46: The Nucleus : Neet-xii-physics

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NEET-XII-Physics

46: The Nucleus

with Solutions - page 2

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Qstn #12
If three helium nuclei combine to form a carbon nucleus, energy is liberated. Why can’t helium nuclei combine on their own and minimise the energy?
Ans : When three helium nuclei combine to form a carbon nucleus, energy is liberated. This energy is greater than the that liberated when these nuclei combine on their own. Hence, formation of carbon nucleus leads to much more stability as compared to the combination of three helium nuclei.
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Section : ii

Qstn #1
The mass of a neutral carbon atom in ground state is
(a) exact 12 u
(b) less than 12 u
(c) more than 12 u
(d) depends on the form of carbon such as graphite of charcoal.
digAnsr: a
Ans : (a) exact 12 u
In nuclear physics, a unit used for measurement of mass is unified atomic mass unit, which is denoted by u.
It is defined such that
1 u = `` \frac{1}{12}\times `` (Mass of neutral carbon atom in its ground state)
Mass of neutral carbon atom in its ground state = 12 × 1 u = 12 u
Thus, the mass of neutral carbon atom in its ground state is exactly 12 u.
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Qstn #2
The mass number of a nucleus is equal to
(a) the number of neutrons in the nucleus
(b) the number of protons in the nucleus
(c) the number of nucleons in the nucleus
(d) none of them.
digAnsr: c
Ans : (c) the number of nucleons in the nucleus
Mass number of a nucleus is defined as the sum of the number of neutron and protons present in the nucleus, i.e. the number of nucleons in the nucleus.
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Qstn #3
As compared to ¹²C atom, ¹⁴C atom has
(a) two extra protons and two extra electrons
(b) two extra protons but no extra electrons
(c) two extra neutrons and no extra electron
(d) two extra neutrons and two extra electron.
digAnsr: c
Ans : (c) two extra neutrons and no extra electron
¹²C and ¹⁴C are the two isotopes of carbon atom that have same atomic number, but different mass numbers. This means that they have same number of protons and electrons, but different number of neutrons. Therefore, ¹²C has 6 protons, 6 electrons and 6 neutrons, whereas ¹⁴C has 6 electrons, 6 protons and 8 neutrons.
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Qstn #4
The mass number of a nucleus is
(a) always less than its atomic number
(b) always more than its atomic number
(c) equal to its atomic number
(d) sometimes more than and sometimes equal to its atomic number.
digAnsr: d
Ans : (d) sometimes more than and sometimes equal to its atomic number
Mass number of a nucleus is defined as the sum of the number of neutron and protons present in the nucleus, i.e. the number of nucleons in the nucleus, whereas atomic number is equal to the number of protons present. Therefore, the atomic number is smaller than the mass number. But in the nucleus (like that of hydrogen ¹H₁), only protons are present. Due to this, the mass number is equal to the atomic number.
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Qstn #5
The graph of ln(R/R₀) versus ln A(R = radius of a nucleus and A = its mass number) is
(a) a straight line
(b) a parabola
(c) an ellipse
(d) none of them.
digAnsr: a
Ans : (a) a straight line
The average nuclear radius (R) and the mass number of the element (A) has the following relation:
`` R={R}_{o}{A}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}``
`` \raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{${R}_{o}$}\right.={A}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}``
`` \,\mathrm{\,ln\,}\left(\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{${R}_{o}$}\right.\right)=\frac{1}{3}\,\mathrm{\,ln\,}A``
`` ``
Therefore, the graph of ln(R/R0) versus ln A is a straight line passing through the origin with slope 1/3.
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Qstn #6
Let F_pp, F_pn and F_nn denote the magnitudes of the net force by a proton on a proton, by a proton on a neutron and by a neutron on a neutron respectively. neglect gravitational force. When the separation is 1 fm.
(a) F_pp > F_pn = F_nn
(b) F_pp = F_pn = F_nn
(c) F_pp > F_pn > F_nn
(d) F_pp < F_pn = F_nn
digAnsr: d
Ans : (d) Fpp < Fpn = Fnn
Protons and neutrons are present inside the nucleus and they exert strong attractive nuclear forces on each other, which are equal in magnitude. Due to their positive charge, protons repel each other. Hence the net attractive force between two protons gets reduced, but the nuclear force is stronger than the electrostatic force at a separation of 1 fm.
∴ Fpp < Fpn = Fnn
Here, Fpp, Fpn and Fnn denote the magnitudes of the net force by a proton on a proton, by a proton on a neutron and by a neutron on a neutron, respectively.
