Atomic and Nuclear Physics MCQ Class XII.pptx

MCQs in Atomic and Nuclear Physics ( Unit - IX TN State Board Syllabus)
May be asked in NEET

1. Atomic and Nuclear physics MCQ Class-XII (TN State Board) (With explanation)

2. By Dr M. Arunachalam Head, Department of Physics ( Rtd .) Sri SRNM College, Sattur

3. 1. Suppose an alpha particle accelerated by a potential of V volt is allowed to collide with a nucleus of atomic number Z , then the distance of closest approach of alpha particle to the nucleus is (a) 14.4(Z/V)A (b) 14.4(V/Z)A (c) 1.44(Z/V)A (d) 1.44(V/Z)A The distance of closest approach = r = (1/4 πε ) (2Ze 2 /E K ) e –Charge of proton = 1.6x10 -19 C 1/4 πε = 9 x10 9 E K – Kinetic energy of the accelerated α particle = (1/2)mv 2 = qV

4. q – Charge of α particle = 2e ie . E K = 2eV Accelerated potential = V volt Atomic number of the nucleus = Z Substituting the values in the equation for r , we get r = (9 x10 9 x 2 Ze 2 / 2 eV) = (9 x10 9 xZx 1.6x10 -19 /V) = (14.4Z/V) x10 -10 m ie . r = 14.4(Z/V)A 0 : Ans: a

5. 2. In a hydrogen atom, the electron revolving in the fourth orbit, has angular momentum equal to (a) h (b) h / π (c) 4h / π (d) 2h / π The angular momentum of the electron is l = nh /2 π For the fourth orbit, n = 4, ie . l = 4h/2 π = 2h/ π Ans: d

6. 3. Atomic number of H-like atom with ionization potential 122.4 V for n = 1 is (a) 1 (b) 2 (c) 3 (d) 4 Ionisation potential E = 13.6Z 2 ie . 122.4 = 13.6Z 2 ie . Z 2 =122.4/13.6 = 9 ie . Z =. 9 = 3 ie . Z = 3 Ans: c  

7. 4. The ratio between the radius of first three orbits of hydrogen atom is (a) 1:2:3 (b) 2:4:6 (c) 1:4:9 (d) 1:3:5 For hydrogen atom, Radius of the orbit r n = a n 2 ie . r n n 2 Hence the ratio is 1 2 :2 2 :3 2 = 1:4:9 Ans: c  

8. 5. The charge of cathode rays particle is (a) positive (b) negative (c) neutral (d) not defined Cathode rays contain negatively charged particles. Ans: b

9. 6. In J.J. Thomson e/m experiment, electrons are accelerated through 2.6 kV enter the region of crossed electric field and magnetic field of strength 3.0 × 10 4 Vm –1 and 1.0 × 10 –3 T, respectively, and pass through it and undeflected, then the specific charge is (a) 1.6 × 10 10 Ckg – 1 (b) 1.7 × 10 11 Ckg – 1 (c) 1.5 × 10 11 Ckg – 1 (d) 1.8 × 10 11 Ckg – 1 The specific charge of an electron e/m = (1/2V)E 2 /B 2 Electric field E = 3.0×10 4 Vm –1 M agnetic field B = 1.0×10 –3 T ,

10. Contd … Accelerating Potential V = 2.6 kV = 2.6×10 3 volt Substituting the values in the equation, we get, e/m = (1/2x 2.6×10 3 )(3.0×10 4 ) 2 /(1.0×10 –3 ) 2 ie . e/m = (1/5.2 ×10 3 )(9×10 8 )/1.0×10 –6 e/m = (9/5.2) ×10 11 ie . e/m = 1 .7 × 10 11 Ckg – 1 Ans: b

11. 7. The ratio of the wavelengths radiation emitted for the transition from n =2 to n = 1 in Li ++, He + and H is (a)1: 2: 3 (b) 1: 4: 9 (c) 3:2:1 (d) 4: 9: 36 Energy of the electron in the n th orbit is E n = hc / λ = - me 4 Z 2 /8 ε 2 h 2 n 2 This equation implies that 1 / λ α Z 2 Z - Atomic number Z for Li=3 Z for He =2 Z for H = 1 Ratio of Z is 3:2:1 Ratio of Z 2 is 9:4:1. Hence the ratio of wavelength λ is 1:4:9 Ans: b

12. 8. The electric potential of an electron is given by V =V ln (r/ r ), where r is a constant. If Bohr atom model is valid, then variation of radius of n th orbit r n with the principal quantum number n is (a) r (b) r n (c) r 2 (d) r n 2 Electric potential energy of an electron is given by U = eV =eV ln (r/r ), Force due to electric field is F = - dU / dr = eV /r = mv 2 /r --- Centripetal force  

13. Contd … mv 2 / r =eV / r ie . v 2 = eV /m ie . v = eV /m ---- 1 By Bohr’s postulates, Angular momentum = mvr = nh /2π ---- 2 Dividing equation 2 by 1 we get, mvr /v = mr = ( nh /2π)/ eV /m ie r = ( nh /2π)/ eV /m ie r Ans: b  

