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Q 13.1 (a) Lithium has o stable isotopes 3Li and �Li have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium. (b) Two stable isotopes of Boron , �OB and 1 B . Their respective masses are 1 0.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of i 0 B and 1 B. A1: (a) Mass of 3Li li t hium isotope, m1 = 6 01512 u Mass of �Li lithium isotope, m = 7.01600 u Abundance of 3Li, n1= 7.5% Abundance of �Li, n2= 92.5% The atomic mass of lithium atom is given as: m _ m1n1+n2 m _ 6.01512x7.5+7.1600x92.5 - n1+1t2 - 92.5+7.5 = 6.940934 u (b) Mass of boron isotope �OB m = 10.01294 u Mass of boron isotope � 1 B m = 11.00931 u Abundance of i 0 B, n1 = x% Abundance of � 1 B, n2= (100 - x)o Atomic mass of boron, m = 10.811 u The atomic mass of boron atom is given as: m = m,n,+m,n, l O.S l l = 10.01294xz+ll.)31x(1-z) n1+"2 z+l-z 1081.11 = 10.01294X + 1100.931 -11.00931 X X = 19. 821/0. 99637 = 19.89 % And 100 - x = 80.11% Hence, the abundance of i 0 B is 19.89% and that of � 1 B is 80.11%. Q 13.2 :The three stable isotopes of neon: �Ne, nNe and �Ne have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon. Answer Atomic mass of �Ne, m 1= 19.99 u Abundance of �Ne, n1 = 90 51% Atomic mass of ?5Ne, m2 = 20.99 u Abundance of ?5Ne, n2 = 0.27% Atomic mass of �Ne, m3 = 21.99 u Abundance of ?5Ne, n3 = 9.22% The average atomic mass of neon is given as: m = m,n1+m2n2+n13n3 n1+n3 19.cx90.51+20.cx0.27+21.99x9.22 90.51+0.27+9.22 = 20.1771 u Q 13.3: Obtain the binding energy in MeV of a nitrogen nucleus � 4 N, given m(} 4 NJ=14.00307 u Ans: Atomic mass of nitrogen � 4 N, m = 14 00307 u A nucl eus of } 4 Nnitrogen contains 7 neutrons and 7 protons. Hence, the mass defect of this nucleus, m = 7mH + 7mn - m Where, Mass of a proton, mH = 1.007825 u NCERT SOLUTIONS CLASS-XII PHYSICS CHAPTER-13 NUCLEI

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Page 1: NCERT SOLUTIONS CLASS-XII PHYSICS CHAPTER-13 NUCLEI · CLASS-XII PHYSICS CHAPTER-13 NUCLEI. Created Date: 6/1/2018 4:22:01 PM

Q 13.1 (a) Lithium has rwo stable isotopes 3Li and �Li have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium.

(b) Two stable isotopes of Boron , �OB and �1 B . Their respective masses are 10.01294 u and 11 .00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of i0B and �1 B.

A1:

(a) Mass of 3Li lithium isotope, m1 = 6 01512 u

Mass of �Li lithium isotope, m = 7.01600 u

Abundance of 3Li, n1= 7.5%

Abundance of �Li, n2= 92.5%

The atomic mass of lithium atom is given as: m

_ m1n1+ffi-2n2 m

_ 6.01512x7.5+7.1600x92.5 -n1+1t2

-92.5+7.5

= 6.940934 u

(b) Mass of boron isotope �OB m = 10.01294 u

Mass of boron isotope �1B m = 11.00931 u

Abundance of i0B, n1 = x%

Abundance of �1B, n2= (100 - x)o/o

Atomic mass of boron, m = 10.811 u The atomic mass of boron atom is given as:

m = m,n,+m,n, lO.Sll = 10.01294xz+ll.00931x(100-z) n1+"2 z+lOO-z

1081.11 = 10.01294X + 1100.931 -11.00931 X

X = 19.821/0.99637 = 19.89 %

And 100 - x = 80.11%

Hence, the abundance of i0B is 19.89% and that of �1 B is 80.11%.

Q 13.2 :The three stable isotopes of neon: �Ne, nNe and �Ne have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.

Answer

Atomic mass of �Ne, m 1= 19.99 u

Abundance of �Ne, n1 = 90 51%

Atomic mass of ?5Ne, m2 = 20.99 u

Abundance of ?5Ne, n2 = 0.27%

Atomic mass of �Ne, m3 = 21.99 u

Abundance of ?5Ne, n3 = 9.22%

The average atomic mass of neon is given as:

m = m,n1+m2n2+n13n3 n1-+T12+n3

19.99x90.51+20.99x0.27+21.99x9.22

90.51+0.27+9.22

= 20.1771 u

Q 13.3: Obtain the binding energy in MeV of a nitrogen nucleus �4N, given m(}4NJ=14.00307 u

Ans:

Atomic mass of nitrogen �4N, m = 14 00307 u

A nucleus of }4Nnitrogen contains 7 neutrons and 7 protons.

Hence, the mass defect of this nucleus, tim = 7mH + 7mn - m

Where,

Mass of a proton, mH = 1.007825 u

NCERT SOLUTIONS CLASS-XII PHYSICS

CHAPTER-13 NUCLEI

Page 2: NCERT SOLUTIONS CLASS-XII PHYSICS CHAPTER-13 NUCLEI · CLASS-XII PHYSICS CHAPTER-13 NUCLEI. Created Date: 6/1/2018 4:22:01 PM
Page 3: NCERT SOLUTIONS CLASS-XII PHYSICS CHAPTER-13 NUCLEI · CLASS-XII PHYSICS CHAPTER-13 NUCLEI. Created Date: 6/1/2018 4:22:01 PM
Page 4: NCERT SOLUTIONS CLASS-XII PHYSICS CHAPTER-13 NUCLEI · CLASS-XII PHYSICS CHAPTER-13 NUCLEI. Created Date: 6/1/2018 4:22:01 PM
Page 5: NCERT SOLUTIONS CLASS-XII PHYSICS CHAPTER-13 NUCLEI · CLASS-XII PHYSICS CHAPTER-13 NUCLEI. Created Date: 6/1/2018 4:22:01 PM
Page 6: NCERT SOLUTIONS CLASS-XII PHYSICS CHAPTER-13 NUCLEI · CLASS-XII PHYSICS CHAPTER-13 NUCLEI. Created Date: 6/1/2018 4:22:01 PM
Page 7: NCERT SOLUTIONS CLASS-XII PHYSICS CHAPTER-13 NUCLEI · CLASS-XII PHYSICS CHAPTER-13 NUCLEI. Created Date: 6/1/2018 4:22:01 PM
Page 8: NCERT SOLUTIONS CLASS-XII PHYSICS CHAPTER-13 NUCLEI · CLASS-XII PHYSICS CHAPTER-13 NUCLEI. Created Date: 6/1/2018 4:22:01 PM
Page 9: NCERT SOLUTIONS CLASS-XII PHYSICS CHAPTER-13 NUCLEI · CLASS-XII PHYSICS CHAPTER-13 NUCLEI. Created Date: 6/1/2018 4:22:01 PM
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