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Qstn #7
Let F_pp, F_pn and F_nn denote the magnitudes of the nuclear force by a proton on a proton, by a proton on a neutron and by a neutron on a neutron respectively. When the separation is 1 fm,
(a) F_pp > F_pn = F_nn
(b) F_pp = F_pn = F_nn
(c) F_pp > F_pn > F_nn
(d) F_pp < F_pn = F_nn
digAnsr: b
Ans : (b) F_pp = F_pn = F_nn
Protons and neutrons are present inside the nucleus and they exert strong attractive nuclear force on each other. These forces are equal in magnitude, irrespective of the charge present on the nucleons.
∴ F_pp = F_pn = F_nn
Here, F_pp, F_pn and F_nn denote the magnitudes of the nuclear force by a proton on a proton, by a proton on a neutron and by a neutron on a neutron, respectively.
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Qstn #8
Two protons are kept at a separation of 10 nm. Let F_n and F_e be the nuclear force and the electromagnetic force between them.
(a) F_e = F_n
(b) F_e >> F_n
(c) F_e << F_n
(d) F_e and F_n differ only slightly.
digAnsr: b
Ans : (b) F_e >> F_n
Two protons exert strong attractive nuclear force and repulsive electrostatic force on each other. Nuclear forces are short range forces existing in the range of a few fms. Therefore, at a separation of 10 nm, the electromagnetic force is greater than the nuclear force, i.e. F_e >> F_n.
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Qstn #9
As the mass number A increases, the binding energy per nucleon in a nucleus
(a) increases
(b) decreases
(c) remains the same
(d) varies in a way that depends on the actual value of A.
digAnsr: d
Ans : (d) varies in a way that depends on the actual value of A
Binding energy per nucleon in a nucleus first increases with increasing mass number (A) and reaches a maximum of 8.7 MeV for A (50-80). Then, again it slowly starts decreasing with the increase in A and drops to the value of 7.5 MeV.
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Qstn #10
Which of the following is a wrong description of binding energy of a nucleus?
(a) It is the energy required to break a nucleus into its constituent nucleons.
(b) It is the energy made available when free nucleons combine to form a nucleus.
(c) It is the sum of the rest mass energies of its nucleons minus the rest mass energy of the nucleus.
(d) It is the sum of the kinetic energy of all the nucleons in the nucleus.
digAnsr: d
Ans : (d) It is the sum of the kinetic energies of all the nucleons present in the nucleus.
Binding energy of a nucleus is defined as the energy required to break the nucleus into its constituents. It is also measured as the Q-value of the breaking of nucleus, i.e. the difference between the rest energies of reactants (nucleus) and the products (nucleons) or the difference between the kinetic energies of the products and the reactants.
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Qstn #11
In one average-life,
(a) half the active nuclei decay
(b) less than half the active nuclei decay
(c) more than half the active nuclei decay
(d) all the nuclei decay.
digAnsr: c
Ans : (c) more than half the active nuclei decay
The average life is the mean life time for a nuclei to decay.
It is given as
`` \tau =\frac{1}{\lambda }=\frac{{{\rm T}}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{0.693}``
Here, `` \tau ,\lambda \,\mathrm{\,and\,}{{\rm T}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}`` are the average life, decay constant and half life-time of the active nuclei, respectively. The value of the average lifetime comes to be more than the average lifetime. Therefore, in one average life, more than half the active nuclei decay.
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Qstn #12
In a radioactive decay, neither the atomic number nor the mass number changes. Which of the following particles is emitted in the decay?
(a) Proton
(b) Neutron
(c) Electron
(d) Photon
digAnsr: d
Ans : (d) Photon
The atomic number and mass number of a nucleus is defined as the number of protons and the sum of the number of protons and neutrons present in the nucleus, respectively. Since in the decay, neither the atomic number nor the mass number change, it cannot be a beta-decay (release of electron, proton or neutron). Hence, the particle emitted can only be a photon.
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Qstn #13
During a negative beta decay,
(a) an atomic electron is ejected
(b) an electron which is already present within the nucleus is ejected
(c) a neutron in the nucleus decays emitting an electron
(d) a proton in the nucleus decays emitting an electron.
digAnsr: c
Ans : (c) a neutron in the nucleus decays emitting an electron
Negative beta decay is given as
`` n\to p+{e}^{-}+\stackrel{¯}{\nu }``
Neutron decays to produce proton, electron and anti-neutrino.
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