14. 9. If the nuclear radius of 27 Al is 3.6 fermi, the approximate nuclear radius of 64 Cu in fermi is (a) 2.4 (b) 1.2 (c) 4.8 (d)3.6 Radius of the nucleus R = R A 1/3 For 27 Al: Radius R 1 = R (27) 1/3 = 3.6 fermi ---- 1 For 64 Cu: Radius R 2 = R (64) 1/3 ---- 2 Dividing equation 2 by 1 we get R 2 /R 1 = R (64) 1/3 / R (27) 1/3

15. Contd … R 2 /3.6= R (4) / R (3) R 2 = 3.6x4 / 3 ie . R 2 = 4.8 fermi Ans: c

16. 10. The nucleus is approximately spherical in shape. Then the surface area of nucleus having mass number A varies as (a) A 2/3 (b) A 4/3 (c) A 1/3 (d) A 5/3 Surface area of the spherical nucleus = 4 π R 2 = a Radius of the nucleus R = R A 1/3 ie . a = 4 π R 2 = 4 π ( R A 1/3 ) 2 a = 4 π R A 2/3 ie . Surface area of the nucleus varies as A 2/3 Ans: a

17. 11. The mass of a 3 Li 7 nucleus is 0.042 u less than the sum of the masses of all its nucleons. The average binding energy per nucleon of 3 Li 7 nucleus is nearly (a) 46 MeV (b) 5.6 MeV (c) 3.9MeV (d)23 MeV Mass defect of 3 Li 7 nucleus = 0.042 u Energy equivalence of 1u = 931MeV Mass number of 3 Li 7 nucleus = A =7  

18. Contd … Binding Energy BE = c 2 0.042x931MeV BE/ Nucleon = BE/A = 0.042x931/ 7 = 0.006x931 = 5.586 Binding energy per nucleon of 3 Li 7 nucleus = 5.586 MeV Ans: b  

19. 12. Mp denotes the mass of the proton and Mn denotes mass of a neutron. A given nucleus of binding energy B, contains Z protons and N neutrons. The mass M(N,Z) of the nucleus is given by(where c is the speed of light) (a) M(N,Z) = N Mn + ZMp - Bc 2 (b) M(N,Z) = N Mn + ZMp + Bc 2 (c) M(N,Z) = N Mn + ZMp - B/c 2 (d) M(N,Z) = N Mn + ZMp + B/c 2

20. Solution M(N,Z) – Mass of the nucleus Z- Atomic number (No. of protons in the nucleus ) N – Number of neutrons in the nucleus Mp - Mass of one proton Mn – Mass of one neutron

21. Solution We know that, But binding energy B = c 2 ie . c 2 this in equation 1 we get c 2 Ans: c  

22. 13. A radioactive nucleus (initial mass number A and atomic number Z) emits two α-particles and 2 positrons. The ratio of number of neutrons to that of proton in the final nucleus will be (a) ( A – Z − 4 )/(Z – 2 ) (b) ( A – Z − 2)/(Z – 6) ( c ) ( A – Z − 4 )/(Z – 6) ( d ) ( A – Z − 12)/(Z – 4) The ratio of neutrons to protons before emission = (A-Z)/Z Emission of two α-particles ( ie . 2x 2 He 4 ), the number of protons Z ( ie positive charges) is reduced by 4

23. Contd … Also number of neutrons (A-Z) is also reduced by 4. Now the ratio is (A-Z-4)/(Z-4) Emission of two positrons , two protons re converted into neutrons. Thus the number of protons is further reduced by 2 (ie.Z-6) and two more neutrons are produced ( ie . A-Z-4+2) ( ie A-Z-2) New ratio is ( A – Z − 2)/(Z – 6) Ans: b

24. 14. The half-life period of a radioactive element A is same as the mean life time of another radioactive element B. Initially both have the same number of atoms. Then ( a) A and B have the same decay rate initially (b) A and B decay at the same rate always (c) B will decay at faster rate than A (d) A will decay at faster rate than B The half-life period of a radioactive element A ( ie . T 1/2 ) is same as the mean life time of another radioactive element B ( ie . τ )

25. Contd … Half life period T 1/2 = 0.693/ λ and Mean life τ = 1/ λ λ – Decay constant By the given condition, we have, 0.693/ λ A = 1/ λ B ie . λ A = λ B x 0.693 ie. λ B λ A Hence B will decay at faster rate than A Ans: c  

26. 15. A radiative element has N number of nuclei at t=0. The number of nuclei remaining after half of a half-life (that is, at time t = (1/2)T 1/2 ) (a) N /2 (b) N / 2 (c) N /4 (d) N /8 In the half life period ( ie . T 1/2 ) the no of nuclei remaining = N /2. In less than half life periond (ie. t = (1/2)T 1/2 ) (ie, half of half life period) definitely more no. of nuclei will remain ie. N ie. N / 2. Ans: b  

27. Thank You