Q 1 :

Binding energy of a certain nucleus is 18×108J. How much is the difference between total mass of all the nucleons and nuclear mass of the given nucleus      [2024]

  • 0.2 μg

     

  • 20 μg

     

  • 10 μg

     

  • 2 μg

     

(2)         

              Δmc2=18×108

               Δm×9×1016=18×108

               Δm=2×10-8kg=20μg

 



Q 2 :

If Mo is the mass of isotope B512Mp and Mn are the masses of proton and neutron, then nuclear binding energy of isotope is                [2024]

  • (Mo-5Mp-7Mn)C2

     

  • (Mo-12Mn)C2

     

  • (Mo-5Mp)C2

     

  • (5Mp+7Mn-Mo)C2

     

(4)         

                 B5125 proton+7 neutron

                  B.E=Δmc2

                 (5Mp+7Mn-Mo)c2

 



Q 3 :

The energy equivalent of 1 g of substance is:                   [2024]

  •  11.2×1024 MeV

     

  •  5.6×1012 MeV

     

  •  5.6 eV

     

  •  5.6×1026 MeV

     

(4)          

                E=10-3×9×1016 J( E=Mc2)

                 E=9×10-13 J 

                 1J=11.6×10-19eV

                 E=9×1013×11.6×10-19 eV=5.6×1026 MeV

 



Q 4 :

The atomic mass of C126 is 12.00000u and that of C136 is 13.003354u. The required energy to remove a neutron from C136, if mass of neutron is 1.008665u, will be                   [2024]

  • 62.5 MeV

     

  • 6.25 MeV

     

  • 4.95 MeV

     

  • 49.5 MeV

     

(3)          

                    C136+EnergyC126+n10

                   Δm=12+1.008665-13.003354

                           =-0.00531u

                    Energy required=0.00531×931.5MeV

                                                          =4.95MeV



Q 5 :

If three helium nuclei combine to form a carbon nucleus then the energy released in this reaction is _________ ×10-2 MeV. (Given 1u = 931 MeV/c2, atomic mass of helium = 4.002603u).                [2024]



(727)              Reaction:  3H24e→ 612C+Q

                       Energy released

                       Q=[3MHe-MC]c2=[3×4.002603-12][931]

                        Q=727×10-2MeV

                       

 



Q 6 :

A star has 100% helium composition. It starts to convert three H4e into one C12 via triple alpha process as H4e+H4e+H4eC12+Q. The mass of the star is 2.0×1032 kg and it generates energy at the rate of 5.808×1030W. The rate of converting these H4e is n×1042 s-1, where n is _______ . 

 

[Take, mass of H4e = 4.0026 u, mass of C12 = 12 u]             [2024]



(5)                H24e+H24e+H24eC612+Q3H24eC612+Q

                    Q=(3mH4e-mC12)c2

                    Q=(3×4.0026-12)(3×108)2=7.266MeV

                    Power generated P=NtQ (N = No. of reactions/sec)

                    Nt=PQ=5.808×10307.266×106×1.6×10-19=5×1042

                     Rate of conversion of H4e in C12=5×1042

 



Q 7 :

In a nuclear fission process, a high mass nuclide (A  236) with binding energy 7.6 MeV/Nucleon dissociated into middle mass nuclides (A  118), having binding energy of 8.6 MeV/Nucleon. The energy released in the process would be _______ meV.              [2024]



(236)             Q=BEProduct-BEReactant

                          =2(118)(8.6)-236(7.6)=236×1=236MeV

 



Q 8 :

The mass defect in a particular reaction is 0.4 g. The amount of energy liberated is n×107 kWh, where n = ______. (speed of light = 3×108 m/s)          [2024]



(1)        E=Δmc2

             =0.4×10-3×(3×108)2=3600×107kWs

              =3600×1073600kWh=1×107kWh

 



Q 9 :

In a nuclear fission reaction of an isotope of mass M three similar daughter nuclei of same mass are formed. The speed of a daughter nuclei in terms of mass defect M will be               [2024]

  • 2cΔMM

     

  • ΔMc23

     

  • c2ΔMM

     

  • c3ΔMM

     

(3)

M=3m

ΔK.E.=energy released due to fission

3(12mv2)=ΔMC2

32mv2=ΔMC2

v2=2ΔMC2M

v=2ΔMC2M=C2ΔMM



Q 10 :

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).                   [2025]

Assertion (A): The binding energy per nucleon is found to be practically independent of the atomic number A, for nuclei with mass numbers between 30 and 170.

Reason (R): Nuclear force is long range.

In the light of the above statements, choose the correct answer from the options given below:

  • (A) is false but (R) is true

     

  • (A) is true but (R) is false

     

  • Both (A) and (R) are true and (R) is the correct explanation of (A)

     

  • Both (A) and (R) are true but (R) is NOT the correct explanation of (A)

     

(2)

From graph between B.E/N and A we can see BE/N is almost constant ⇒ correct

Reason ⇒ incorrect as nuclear forces are short range forces.

