## NUCLEI QUIZ-6

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.

Q1The initial activity of a certain radioactive isotope was measured as 16000 counts min┴(-1)⁡ . Given that the only activity measured was due to this isotope and that its activity after 12 h was 2100 counts min┴(-1)⁡ , its half-life, in hours, is nearest to [Given log_e⁡〖(7.2)=2)〗]
•  9.0
•  6.0
•  4.0
•  3.0

Q2. Some radioactive nucleus may emit
•   Only one α,β or γ at a time
•   All the three α,β and γ one after another
•   All the three α,β and γ simultaneously
•   Only α and β simultaneously
Solution (a) No radioactive substance emits both α and β particles simultaneously. Some substances emit α-particles and some other emits β-particles, γ-rays are emitted along with both α and β-particles

Q3A radioactive sample consists of two distinct species having equal number of atoms initially. The mean lifetime of one species is τ and that of the other is 5τ. The decay products in both cases are stable. A plot is made of the total number of radioactive nuclei as a function of time. Which of the following figures best represents the form of this plot?

Q4 Atomic mass number of an element is 232 and its atomic number is 90. The end product of this radioactive element is an isotope of lead (atomic mass 208 and atomic number 82). The number of α- and β-particles emitted are
•   α=3,β=3
•   α=6,β=4
•   α=6,β=0
•   α=4,β=6
Solution (b) Number of α-particles emitted =(232-208)/4=6 Decrease in charge number due to α-emission = 12 But actual decrease in charge number =90-82=8 Clearly, four β-particles are emitted

Q5. The compound unstable nucleus _92^236 U often decays in accordance with the following reaction _92^236 U→ _54^140 Xe+ _38^94 Sr+ order particles In the nuclear reaction presented above, the ‘other particle’ might be

•   An alpha particle, which consist of two protons and two neutrons
•   Two protons
•   One proton and one neutron
•   Two neutrons
Solution (d) Nuclear reactions conserve total charge, and also conserve the total approximate mass. The other particles in the reaction will have mass =236-140-94=2 The other particles are two neutrons. Hence, (a) is not correct. For nuclei, number of protons tells the charge. So, the other particles must have charge Z such that 92=54+38+Z ∴Z=0 Therefore, the other particles have a total atomic mass 2 and total charge 0. Hence, only (d) is correct

Q6 In which of the following processes, the number of protons in the nucleus increase?
•   α-decay
•   β^--decay
•   β^+-decay
•   k-capture

Q7. Consider α-particles, β-particles and γ-rays, each having an energy of 0.5 MeV. In increasing order of penetrating powers, the radiations are:

•   α,β,γ
•   α,γ,β
•   β,γ,α
•   γ,β,α
Solution (c) The penetrating power is dependent on velocity. For a given energy, the velocity of γ-radiation is highest and α-particle is least

Q8. If in nature there may not be an element for which the principle quantum number n>4, then the total possible number of elements will be
•  60
•  32
•  4
•  64
Solution (a) For n=1, maximum number of states =2n^2=2 and for n=2,3,4, maximum number of states would be 8, 18, 32 respectively, Hence number of possible elements =2+8+18+32=60

Q9A freshly prepared radioactive source of half-life 2 h emits radiation of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with the source is
•  6 h
•  12 h
•  24 h
•  128 h

Q10. In the nuclear reaction given by _2 He^4+〖 _7 N〗^14→〖 _1 H〗^1+X, the nucleus X is
•  Nitrogen of mass 16
•  Nitrogen of mass 17
•  Oxygen of mass 16
•  Oxygen of mass 17

Solution
(d) Use mass balance and balance of atomic number #### Written by: AUTHORNAME

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