078 amu $$;
Mass of neutron $$m_{N}=$$ $$1.0087 amu$$; atomic number of Uranium $$Z $$$$=$$ 92; Mass Number of Uranium $$A$$ $$=$$ 238)
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$$7.580$$
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$$6.216$$
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$$5.162$$
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$$3.146$$
Explanation
No. of protons $$=$$ $$92$$
No. of neutrons $$= 238 - 92 = 146$$
$$\Delta m = [92\times 1.0078 + 146\times 1.0087 - 238.0508] amu$$
$$= 1.937 amu$$
$$BE = 1.937\times 931.478 MeV$$
$$= 1804.27 MeV$$
$$BE/per\ nucleon = \dfrac{1804.27}{238}MeV$$
$$= 7.58MeV$$
The energy required to split $$^{16}_{8}O$$ nucleus into four $$\alpha $$ particles is $$\underline{\hspace{0.5in}}$$.
(The mass of an $$\alpha $$ particle is $$4.0026.03 amu$$ and that of oxygen is $$15.994915 amu$$)
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$$11.6 MeV$$
0%
$$13.7 MeV$$
0%
$$14.4 MeV$$
0%
$$15.8 MeV$$
Explanation
$$^{16}_8O \rightarrow 4 ^4_2{He}$$
$$\Delta m = [4\times 4.002603 - 15.994915] \mu$$
$$= 0.015497 \mu$$
$$Q = \Delta m C^2$$
$$= 0.015497\times 931.478\ MeV$$
$$= 14.44 \ MeV$$
The kinetic energy of the $$\alpha $$ - particle emitted in the decay $$^{238}_{94}Pu \rightarrow ^{234}_{92}U + $$ $$^4_2He$$.
(The atomic masses of $$^{238}Pu, ^{234}U$$ and $$\alpha$$ particle are 238.04955u, 234.04095 u and 4.002603u respectively. Neglect any recoil of the nucleus)
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4.392 MeV
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6.259 MeV
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5.592 MeV
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4.952 MeV
Explanation
Assuming that energy librated by mass defect is converted into KE of $$\alpha -$$particle.
$$\Delta m = (238.04955 - 234.04095 - 4.002603) \mu$$
$$= 0.005997 \mu$$
$$KE = E = 0.005997\times 931.478\ MeV$$
$$KE = {5.586} \ MeV$$
The energy released in the following process is :
$$A + B $$ $$\rightarrow $$$$ C + D +Q$$
(mass of $$A$$ is $$1.002$$ amu ; mass of $$B$$ is $$1.004$$ amu ;mass of $$C$$ is $$1.001$$ amu ; mass of $$D$$ is $$1.003$$ amu)
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1.254 MeV
0%
0.931 MeV
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0.465 MeV
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1.862 MeV
Explanation
$$\Delta m = [1.002 + 1.004 - 1.001 - 1.003]$$
$$ = 0.002 \mu$$
$$E = Q = 0.002\times 931.478 MeV$$
$$= 1.862 MeV$$
In nuclei with mass number greater than $$20$$, the average binding energy is:
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$$8MeV$$
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$$0.8MeV$$
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$$80MeV$$
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$$0.08MeV$$
Explanation
The binding energy per nucleon Vs number of nucleon curve suggests that average binding is almost $$8\ MeV$$ for the nuclei having mass number greater than $$20$$.
The masses of neutron and proton are $$1.0087$$ and $$1.0073$$ amu respectively. If the neutrons and protons combine to form a Helium nucleus of mass $$4.0015 amu$$, the binding energy of the Helium nucleus will be :
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0%
$$28.4 MeV$$
0%
$$20.8 MeV$$
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$$27.3 MeV$$
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$$14.2 MeV$$
Explanation
Mass of neutron $$ = 1.0087 amu $$
Mass of proton $$= 1.0073 amu $$
In a helium nucleus,
protons $$ = 2 $$
neutrons$$ = 2 $$
Mass(theoritically) $$ = (2\times1.0087 + 2\times1.0073)$$
$$ =4.032 $$
Mass defect $$ = (4.032- 4.0015)\ amu $$
$$=0.0305\ amu$$
So, $$1\ amu$$ gives $$931.5\ MeV $$ of energy.