 



Q 11 :

Energy released when two deuterons (H21) fuse to form a helium nucleus (He42) is:             

(Given: Binding energy per nucleon of H21=1.1MeV and binding energy per nucleon of He42=7.0MeV)                [2025]

  • 8.1 MeV

     

  • 5.9 MeV

     

  • 23.6 MeV

     

  • 26.8 MeV

     

(3)

H2+H21He421

Q=BEProduct-BEReactant

Q=(4×7-4×1.1)MeV=23.6 MeV



Q 12 :

The mass of proton, neutron and helium nucleus are respectively 1.0073 u, 1.0087 u and 4.0015 u. The binding energy of helium nucleus is          [2023]

  • 14.2 MeV

     

  • 28.4 MeV

     

  • 56.8 MeV

     

  • 7.1 MeV

     

(2)

B.E. of Helium=(2mp+2mN-mHe)c2=28.4 MeV



Q 13 :

A92238B90234+D24+Q

In the given nuclear reaction, the approximate amount of energy released will be:              [2023]

[Given, mass of A92238=238.05079×931.5 MeV/c2

mass of B90234=234.04363×931.5 MeV/c2

mass of D24=4.00260×931.5 MeV/c2]

  • 3.82 MeV

     

  • 5.9 MeV

     

  • 2.12 MeV

     

  • 4.25 MeV

     

(4)

Q=(mA-mB-mD)×931.5 MeV

     =(238.05079-234.04363-4.00260)×931.5

 4.25 MeV
  



Q 14 :

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R                   [2023]

Assertion A: The binding energy per nucleon is practically independent of the atomic number for nuclei of mass number in the range 30 to 170.

Reason R: Nuclear force is short ranged.

In the light of the above statements, choose the correct answer from the options given below:

  • Both A and R are true but R is NOT the correct explanation of A

     

  • A is true but R is false

     

  • A is false but R is true

     

  • Both A and R are true and R is the correct explanation of A

     

(4)

Binding energy per nucleon is almost same for nuclei of mass number ranging 30 to 170.



Q 15 :

Nucleus A having Z = 17 and equal number of protons and neutrons has 1.2 MeV binding energy per nucleon. Another nucleus B of Z = 12 has total 26 nucleons and 1.8 MeV binding energy per nucleon. The difference of binding energy of B and A will be ______ MeV.                 [2023]



(6)

For A mass number =34

Total binding energy=1.2×34=40.8 MeV

For B mass number =26

Total binding energy=1.8×26 MeV=46.8 MeV

Difference of BE=6 MeV



Q 16 :

If the binding energy of ground state electron in a hydrogen atom is 13.6 eV, then, the energy required to remove the electron from the second excited state of Li2+ will be: x×10-1 eV. The value of x is _______ .              [2023]



(136)

EH=13.6

ELi2=13.6Z2n2=13.6×99=13.6 eV=136×10-1 eV



Q 17 :

A nucleus with mass number 242 and binding energy per nucleon as 7.6 MeV breaks into two fragment each with mass number 121. If each fragment nucleus has binding energy per nucleon as 8.1 MeV, the total gain in binding energy is ________ MeV.          [2023]



(121)

Initial binding energy=242×7.6 MeV

Final binding energy=121×8.1 MeV+121×8.1 MeV=242×8.1 MeV

Total gain in binding energy=242(8.1-7.6)=121 MeV



Q 18 :

A common example of alpha decay is

            U92238Th90234+He42+Q

Given:  U92238=238.05060 u

            Th90234=234.04360 u

            He24=4.00260 u  and 1u=931.5 MeVe2

The energy released (Q) during the alpha decay of U92238 is ________ MeV.                      [2023]



(4)

Energy released=(Δm)amu×931.5 MeV

                                =(mu-mTh-mHe)amu×931.5 MeV

                                =0.0044×931.5 MeV=4.0986 MeV



Q 19 :

Given below are two statements:

Statement I: For all elements, greater the mass of the nucleus, greater is the binding energy per nucleon.

Statement II: For all elements, nuclei with less binding energy per nucleon transform to nuclei with greater binding energy per nucleon.

In the light of the above statements, choose the correct answer from the options given below:                       [2026]

  • Statement I is true but Statement II is false

     

  • Both Statement I and Statement II are false

     

  • Statement I is false but Statement II is true

     

  • Both Statement I and Statement II are true

     

(3)

Theoretical



Q 20 :

The minimum frequency of photon required to break a particle of mass 15.348 amu into 4 α particles is _______ kHz.

[Mass of He nucleus = 4.002 amu, 1 amu = 1.66×10-27 kg, h=6.6×10-34 J.s and c=3×108 m/s]             [2026]

  • 14.94×1019

     

  • 9×1019

     

  • 9×1020

     

  • 14.94×1020

     

(1)

hν=(4×4.002-15.348)×1.66×10-27×(3×108)2

ν=14.94×1019 kHz



Q 21 :

The binding energy for the following nuclear reactions are expressed in MeV.

He23+n01He24+20 MeV

He24+n01He25-0.9 MeV

If X3,X4,X5 denote the stability of He23,He24 and He25 respectively, then the correct order is:             [2026]

  • X4>X5>X3

     

  • X4<X5<X3

     

  • X4>X5<X3

     

  • X4=X5=X3

     

(1)

BEHe4-BEHe3=20 MeV  (1)

BEHe5-BEHe4=-0.9 MeV  (2)

From eq (1) & (2)

BEHe4>BEHe5>BEHe3

X4>X5>X3



Q 22 :

The average energy released per fission for the nucleus of U92235 is 190 MeV. When all the atoms of 47 g pure U92235 undergo fission process, the energy released is α×1023 MeV. The value of α is ______.

(Avogadro Number =6×1023 per mole)  [2026]



(228)

Total number of U-235 atoms is

47 g=47235 moles=15 moles

  Total energy released=15×6×1023×190 MeV

=228×1023 MeV