Q 1 :

From the statements given below:

(A) The angular momentum of an electron in nth orbit is an integral multiple of h.

(B) Nuclear forces do not obey inverse square law.

(C) Nuclear forces are spin dependent.

(D) Nuclear forces are central and charge independent.

(E) Stability of nucleus is inversely proportional to the value of packing fraction.

Choose the correct answer from the options given below:                          [2024]

  • (A), (B), (C), (D) only

     

  • (A), (C), (D), (E) only

     

  • (A), (B), (C), (E) only

     

  • (B), (C), (D), (E) only

     

(3)

 



Q 2 :

The mass number of nucleus having radius equal to half of the radius of nucleus with mass number 192 is            [2024]

  • 24

     

  • 32

     

  • 40

     

  • 20

     

(1)              

                  R1=R22

                   R0(A1)1/3=R02(A2)1/3

                   A1=18A2

                   A1=1928=24

 



Q 3 :

A nucleus has mass number A1 and volume V1. Another nucleus has mass number A2 and volume V2. If relation between mass number is A2=4A1, then V2/V1= ___________ .                [2024]



(4)          For a nucleus

               Volume: V=43πR3

               R=R0(A)1/3

                V=43πR03AV2V1=A2A1=4



Q 4 :

The radius of a nucleus of mass number 64 is 4.8 fermi. Then the mass number of another nucleus having radius of 4 fermi is 1000x, where x is ______ .      [2024]



(27)             R=R0A1/3

                    4.8=R0(64)1/3

                    4.8=R04                                                     ...(i)

                    4=R0(A)1/3                                               ...(ii)

                    (i) divided by (ii),1.2=4(A)1/3

                     A=641.728=1000x

                     x=172864=27



Q 5 :

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

Assertion (A): The density of the copper (C2964u) nucleus is greater than that of the carbon (C612) nucleus.

Reason (R): The nucleus of mass number A has a radius proportional to A1/3.

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

  • (A) is correct but (R) is not correct

     

  • (A) is not correct but (R) is correct

     

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

     

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

     

(2)

ρ=MV=mn×A43πR3=mn×A43πAR03

So ρ is almost constant, density of nucleus is independent of mass number.

R=R0A1/3RA1/3

Radius is proportional to A1/3.

 



Q 6 :

For a nucleus of mass number A and radius R, the mass density of the nucleus can be represented as           [2025]
 

  • A3

     

  • A13

     

  • A23

     

  • Independent of A

     

(4)

R=R0A1/3

Density=Mass numberVolume=A43πR03A=constant

 



Q 7 :

The ratio of the density of oxygen nucleus (O816) and helium nucleus (He24) is                [2023]

  • 4 : 1

     

  • 1 : 1

     

  • 8 : 1

     

  • 2 : 1

     

(2)

Density of any nucleus is same.

 



Q 8 :

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

Assertion A: The nuclear density of nuclides B510,36Li,2656O,1020Ne and Bi83209 can be arranged as

ρBiN>ρFeN>ρNeN>ρBN>ρLiN.

Reason R: The radius R of nucleus is related to its mass number A as R=R0A1/3, where R0 is a constant.

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

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

     

  • A is false but R is true

     

  • A is true but R is false

     

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

     

(2)

R=R0A1/3, using this

ρ=M43πR3=Amp43πR03A=mp43πR03

ρ is independent of mass number.

 A is false.
 



Q 9 :

For a nucleus XZA having mass number A and atomic number Z

A. The surface energy per nucleon (bs)=-a1A2/3

B. The Coulomb contribution to the binding energy bc=-a2Z(Z-1)A4/3

C. The volume energy bv=a3A

D. Decrease in the binding energy is proportional to surface area.

E. While estimating the surface energy, it is assumed that each nucleon interacts with 12 nucleons. (a1,a2 and a3 are constants)

Choose the most appropriate answer from the options given below:                   [2023]

  • B, C, E only

     

  • C, D only

     

  • A, B, C, D only

     

  • B, C only

     

(2)

Surface energy per nucleonr2AA2/3A1A1/3
(Mass number Ar3rA1/3)

A is incorrect

Contribution to binding energy by coulombic forces is

             =-a2Z(Z-1)A1/3

B is incorrect

Volume energyA

C is correct

For (D), if we consider only surface energy contribution then option is correct.

For (E) only 3 interactions contribute to surface energy.



Q 10 :

Assume that protons and neutrons have equal masses. Mass of a nucleon is 1.6×10-27kg and radius of nucleus is 1.5×10-15A1/3m. The approximate ratio of the nuclear density and water density is n×1013. The value of n is _______.              [2023]



(11)

Density of nuclei=mass of nucleivolume of nuclei

ρ=1.6×10-27A43π(1.5×10-15)3A=0.113×1018 kg/m3

ρw=103 kg/m3

Hence  ρρw=11.31×1013



Q 11 :

A nucleus disintegrates into two smaller parts, which have their velocities in the ratio 3 : 2. The ratio of their nuclear sizes will be (x3)1/3. The value of 'x' is ______.    [2023]



(2)

v1v2=32

m1v1=m2v2  m1m2=23

Since nuclear mass density is constant

m143πr13=m243πr23

(r1r2)3=m1m2

r1r2=(23)13    So, x=2



Q 12 :

A nucleus disintegrates into two nuclear parts, in such a way that the ratio of their nuclear sizes is 1:21/3. Their respective speeds have a ratio of n:1. The value of n is ______.                      [2023]



(2)

v1v2=m2m1=A2A1=21



Q 13 :

If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ______ m 
(Atomic number of gold = 79 and 14πε0=9×109 in SI units)    [2026]

  • 3.85×10-16

     

  • 2.95×10-14

     

  • 3.85×10-14

     

  • 2.95×10-16

     

(2)

Energy conservation

Ki+Ui=Kf+Uf

7.7×106×1.6×10-19+0

=0+9×109(1.6×10-19)(79×1.6×10-19)r

r=2.95×10-14



Q 14 :

7.9 MeV α-particle scatters from a target material of atomic number 79. From the given data, the estimated diameter of nuclei of the target material is (approximately) _____ m.


[14πε0=9×109 Nm2/C2 and electron charge =1.6×10-19 C]                      [2026]

  • 1.44×10-13

     

  • 2.88×10-14

     

  • 1.69×10-12

     

  • 5.76×10-14

     

(4)

By mechanical energy conservation

(ME)i=(ME)f

PEi+KEi=PEf+KEf

0+7.9×106×1.6×10-19=k(2e)(Ze)r+0

r=9×109×2×(1.6×10-19)2×797.9×106×1.6×10-19=2.88×10-14 m

For diameter D=2r=5.76×10-14 m



Q 15 :

A nucleus has mass number α and radius Rα. Another nucleus has mass number β and radius Rβ. If β=8α then Rα/Rβ is :   [2026]

  • 1

     

  • 0.5

     

  • 2

     

  • 8

     

(2)

Rα=R0α1/3

Rβ=R0β1/3

RαRβ=(αβ)1/3=12



Q 16 :

Which of the following pair of nuclei are isobars of the element?  [2026]

  • H12 and H13

     

  • U92236 and U92238

     

  • H13 and He23

     

  • Hg80198 and Au79197

     

(3)

Isobars are nuclei that have the same mass number.

13& He have same mass number.23