So, $$0.0305\times931.5 MeV $$ of binding energy $$ = 28.4 \ MeV$$
$$_{3}Li^{7}+_{1}H^{2}\rightarrow _{4}Be^{8}+_{o}n^{1}+Q$$
Mass of $$_{3}Li^{7}=$$ $$7.01823amu$$
Mass of $$_{1}H^{2}=$$ $$2.01474amu$$
Mass of $$_{4}Be^{8}=$$$$8.00785 amu$$
Mass of $$_{o}n^{1}=$$ $$1.00899 amu$$
Then, the value of Q is
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$$5\ MeV$$
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$$10\ MeV$$
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$$15\ MeV$$
0%
$$0$$
Explanation
Mass of reactants is: $$7.01824 + 2.01474 = 9.03297\ amu$$
Mass of products is: $$8.00785 + 1.00899 = 9.01684\ amu$$
The difference in mass $$= 0.01613\ amu$$. (This mass has been converted to energy).
$$1 amu = 1.67 \times 10^{-27} kg$$
Hence, mass difference is $$= 0.01613 \times 1.67 \times 10^{-27} kg = 2.69371 \times 10^{-29} kg$$
Using Einstein's relation:
$$E = mc^{2}$$
$$E = 2.69371 \times 10^{-21} \times (3 \times 10^{8})^{2} = 2.42 \times 10^{-12} J$$
$$E = 15\ MeV$$
The mass defect and binding energy of $$^{12}_{6}$$C nucleus is
(Mass of $$^{12}_{6}$$ C $$=$$12.000000 amu; $$m_{p}=$$1.007825 amu and $$m_{n}=$$1.008665 amu)
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$$0.06522$$ amu; $$72.1$$ MeV
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$$0.09894$$ amu; $$92.1$$ MeV
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$$0.05315$$ amu; $$102.2$$ MeV
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$$0.05315$$ amu; $$82.2$$ MeV
Explanation
C $$=$$ 6 neutrons + 6 protons
$$\Delta m = (6\times 1.007825 + 6\times 1.008665 - 12)$$
$$= 0.09894 \ amu$$
$$E = (0.09894\times 931.478 \dfrac{MeV}{C^2})C^2$$
$$= 92.1 MeV$$
The binding energy per nucleon of $$C^{12}$$ is $$7.68$$ MeV and that of $$C^{13}$$ is $$7.47$$ Mev. The energy required to remove one neutron from $$C^{13}$$ is
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$$495\ MeV$$
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$$49.5\ MeV$$
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$$4.95\ MeV$$
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$$0.495\ MeV$$
Explanation
$$C^{13} \rightarrow C^{12}_1 + ^1n$$
$${BE)_C}^{12} = 12\times 7.68$$$$= 92.16\ MeV$$
$${BE)_C}^{13} = 13\times 7.47$$$$= 97.11\ MeV$$
$$Q = {BE)_C}^{12} - {BE)_C}^{12}$$
$$= 97.11 - 92.16$$
$$={4.95\ MeV}$$
True masses of neutron,proton and deutron in $$a.m.u$$ are $$1.00893, 1.00813$$ and $$2.01473$$ respectively. The packing fraction of the deutron in $$a.m.u$$ is
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0%
$$11.56 \times 10^{-4}$$
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$$23.5 \times 10^{-4}$$
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$$33.5 \times 10^{-4}$$
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$$47.15 \times 10^{-4}$$
Explanation
Deutron $$=\ ^{2}_{1}H$$ i.e., $$1$$ proton and $$1$$ neutron.
