CBSE Class 12 Physics Chapter 13 Nuclei NCERT Solutions

Solution:

(a) To find the atomic mass of lithium, we need to calculate the weighted average of the masses of its isotopes based on their natural abundances.

Given:

• Mass of ⁶₃Li isotope, <math>m_1 = 6.01512 \text{ u}</math>
• Abundance of ⁶₃Li, <math>\eta_1 = 7.5\% = 0.075</math>
• Mass of ⁷₃Li isotope, <math>m_2 = 7.01600 \text{ u}</math>
• Abundance of ⁷₃Li, <math>\eta_2 = 92.5\% = 0.925</math>

The atomic mass of lithium (M) is calculated using the formula for weighted average:

$$M\; =\; \backslash frac\{(m\_1\; \backslash times\; \backslash eta\_1)\; +\; (m\_2\; \backslash times\; \backslash eta\_2)\}\{\backslash eta\_1\; +\; \backslash eta\_2\}$$

Since the abundances are given as percentages, their sum is 100%. We can use the fractional abundances directly:

$$M\; =\; (m\_1\; \backslash times\; \backslash eta\_1)\; +\; (m\_2\; \backslash times\; \backslash eta\_2)$$

Substituting the given values:

$$M\; =\; (6.01512\; \backslash text\{\; u\}\; \backslash times\; 0.075)\; +\; (7.01600\; \backslash text\{\; u\}\; \backslash times\; 0.925)$$

$$M\; =\; 0.451134\; \backslash text\{\; u\}\; +\; 6.489800\; \backslash text\{\; u\}$$

$$M\; =\; 6.940934\; \backslash text\{\; u\}$$

Therefore, the atomic mass of lithium is approximately 6.941 u.

(b) To find the abundances of the boron isotopes, we use the given atomic mass and the masses of the isotopes.

Let the abundance of ¹⁰₅B be <math>x\%</math>. Then the abundance of ¹¹₅B will be <math>(100 - x)\%</math>.

Given:

• Mass of ¹⁰₅B isotope, <math>m_1 = 10.01294 \text{ u}</math>
• Mass of ¹¹₅B isotope, <math>m_2 = 11.00931 \text{ u}</math>
• Atomic mass of boron, <math>M = 10.811 \text{ u}</math>

Using the weighted average formula:

$$M\; =\; \backslash frac\{(m\_1\; \backslash times\; \backslash eta\_1)\; +\; (m\_2\; \backslash times\; \backslash eta\_2)\}\{\backslash eta\_1\; +\; \backslash eta\_2\}$$

Substituting the values and using fractional abundances (x/100 and (100-x)/100):

$$10.811\; =\; \backslash frac\{(10.01294\; \backslash times\; \backslash frac\{x\}\{100\})\; +\; (11.00931\; \backslash times\; \backslash frac\{100-x\}\{100\})\}\{\backslash frac\{x\}\{100\}\; +\; \backslash frac\{100-x\}\{100\}\}$$

The denominator simplifies to <math>\frac{100}{100} = 1</math>.

$$10.811\; =\; \backslash frac\{10.01294x\; +\; 11.00931(100-x)\}\{100\}$$

Multiply both sides by 100:

$$1081.1\; =\; 10.01294x\; +\; 1100.931\; -\; 11.00931x$$

Rearrange the terms to solve for x:

$$1081.1\; -\; 1100.931\; =\; 10.01294x\; -\; 11.00931x$$

$$-19.831\; =\; -0.99637x$$

$$x\; =\; \backslash frac\{-19.831\}\{-0.99637\}\; \backslash approx\; 19.89\backslash \%$$

So, the abundance of ¹⁰₅B is approximately 19.89%.

The abundance of ¹¹₅B is:

$$100\; -\; x\; =\; 100\; -\; 19.89\; =\; 80.11\backslash \%$$

Hence, the abundance of ¹⁰₅B is 19.89% and that of ¹¹₅B is 80.11%.

Solution:

To find the average atomic mass of neon, we calculate the weighted average of the masses of its three stable isotopes, considering their respective abundances.

Given:

• Isotope 1: <math>^{20}_{10}\text{Ne}</math>, Mass <math>m_1 = 19.99 \text{ u}</math>, Abundance <math>\eta_1 = 90.51\% = 0.9051</math>
• Isotope 2: <math>^{21}_{10}\text{Ne}</math>, Mass <math>m_2 = 20.99 \text{ u}</math>, Abundance <math>\eta_2 = 0.27\% = 0.0027</math>
• Isotope 3: <math>^{22}_{10}\text{Ne}</math>, Mass <math>m_3 = 21.99 \text{ u}</math>, Abundance <math>\eta_3 = 9.22\% = 0.0922</math>

The average atomic mass (M) is calculated using the formula for a weighted average:

$$M\; =\; \backslash frac\{(m\_1\; \backslash times\; \backslash eta\_1)\; +\; (m\_2\; \backslash times\; \backslash eta\_2)\; +\; (m\_3\; \backslash times\; \backslash eta\_3)\}\{\backslash eta\_1\; +\; \backslash eta\_2\; +\; \backslash eta\_3\}$$

Since the sum of the fractional abundances is 1 (0.9051 + 0.0027 + 0.0922 = 1.0000), the formula simplifies to:

$$M\; =\; (m\_1\; \backslash times\; \backslash eta\_1)\; +\; (m\_2\; \backslash times\; \backslash eta\_2)\; +\; (m\_3\; \backslash times\; \backslash eta\_3)$$

Substitute the given values:

$$M\; =\; (19.99\; \backslash text\{\; u\}\; \backslash times\; 0.9051)\; +\; (20.99\; \backslash text\{\; u\}\; \backslash times\; 0.0027)\; +\; (21.99\; \backslash text\{\; u\}\; \backslash times\; 0.0922)$$

Calculate each term:

$$M\; =\; 18.092949\; \backslash text\{\; u\}\; +\; 0.056673\; \backslash text\{\; u\}\; +\; 2.027478\; \backslash text\{\; u\}$$

Sum the terms to find the average atomic mass:

$$M\; =\; 20.177099\; \backslash text\{\; u\}$$

Therefore, the average atomic mass of neon is approximately 20.177 u.