Nuclei-Quiz-3

NUCLEI QUIZ-3

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.  A radioactive nucleus A finally transforms into a stable nucleus B. Then, A and B may be

•  Isobars 
•   Isotones 
•   Isotopes 
•   None of these 

Solution

 (c) A and B can be isotopes if number of β-decays is two times the number of α-decays

Q2.  As per Bohr model, the minimum energy (in eV) required to remove an electron from the ground state of doubly ionized Li atom (Z=3) is

•  1.51
•  13.6
•  14.8
•  122.4

Solution

Q3.  Masses of two isobars _29 Cu^64 and _30 Zn^64 are 63.9298 u and 63.9292 u, respectively. It can be concluded from these data that 

•    Both the isobars are stable
•   Zn^64 is radioactive, decaying to Cu^64 through β-decay
•   Cu^64 is radioactive, decaying to Zn^64 through γ-decay
•  Cu^64 is radioactive, decaying to Zn^64 through β-decay

Solution
3 (c) By the conservation of charge and nucleons, only potential is feasible

Q4.  Binding energy per nucleon for C^12 is 7.68 MeV and for C^13 is 7.74 MeV. The energy required to remove a neutron from C^13 is

•   5.49 MeV
•   8.46 MeV
•   9.45 MeV
•   15.49 MeV

Solution
4 (b) The difference in the binding energies is the energy required to add an extra neutron

Q5.  A radioactive sample S1 having an activity of 5μCi has twice the number of nuclei as another sample S2 which has an activity of 10μCi. The half lives of S1 and S2 can be 

•  20 years and 5 years, respectively
•  20 years and 10 years, respectively
•  10 years each
•  5 years each

Solution
 5

 

Q6.The luminous dials of watches are usually made by mixing a zinc sulphide phosphor with an α-particle emitter. The mass of radium (mass number 226, half-life 1620 years) that is needed to produce an average of 10 α-particles per second for this purpose is

•  2.77 mg span>
•  2.77 g
• 2.77×10^(-23) g
  
•  2.77×10^(-13) kg

Solution

Q7. Calculate the binding energy of a deuteron atom, which consists of a proton and a neutron, given that the atomic mass of the deuteron is 2.014102 u

•   0.002388 MeV
•   2.014102 MeV
•   2.16490 MeV
•   2.224 MeV

Solution
7 (d) Atomic mass M(H) of hydrogen and nuclear mass (M_n) are M(H)=1.007825 u and M_n=1.008665 u Mass defect, ∆m=[M(H)+M_n-M(D)] M(D)= mass of deuteron =2.016490 u-2.014102 u=0.002388 u As 1 u corresponds to 931.494 MeV energy, therefore, mass defect corresponds to energy, E_b=0.002388×931.5=2.224 MeV

Q8 A radioactive substance X decays into another radioactive substance Y. Initially, only X was present. λ_x and λ_y are the disintegration constants of X and Y. N_y will be maximum when

•   N_y/(N_x-N_y )=λ_y/(λ_x-λ_y )
•   N_x/(N_x-N_y )=λ_x/(λ_x-λ_y ) c)
•   λ_y N_y=λ_x N_x²
•   λ_y N_x=λ_x N_y³

Solution

 

Q9 A proton and a neutron are both shot at 100 ms^(-1) towards a _6^12 C nucleus. Which particle, if either, is more likely to be absorbed by the nucleus?

•   The proton
•  The neutron
•  The neutron
•   Neither particle will be absorbed

Solution

(b)

Once the neutron gets sufficiently close to the nucleus, the strong nuclear force sucks it in. Same happens with the proton except it is electrostatically repelled by the six protons already inside the carbon nucleus. The repulsion prevents a  proton from getting close enough to the nucleus. Therefore, the answer is (b)

Q10 To determine the half- life of radioactive element, a student plots graph of ln|(dN (t))/dt|versus t. Here (dN (t))/dt is the rate of radioactive decay at time t. If the number of radioactive nuclei of this element decreases by a factor of p after 4.16 yr, the value of p is

 

•  8
•  7
•  4
• 8.5

Solution