Ncert Exercises & Solutions

You may find the following data useful in solving the exercises:

e = 1.6×10^–19C

N = 6.023×10²³ per mole

1/(4πε₀) = 9 × 10⁹ N m²/C²

k = 1.381×10^–23J ⁰K^–1

1 MeV = 1.6×10^–13J

1 u = 931.5 MeV/c²

1 year = 3.154×10⁷ s

m_H = 1.007825 u

m_n = 1.008665 u

m() = 4.002603 u

me = 0.000548 u

13.1 (a) Two stable isotopes of lithium and 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, and . 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 and .

Hint: Atomic mass of an atom is weightage mass of its isotopes.
Step 1: Find the atmoc mass of Lithium atom.
(a) Mass of

{}_{3}^{6}{Li}

lititum isotpe,

m_{1} = 6.01512 u

Mass of

{}_{3}^{7}L

litium isotope,

m_{2} = 7.01600 u

Abundance of

{}_{3}^{7}{Li}, n_{2} = 92.5 %

The atomic mass of lithium atom is given as:

m = \frac{m_{1} n_{1} + m_{2} n_{2}}{n_{1} + n_{2}} = \frac{6.01512 \times 7.5 + 7.01600 \times 92.5}{7.5 + 92.5} = 6.940934 u

Step 2: Find the abundances of isotopes of Boron.
(b)

Mass of

{}_{5}^{10}B

boron isotop,

m_{1} = 10.01294 u

Mass of

{}_{5}^{11}B

boron isotope,

m_{2} = 11.00931 u

Let the abundance of

{}_{5}^{11}B, n_{1} = x %

Therefore, the abundance of

{}_{5}^{11}B, n_{2} = (100 - x) %

The atomic mass of boron = 10.81u. The atomic mass of Boron atom is given as:

m = \frac{m_{1} n_{1} + m_{2} n_{2}}{n_{1} + n_{2}}

\Rightarrow 10.811 = \frac{10.01294 \times x + 11.00931 \times (100 - x)}{x + (100 - x)}

\Rightarrow_{} 1081.1 = 10.01294 x + 1100.931 -11.00 931 x

\Rightarrow x = \frac{19.821}{0.99637} = 19.89 %

Therefore,

100 - x = 100 % - 19.89 % = 80.11 %

Hence, the abundances of

{}_{5}^{10}B is 19.89 % and {}_{5}^{11}B is 80.11 %

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