Packing fraction $$= \dfrac{True \ mass - Mass \ number}{Mass\ number}$$
$$ = \dfrac{(1.00893 + 1.00813) - 2.01473}{2.01473} $$
$$ = 11.56 \times10^{-4} $$
The atomic mass of $$_{7}N^{15}$$ is $$15.000108 amu$$ and that of $$_{8}O^{16}$$ is $$15.994915 amu$$. The minimum energy required to remove the least tightly bound proton is (mass of proton is $$1.007825 amu)$$
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$$0.01 3018 amu$$
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$$12.13 MeV$$
0%
$$13.018 MeV$$
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$$12.13 eV$$
Explanation
$$_8^{16}O \rightarrow ^{15}_7N + ^1_1P$$
$$\Delta m = [15.000108 + 1.007825 - 15.994915]$$
$$= 0.013018 amu$$
$$\Delta E = 0.013018\times 931.478$$ MeV
$$={12.13\ MeV}$$
If $$2g$$ of hydrogen is converted into $$1.986 gm$$ Helium in a thermonuclear reaction, the energy released is
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63 X 10$$^{7}J$$
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63 X 10$$^{10}J$$
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126 X 10$$^{10}J$$
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6.3 X 10$$^{20}J$$
Explanation
$$\Delta m = 2 - 1.986$$ $$= {0.014 gm}$$
$$\Delta E = (\dfrac{0.014}{1000}Kg)\times (3\times 10^8)^2$$$$= 126\times 10^{10} J$$
$$_{7}N^{14} +_{2}He^{4}\rightarrow $$ X+$$_{1}H^{1}$$ ; X is
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$$_{9}F^{18}$$
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$$_{10}Ne^{18}$$
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$$_{8}O^{17}$$
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$$_{6}C^{15}$$
Explanation
$$_7N^{14} + _2He^4 \rightarrow _Z^A{X} + _1H^1$$
Change balance: $$7 + 2 = Z + 1$$
$$Z = 8$$
Mass balance: $$14 + 4 = A + 1$$
$$A = 17$$
Bombardment of lithium with proton gives rise to the following reaction :
$$^{7}_{3}Li+^{1}_{1} H \rightarrow $$2($$^{4}_{2}He) + Q $$
The Q-value is
(atomic masses of lithium, proton and helium are $$7.016$$ amu, $$1.008$$ amu and $$4.004$$ amu respectively)
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$$14.904$$ MeV
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$$16.774$$ MeV
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$$10.634$$ MeV
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$$18.633$$ MeV
Explanation
$$^7_3{Li} + ^1_1H \rightarrow 2 ^4_2{He} + Q$$
Mass defect $$= (7.016 + 1.008 - 2\times 4.004)\mu$$$$= 0.016 amu$$
$$1amu = 931.478 \dfrac{MeV}{C^2}$$
$$E = 0.016\times 931.478 \dfrac{MeV}{C^2}\times C^2$$$$=14.903 MeV$$
$$4.6\times 10^{22}$$ atoms of an element weight $$13.8$$ g. What is the atomic mass of the element?
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290 u
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180.6 u
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34.4 u
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104 u
Explanation
$$1$$ mole of any substance contains $$6.02 \times 10^{23}$$ atoms.
Thus, $$4.6 \times 10^{22}$$ atoms corresponds to $$\dfrac{4.6 \times 10^{22}}{6.022 \times 10 ^{23}}=0.0764$$ moles.
$$0.0764$$ moles weighs $$13.8$$ g.
Thus, $$1$$ mole will weigh $$\dfrac{13.8}{0.0764}= 180.6 \ g$$.
Hence, the atomic mass of the element will be 180.6 u.
A $$\gamma $$ -ray photon creates an electron-positron pair. The rest mass equivalent of electron is 0.5 MeV. KE of the electron - positron system is 0.78 MeV. Then the energy of $$\gamma $$ -ray photon is
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$$0.28 MeV$$
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$$1.78 MeV$$
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$$1.28 MeV$$
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$$0.14 MeV$$
Explanation
The rest mass of electron is $$= 0.5\ MeV$$
As, the momentum is conserved the rest mass of positron should also be $$ 0.5\ MeV$$.
The kinetic energy of the pair is $$0.78\ MeV$$.
$$\therefore$$ The energy of $$\gamma$$ ray photon is $$= (0.5+0.5+0.78) \ MeV= 1.78 \ MeV$$
Assertion (A) : All the radioactive elements are ultimately converted into lead
Reason (R): All the elements above lead are unstable
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Both A and R are true and R is correct explanation of A
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Both A and R are true and R is not correct explanation of A
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A is true but R is false
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A is false but R is true
Explanation
For all $$4n, 4n+2$$ series elements, on emission of different kind of rays, they form different isotopes of lead and also all element above lead are unstable .
But, $$4n+1$$ series elements (Neptunium series), end in $$^{209}_{83}Bi $$ (Bismuth not lead)
Assertion (A): Binding energy per nucleon is the measure of the stability of the nucleus.
