NUCLEI QUIZ-19
Dear Readers,
JEE Advanced Physics Syllabus can be referred by the IIT aspirants to get a detailed list of all topics that are important in cracking the entrance examination. JEE Advanced syllabus for Physics has been designed in such a way that it offers very practical and application-based learning to further make it easier for students to understand every concept or topic by correlating it with day-to-day experiences. In comparison to the other two subjects, the syllabus of JEE Advanced for physics is developed in such a way so as to test the deep understanding and application of concepts.
Q1.Statement 1: Light nuclei are most stable if N=Z, while heavy nuclei are more stable if N>Z. [N→ number of neutrons, Z→ number of protons]
Statement 2: As the number of protons increases in a nucleus, the Coulomb’s repulsive, force increases, which tends to break the nucleus apart. So, to keep the nucleus stable, more number of neutrons are needed which are neutral in nature
Q2. Statement 1: 〖 _z X〗^4 undergoes 2 α-decays, 2 β-decays (negative β) and 2 γ-decays. As a result, the daughter product is 〖 _(z-2) Y〗^(A-B)
Statement 2: In α-decay, the mass number decreases by 4 unit and atomic number decreases by 2 unit. In β-decay (negative β), the mass number remains unchanged and atomic number increases by 1 unit. In γ-decay, mass number and atomic number remain unchanged
Solution
(a)
Statement II, is true by definition and correctly explains Statement I, namely, 〖 _z X〗^4 undergoes 2 α-decay, 2 β-decays (negative β) and 2 γ-decays. As a result, the daughter product is 〖 _(Z-2) X〗^(A-B)
Q3.Statement 1: Density of all the nuclei is same
Statement 2: Radius of nucleus is directly proportional to the cube root of mass number
Solution
Q4.Statement 1: Electron capture occurs more often than positron emission in heavy elements Statement 2:Heavy elements exhibit radioactivity
Solution
(b)
Electron capture occurs more often than positron emission in heavy elements. This is because if positron emission is energetically allowed, electron capture is necessarily allowed, but the reverse is not true, i.e., when electron capture is energetically allowed, positron emission is not necessarily allowed
Q5.
Statement 1: The ionisation potential of hydrogen to be 13.6 eV, the ionised potential of doubly ionized lithium is 122.4 eV.
Statement 2: Energy in nth state of hydrogen atom is E_n=-13.6/n^2
Solution
(a)
From Bohr’s theory the energy of hydrogen atom in the n^thstate is given by E_n=13.6/n^2 eV. For an atom of atomic number Z, with one electron in the outer orbit (singly ionised He or double ionised lithium) we use E_n=-(13.6Z^2)/n^2 eV,where Z is atomic number. Hence, ground state energy of doubly ionised lithium is (-13.6×9)/1^2 =-122.4 eV
Ionisation potential (potential to be applied to electron to overcome this energy) is 122.4V.
Q6.Statement 1: Heavy nuclides tend to have more number of neutrons than protons
Statement 2: In heavy nuclei, as there is coloumbic repulsion between protons, so excess of neutrons are preferable
Solution
(a)
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I
Q7.Statement 1: All nuclei are not of same size
Statement 2: Size depends on atomic mass
Solution
(a)
The radius of nucleus is given by R=R_0 A^(1/3) where R_0 is a constant =1.1×10^(-15)m. For different nuclei mass number A is different, therefore R is different
Q8. Statement 1: The mass of β-particles when they are emitted is higher than the mass of electrons obtained by other mean
Statement 2: β-particle and electron, both are similar particles
Solution
(b)
β-particles are emitted with very high velocity (up to 0.99 c). So, according to Einstein’s theory of relatively, the mass of a β-particle is much higher compared to its rest mass (m_0). The velocity of electrons obtained by other means is very small compared to c (velocity of light). So its mass remains nearly m_0. But β-paricle and electron both are similar particles
Q9. Statement 1: Cobalt-60 is useful in cancer therapy
Statement 2: Cobalt-60 is source of γ-radiations capable of killing cancerous cell
Solution
(a)
Factual
Q10. Statement 1: 4_1^1 H → _2^4 He^(2+)+2e^++26 MeV,represents fusion.
Statement 2: The above case is a β-decay.
Solution
(c)
From the reaction hydrogen is converted into helium, with the nucleus releasing two positions and energy. Because of positron emission it cannot be β-decay. The energy emitted and participation of light nuclei correspond to the fusion reaction.
