NCERT Solutions Class 12 Physics Chapter 13 Nuclei
1. (a) Lithium has two stable isotopes \(_{3}^{6}{Li}\) and \(_{3}^{7}{Li}\) 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) Boron has two stable isotopes, \(_{5}^{10}{B}\) and \(_{5}^{11}{B}\) . 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 \(_{5}^{10}{B}\) and \(_{5}^{11}{B}\).
Answer:
(a) Mass of \(_{3}^{6}{Li}\) lithium isotope , m1 = 6.01512 u
Mass of \(_{3}^{7}{Li}\) lithium isotope , m2 = 7.01600 u
Abundance of \(_{3}^{6}{Li}\) , n1= 7.5%
Abundance of \(_{3}^{7}{Li}\) , n2= 92.5%
The atomic mass of lithium atom is:
\(m =\frac{ m_1 n_1 + m_2 n_2 }{ n_1 + n_2 }\)
\(m =\frac{ 6.01512 \times 7.5 + 7.01600 \times 92.5 }{ 7.5 + 92.5 }\)
= 6.940934 u
(b) Mass of boron isotope \(_{5}^{10}{B}\) m1 = 10.01294 u
Mass of boron isotope \(_{5}^{11}{B}\) m2 = 11.00931 u
Abundance of \(_{5}^{10}{B}\), n1 = x%
Abundance of \(_{5}^{11}{B}\), n2= (100 − x)%
Atomic mass of boron, m = 10.811 u
The atomic mass of boron atom is:
\(m =\frac{ m_1 n_1 + m_2 n_2 }{ n_1 + n_2 }\)
\(10.811 =\frac{ 10.01294 \times x + 11.00931 \times ( 100 – x ) }{ x + 100 -x }\)
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 \(_{5}^{10}{B}\) is 19.89% and that of \(_{5}^{11}{B}\) is 80.11%.
2. The three stable isotopes of neon: \(_{10}^{20}{Ne}\), \(_{10}^{21}{Ne}\) and \(_{10}^{22}{Ne}\) 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 \(_{10}^{20}{Ne}\), m1= 19.99 u
Abundance of \(_{10}^{20}{Ne}\), n1 = 90.51%
Atomic mass of \(_{10}^{21}{Ne}\), m2 = 20.99 u
Abundance of \(_{10}^{21}{Ne}\), n2 = 0.27%
Atomic mass of \(_{10}^{22}{Ne}\), m3 = 21.99 u
Abundance of \(_{10}^{22}{Ne}\), n3 = 9.22%
Below is the average atomic mass of neon:
\(m=\frac{ m_1 n_1 + m_2 n_2 + m_3 n_3 }{ n_1 + n_2 + n_3 }\)
\(= \frac{ 19.99 \times 90.51 + 20.99 \times 0.27 + 21.99 \times 9.22 }{ 90.51 + 0.27 + 9.22 }\)
= 20.1771 u
3. Obtain the binding energy in MeV of a nitrogen nucleus \(_{7}^{14}{N}\), given m(\(_{7}^{14}{N}\)) = 14.00307 u
Answer:
Atomic mass of nitrogen \(_{7}^{14}{N}\), m = 14.00307 u
A nucleus of \(_{7}^{14}{N}\) nitrogen contains 7 neutrons and 7 protons.
∆m = 7mH + 7mn − m is the mass defect the nucleus
Where,
Mass of a proton, mH = 1.007825 u
Mass of a neutron, mn = 1.008665 u
∆m = 7 × 1.007825 + 7 × 1.008665 − 14.00307
= 7.054775 + 7.06055 − 14.00307
= 0.11236 u
But 1 u = 931.5 MeV/c2
∆m = 0.11236 × 931.5 MeV/c2
Eb = ∆mc2 is the binding energy of the nucleus
Where, c =Speed of light
\(E_b = 0.11236 \times 931.5 \left ( \frac{MeV}{c^{2}} \right ) \times c^{2}\)
= 104.66334 Mev
Therefore, 104.66334 MeV is the binding energy of the nitrogen nucleus.
4. Obtain the binding energy of the nuclei \(_{26}^{56}{Fe}\) and \(_{83}^{209}{Bi}\) in units of MeV from the following
data:
m (\(_{26}^{56}{Fe}\)) = 55.934939 u
m(\(_{83}^{209}{Bi}\)) = 208.980388 u
Answer:
Atomic mass of \(_{26}^{56}{Fe}\), m1 = 55.934939 u
\(_{26}^{56}{Fe}\) nucleus has 26 protons and (56 − 26) = 30 neutrons
Hence, the mass defect of the nucleus, ∆m = 26 × mH + 30 × mn − m1
Where,
Mass of a proton, mH = 1.007825 u
Mass of a neutron, mn = 1.008665 u
∆m = 26 × 1.007825 + 30 × 1.008665 − 55.934939
= 26.20345 + 30.25995 − 55.934939
= 0.528461 u
But 1 u = 931.5 MeV/c2
∆m = 0.528461 × 931.5 MeV/c2
Eb1 = ∆mc2 is the binding energy of the nucleus.
Where, c = Speed of light
\(E_{b_{1}} = 0.528461 \times 931.5 \left ( \frac{MeV}{c^{2}} \right ) \times c^{2}\)
= 492.26 MeV
Average binding energy per nucleon = \(\frac{492.26}{56}\) = 8.79 MeV
Atomic mass of \(_{83}^{209}{Bi}\), m2 = 208.980388 u
\(_{83}^{209}{Bi}\) nucleus has 83 protons and (209 − 83) 126 neutrons.
The mass defect of the nucleus is given as:
∆m’ = 83 × mH + 126 × mn − m2
Where,
Mass of a proton, mH = 1.007825 u
Mass of a neutron, mn = 1.008665 u
∆m’ = 83 × 1.007825 + 126 × 1.008665 − 208.980388
= 83.649475 + 127.091790 − 208.980388
= 1.760877 u
But 1 u = 931.5 MeV/c2
∆m’ = 1.760877 × 931.5 MeV/c2
Eb2 = ∆m’c2 is the binding energy of the nucleus.
= 1.760877 × 931.5 \(\left ( \frac{MeV}{c^{2}} \right ) \times c^{2}\)
= 1640.26 MeV
Average binding energy per nucleon = 1640.26/209 = 7.848 MeV