Reason (R): Binding energy per nucleon is more for heavier nuclides.
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Both A and R are true and R is correct explanation of A.
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Both A and R are true and R is not the correct explanation of A.
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A is true but R is false
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A is false but R is true
Explanation
The binding energy is the energy that is released when a nucleus is assembled from its constituent nucleons Hence, the energy equivalent of the mass-defect is called the binding-energy of the nucleus
.
The heavier the nucleus, the greater the internal repulsive forces due to the greater number of protons and less energy must be applied to remove a nucleon from the nucleus, hence the binding energy is lower. Thus for lighter nuclei binding energy is more. The greater the binding energy, the more stable the nucleus is.
In the nuclear fusion reaction : $$^{2}_{1}H+ ^{3}_{1}H \rightarrow $$ $$^{4}_{2}He+n $$ , given that the repulsive potential energy between the two nuclei is $$\sim 7.7 \times 10^{-14}$$J, the temperature to which the gases must be heated to initiate the reaction is nearly
(Boltzmann's constant $$k=$$1.38 x 10$$^{-23}$$J)
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$$10^{7} K$$
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$$10^{5} K$$
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$$10^{3} K$$
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$$10^{9}K$$
Explanation
The repulsive potential energy $$=\dfrac{3}{2}kT$$
$$7.7\times 10^{-14}=\dfrac{3}{2}\times 1.38\times\times 10^{-23} \times T$$
$$\rightarrow \ T = 3.7\times 10^9 K \sim 10^9 K$$
Assertion (A) : If a heavy nucleus is split into two medium sized parts,each of the new nuclei will have more binding energy per nucleon than the original nucleus.
Reason (R): Joining two light nuclei together to give a single nucleus of medium size means more binding energy per nucleon in the new nucleus.
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0%
Both A and R are true and R is correct explanation of A
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Both A and R are true and R is not correct explanation of A
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A is true but R is false
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A is false but R is true
Explanation
If a heavy nucleus splits, it forms smaller medium size particles, which are more stable as compared to parent nuclei thus having higher
Binding energy. And this is the principle for nuclear fission and in case of nuclear fusion the two light nuclei join to form a more stable medium sized nuclei to get the optimum n/p ratio as close and near to Fe which has most stable nuclei in terms of n/p ratio.
A nuclear transformation is denoted by X(n,$$\alpha $$)$$\rightarrow _{3} Li^{7}$$. The nucleus of element X is
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$$_{6}C^{12}$$
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$$_{5}B^{10}$$
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$$_{5}B^{9}$$
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$$_{4}Be^{11}$$
Explanation
The notation means that on being bombarded with neutrons the element breaks into Lithium and an $$\alpha-particle.$$
$$^{Z}_{A}X + ^{1}_{0}n \Rightarrow ^{Z-3}_{A-2}Y + ^{4}_{2}He$$
Now,
$$^{Z-3}_{A-2}Y = ^{7}_{3}Li$$
$$\therefore Z = 10 , A= 5 $$
So, The element is Boron and X is $$_{5}B^{10} $$
Assertion (A) : Isotopes of an element can be separated by using a mass spectrometer.
Reason (R) : Separation of isotopes is possible due to difference in electron number of isotopes.
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0%
Both A and R are true and R is correct explanation of A
0%
Both A and R are true and R is not correct explanation of A
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A is true but R is false
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A is false but R is true
Explanation
This is fact that isotopes can be separated using mass spectrometer and 2nd statement is false because in isotopes the number of neutrons differs not electrons.
Assertion (A): Nuclear fusion reactions are considered as thermo-nuclear reactions
Reason (R): The source of stellar energy is nuclear fusion
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0%
Both A & R are true and R is the correct explanation of A
0%
Both A & R are true and R is not correct explanation of A
0%
A is true but R is false
0%
A is false but R is true
Explanation
Nuclear fusion reaction are consider as thermo-nuclear reaction because for the nuclear fusion reaction to start we have to provide lot of temperature and heat energy to overcome the potential energy before fusion and the source of energy of stars is nothing but nuclear energy.
Consider the following statements (A) and (B) and identify the correct answer given below :
Statement A: Positive values of packing fraction implies a large value of binding energy
Statement B: The difference between the mass of the nucleus and the mass number of the nucleus is called the packing fraction
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A and B are correct
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A and B are false
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A is true, B is false
0%
A is false, B is true
Explanation
Packing fraction $$= \dfrac{Actual \ isotopic \ mass - mass \ number}{ Mass\ number}$$
A negative packing fraction indicates stability of the nucleus thus a large Binding energy and vice versa.
So, we easily see that statement A and b are false.
When the mass of an electron becomes equal to thrice its rest mass, its speed is
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0%
$$\dfrac{2\sqrt{2}}{3}c$$
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$$\dfrac{2}{3}c$$
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$$\dfrac{1}{3}c$$
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$$\dfrac{1}{4}c$$
Explanation
The mass of a moving body $$m$$ is given by special theory of relativity is given by:
$$m=\dfrac{m_0}{\sqrt{1-\dfrac{v^2}{c^2}}}$$,
where, $$m_0$$ is the rest mass of the body.
According to the question, the mass of an electron becomes equal to thrice its rest mass,
So, $$m=3m_0$$.
Using this, the speed of the moving body will be: $$v=\dfrac{2\sqrt {2}}{3} c$$
An electron moves with a speed of $$\dfrac{\sqrt{3}}{2}c$$. Then its mass becomes_____ times its rest mass. (Given velocity of light $$=c$$)
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$$2$$
0%
$$3$$
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$$3/2$$
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$$4$$
Explanation
Mass of on electron moving with speed $$V$$ is $$\dfrac{m_o}{\sqrt{1-\dfrac{v^{2}}{c^{2}}}}$$
where, $$m_o$$ is rest mass
$$V$$ is speed of electron
$$C$$ is speed of light
So, $$m$$ $$=\dfrac{m_o}{\sqrt{1-\dfrac{v^{2}}{c^{2}}}}$$
$$=\dfrac{m_o}{\sqrt{1-\dfrac{\left ( \frac{\sqrt{3C}}{2} \right )^{2}}{C^{2}}}}$$
$$=\dfrac{m_o}{\sqrt{1-\dfrac{3}{4}}}$$
$$=\dfrac{m_o}{\sqrt{\frac{1}{4}}}$$
$$=\dfrac{m_o}{\dfrac{1}{2}}$$
$$=2m_o$$
So, mass becomes 2 times of its rest mass.
To obtain an isotope of a given radioactive atom, the atom must emit
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one alpha and one beta particle
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one alpha and two beta particle
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two alpha and two beta particle
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three alpha and four beta particle
Explanation
Emission of-particle decreases the atomic number of the product by two and each $$\beta$$ - emission increases the atomic number by one. So the atomic number remains the same and the mass number is different. after emitting one alpha and 2 beta.
The atomic mass of $$_{8}O^{16}$$ is $$15.9949 amu$$. Mass of one neutron and one proton is $$2.016490 amu$$ and the mass of an electron is $$0.00055 amu$$. The binding energy per nucleon of oxygen atom is
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$$7.97 MeV$$
0%
$$11.5 MeV$$
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$$22.8 MeV$$
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$$82.3 MeV$$
Explanation
The number of proton $$p=8$$ and number of neutron, $$n=16-8=8 $$
Total number of nucleon $$N=16$$
Combined mass of nuclei $$= $$ total mass of proton + total mass of neutron $$=8\times 2.016490=16.13192 $$ amu
Mass defect $$\Delta m =16.13192-15.9949=0.13702 $$ amu
Binding energy per nucleon $$=\Delta m c^2/N=(0.137\times 931.49)/16= 127.6141/16=7.975MeV$$ where $$ 1 amu =931.49/c^2 MeV$$
The mass of a $$^7_3Li $$ nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of
$$^7_3Li $$
nucleus is nearly
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0%
46 MeV
0%
5.4 MeV
0%
3.9 MeV
0%
23 MeV
Explanation
$$\begin{aligned}\Delta m &=0.042 \\&=0.042+2\times 1.66 \times 10^{-27}\mathrm{~kg} \\\text { Binding energy }&=\Delta \mathrm{m} c^{2} \\\text { Binding energy per nucleon } \\=& \frac{\Delta m c^{2}}{7} \\&[\operatorname{mass number} \text { of } l i=7]\end{aligned}$$
$$\begin{array}{l}=\frac{0.042 \times 1.66\times 10^{-27} \times\left(3 \times 10^{8}\right)^{2}}{7} \mathrm{~J} \\\text { To convert the energy in ev from } \\\text { joule, we will divide it by } 1.6 \times 10^{-19} \\=\frac{0.042 \times 1.66 \times 10^{-27} \times\left(3 \times 10^{8}\right)^{2}}{7 \times 1.6 \times 10^{-19}}\mathrm{ev} \\=5.4 \times 10^{6}\mathrm{ev} \\\therefore \text { Energy }\operatorname{lost} =5.4\mathrm{Mev}\end{array}$$
The mass of proton is 1.0073 u and that of neutron is 1.0087 u (u $$=$$ atomic mass unit). The binding energy of $$ ^{4}_{2}He$$ is (Given : helium
nucleus mass $$\approx $$ 4.0015 u)
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0.0305 J
0%
0.0305 erg
0%
28.4 MeV
0%
0.061 u
Explanation
$$BE=\Delta m\times 931$$
$$=[2(1.0087+1.0073)-4.0015]\times 931$$
$$=28.4 MeV$$
Which of the following is true for the following isotopes of uranium
U$$^{235}$$ and U$$^{238}$$?
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both contain the same number of neutrons
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both contain the same number of protons, electrons and neutrons
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both contain the same number of protons and electrons but U$$^{238}$$ contains three more neutrons than U$$^{235}$$
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U$$^{238}$$ contain three less neutrons than U$$^{235}$$
Explanation
Natural Uranium contains $$3$$ radioactive isotopes $${ U }^{ 234 },{ U }^{ 235 }$$ and $${ U }^{ 238 }$$
$${ U }^{ 238 }$$ has a mass number
(proton $$92+$$neutron $$146$$) $$=238$$
$${ U }^{ 235 }$$ has mass number (proton $$92+$$ neutron$$143$$) $$=235$$
Hence we can say that both contain the same number of protons and electrons but $${ U }^{ 238 }$$ contains three more neutrons than $${ U }^{ 235 }$$
In the process of nuclear fusion
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only heavy nucleus break into light nuclei
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fusion of light nuclei at normal temperature takes place
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fusion of light nuclei at high pressure and low temperature
takes place
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fusion of light nuclei at high pressure and high temperature
takes place
Explanation
Nuclear fusion is a reaction in which two or more atomic nuclei come close enough to form one or more different atomic nuclei and subatomic particles (neutrons or protons)
Inside the sun fusion reaction take place at very high temperature and enormous gravitational pressure.
When four hydrogen nuclei fuse together to form a helium nucleus, then in this process
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energy is absorbed.
0%
energy is liberated.
0%
absorption and liberation of energy depends upon the temperature.
0%
energy is neither liberated nor absorbed.
Explanation
In the basic hydrogen fusion cycle four hydrogen nuclei (photons) come together to make a helium nucleus. This fusion cycle releases energy in the core of the star i.e., energy is liberated in this process
The binding energy per nucleon in deutorium 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 -
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0%
2.2 MeV
0%
28.0 MeV
0%
30.2 MeV
0%
23.6 MeV
For a nuclear fusion process, suitable nuclei are
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Any Nuclei
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Heavy Nuclei
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Light Nuclei
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Nuclei lying in the middle of Periodic table
Explanation
The process in which two or more light, nuclei are combined into a single nucleus with the release of tremendous amount of energy is called as nuclear fusion
Hence light nuclei are suitable for the nuclear fusion process
The heavier nuclei tend to have larger N/Z ratio because
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a neutron is heavier than a proton
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a neutron is an unstable particle
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a neutron does not exert electric repulsion
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Coulomb force has longer range as compared to the nuclear force
Atoms of same element having same atomic number but different mass number is
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Isobars
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Isotone
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Isotope
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None of these
Explanation
$$\begin{array}{l}\text { Atoms having same atomic number but different mass } \\\text { number is called Isotope. } \\\therefore \text { Isotope is correct }\end{array}$$
Principle of Hydrogen bomb is .................. reactions.
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Uncontrolled fusion
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Fission
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Controlled fusion
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None of these
Explanation
A hydrogen bomb is based on the principle of uncontrollable nuclear fusion . Nuclear fusion is a process where nuclei of two light atoms combine to form a new nucleus.
Hence, the correct option is A.
One a.m.u. or one 'u' is equal to :
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$$1.66053892 10^{-27} kg$$
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$$1.66053892 10^{-29} kg$$
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$$1.23457656 10^{-27} kg$$
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$$1.23457656 10^{-29} kg$$
Explanation
One a.m.u.(atomic mass unit) or one 'u' is
$$\dfrac{1}{12}$$ of the mass of one carbon-$$12$$ atom.
It is equal
$$1.6605389210 \times 10^{-24} g $$ or $$1.6605389210 \times 10^{-27} kg$$.
The modern atomic mass unit is based on the mass of :
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C-12 isotope
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hydrogen
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oxygen
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nitrogen
Explanation
One a.m.u. or one 'u' is equal to $$1.66053892 10^{-24}\ g $$ or $$1.66053892 10^{-27}\ kg$$. It is equal to $$\dfrac{1}{12}$$ of the mass of an atom of carbon-12.
It is used as a standard. The masses of all other atoms are determined relative to the mass of an atom of carbon-12.
The binding energies energy per nucleon for $$C^{12}$$ is $$7.68\ MeV$$ and that for $$C^{13}$$ is $$7.5\ MeV.$$ The energy required to remove a neutron from $$C^{13}$$ is
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$$5.34\ MeV$$
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$$5.5\ MeV$$
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$$9.5\ MeV$$
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$$9.34\ MeV$$
Explanation
Total binding energy of $$C^{13}$$ is $$E_1=7.5\times 13=97.5\ MeV$$
Total binding energy of $$C^{12}$$ is $$E_2=7.68\times 12=92.16\ MeV$$
The energy required to remove a neutron is $$E_1-E_2=5.34\ MeV$$
What is the mass of one atom of $$C-12$$ in grams?
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1.992$$\displaystyle \times 10^{-23}$$ gm
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1.989$$\displaystyle \times 10^{-23}$$ gm
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1.892$$\displaystyle \times 10^{-23}$$ gm
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1.965$$\displaystyle \times 10^{-23}$$ gm
Explanation
Mass of $$1$$ mole = $$12\,gm$$
Mass of $$6.022 \times 10^{23}\,atom = 12\,gm$$
mass of $$1$$ atom = $$\dfrac{12}{6.023 \times 10^{23}} = 1.993 \times 10^{-23}\,gm$$
Find the uncertainty in mass.
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$$1.4\times10^{-35}$$ $$kg$$
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$$2.6\times10^{-11}$$ $$kg$$
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$$2.6\times10^{-13}$$ $$kg$$
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$$8\times10^{-14}$$ $$kg$$
Explanation
$$\triangle E.\triangle t$$ = $$h/2\pi$$
$$\triangle E$$$$=\displaystyle\ \frac{h}{2\pi \triangle t}$$ = $$\displaystyle\ \frac
{6.62\times10^{-34}}{2\pi\times8.4\times10^{-17}}$$$$=1.24\times10^{-18}$$
$$\triangle mc^{2}$$ = $$\triangle E$$
Uncertainty in mass is $$\triangle m$$ $$=\displaystyle\ \frac{1.24\times10^{-18}}{9\times10^{16}}$$$$=1.38\times10^{-35}\approx 1.4\times10^{-35}$$ $$kg$$
Isotopes are the atoms of the same element which contain equal number of
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nucleons
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neutrons
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protons
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neutrons and protons
Explanation
Isotopes of the element are atoms of the element that have the same number of protons, electrons but different number of neutrons.
Option C is correct.
Find the binding energy of $$_{28}^{62}\textrm{Ni}$$, Given $$m_{H}$$ = $$1.008$$ $$u$$, $$m_{n}$$ = $$1.0087$$ $$u$$, $$_{28}^{62}\textrm{m}$$ = $$62.9237$$ $$u$$
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$$545.3$$ $$MeV$$
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$$595.3$$ $$MeV$$
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$$645.3$$ $$MeV$$
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$$695.3$$ $$MeV$$
Explanation
Number of protons in $$_{28}^{62}Ni$$, $$p=28$$
Number of neutrons in $$_{28}^{62}Ni$$, $$n=28$$
Given: $$m_{H}=1.008u , m_{n}=1.0087u$$ ,
If $$_{28}^{62}m=61.9237u$$
Then mass defect,
$$\Delta m=28m_{H}+34m_{n}-_{28}^{34}m$$
$$\Delta m=28\times1.008+34\times1.0087-61.9237$$
$$\Delta m=0.5961u$$
Hence binding energy,
$$B.E.=\Delta m\times931$$
$$B.E.=0.5961\times930=554.3\ MeV$$
The kinetic energy of $$\alpha$$-particles at a distance $$5 \times 10^{-14}$$m from the nucleus will be(in Joules)
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$$6.4 \times 10^{-13}$$
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$$4.3 \times 10^{-13}$$
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$$2.1 \times 10^{-13}$$
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$$3.4 \times 10^{-14}$$
Explanation
The charge on an alpha particle is $$+2$$.
Hence $$q = 2e = 3.2 \times 10^{-19} C$$
This alpha particle is accelerated by a potential of $$2 \times 10^{6} V$$
Hence Energy of the particle $$ = qV = 3.2 \times 10^{-19} \times 2 \times 10^{6} = 6.4 \times 10^{-13} J$$
This is the total energy.
Potential energy when it is at a distance of $$5 \times 10^{-14} m $$ from the nucleus
$$PE = \dfrac{1}{4\pi \epsilon_{0}}\dfrac{q_{1}q_{2}}{d}$$
Nucleus contains 47 protons.
Hence, $$q_{2} = 47e = 47 \times 1.6 \times 10^{-19} C$$
Substituting these values:
We get PE = $$ 4.3 \times 10^{-13} J$$
$$KE + PE = TE$$
$$KE = TE - PE = (6.4 - 4.3) \times 10^{-13} J = 2.1 \times 10^{-13}J$$
Find the fraction of the mass.
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$$5.28\times10^{-6}$$
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$$5. 84\times10^{-7}$$
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$$5.28\times10^{-8}$$
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$$7.8\times10^{-9}$$
Explanation
Given : $$\Delta t = 8.4\times 10^{-17}$$ s
As we know the relation $$\triangle E.\triangle t$$ $$=\dfrac{h}{2\pi}$$
$$\therefore $$ $$\triangle E$$$$=\displaystyle\ \frac{h}{2\pi \triangle t}$$ = $$\displaystyle\ \frac
{6.62\times10^{-34}}{2\pi\times8.4\times10^{-17}}=1.24\times10^{-18}$$ J
From mass-energy equivalance relation : $$\triangle mc^{2}$$ = $$\triangle E$$
Uncertainty in mass $$\triangle m$$ $$ = \dfrac{\Delta E}{c^2} =\displaystyle\ \frac{1.24\times10^{-18}}{(3\times10^{8})^2}=1.38\times10^{-35}\approx 1.4\times10^{-35}$$ $$kg$$
Mass of neutral pion $$m_{pion} = 264 m_{electron} = 264\times 9.1\times 10^{-31}$$ kg
$$\therefore$$ Fraction of mass $$\displaystyle\ \frac{\triangle m}{m_{pion}}$$ = $$\displaystyle\
\frac{1.38\times10^{-35}}{264\times9\times10^{-31}}=5.28\times10^{-8}$$
The average $$\ KE $$ of molecules in a gas at temperature $$\ T $$ is $$\displaystyle \frac {3}{2}kT $$. Find the temperature at which the average KE of molecules equal to binding energy of its atoms.
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$$\ 1.05 \times 10^{5}K $$
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$$\ 1.05 \times 10^{4}K $$
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$$\ 1.05 \times 10^{3}K $$
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none of these
Explanation
Binding energy of atoms $$=13.6\ eV=2.1789\times 10^{-18}J$$
Now since this energy is equal to $$\dfrac{3}{2}kT$$
$$T=\dfrac{2}{3}\dfrac{2.1789\times10^{-18}}{k}=1.05\times 10^5K$$
As the mass number $$A$$ increases, the binding energy per nucleon in a nucleus
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increases
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decreases.
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remains the same.
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varies in a way that depends on the actual value of $$A$$.
Explanation
From binding energy curve,
The binding energy per nucleon varies with number of nucleons, A.
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