CBSE Class 12 Physics Chapter 24: Nuclei NCERT Solutions

Solution:

We are given the average atomic mass of magnesium and the masses and abundance of its isotopes. We need to find the abundances of the other two isotopes.

Given:

• Average atomic mass, $m = 24.312$ u
• Mass of $^{24}_{12}$Mg isotope, $m_1 = 23.98504$ u
• Mass of $^{25}_{12}$Mg isotope, $m_2 = 24.98584$ u
• Mass of $^{26}_{12}$Mg isotope, $m_3 = 25.98259$ u
• Abundance of $^{24}_{12}$Mg, $\eta_1 = 78.99\%$

Let the abundance of $^{25}_{12}$Mg be $x\%$. Then, the abundance of $^{26}_{12}$Mg will be $(100 - 78.99 - x)\% = (21.01 - x)\%$.

The formula for the average atomic mass ($m$) based on the masses ($m_i$) and abundances ($\eta_i$) of isotopes is:

$$m\; =\; \backslash frac\{m\_1\; \backslash eta\_1\; +\; m\_2\; \backslash eta\_2\; +\; m\_3\; \backslash eta\_3\}\{100\}$$

Substituting the given values:

$$24.312\; =\; \backslash frac\{(23.98504\; \backslash times\; 78.99)\; +\; (24.98584\; \backslash times\; x)\; +\; (25.98259\; \backslash times\; (21.01\; -\; x))\}\{100\}$$

Multiply both sides by 100:

$$2431.2\; =\; (23.98504\; \backslash times\; 78.99)\; +\; (24.98584\; \backslash times\; x)\; +\; (25.98259\; \backslash times\; 21.01)\; -\; (25.98259\; \backslash times\; x)$$

Calculate the products:

$$2431.2\; =\; 1894.5783096\; +\; 24.98584x\; +\; 545.8942159\; -\; 25.98259x$$

Combine the constant terms and the terms with $x$:

$$2431.2\; =\; (1894.5783096\; +\; 545.8942159)\; +\; (24.98584\; -\; 25.98259)x$$

$$2431.2\; =\; 2440.4725255\; -\; 0.99675x$$

Rearrange the equation to solve for $x$:

$$0.99675x\; =\; 2440.4725255\; -\; 2431.2$$

$$0.99675x\; =\; 9.2725255$$

$$x\; =\; \backslash frac\{9.2725255\}\{0.99675\}\; \backslash approx\; 9.300\backslash \%$$

So, the abundance of $^{25}_{12}$Mg is approximately $9.30\%$.

Now, calculate the abundance of $^{26}_{12}$Mg:

$$\backslash eta\_3\; =\; (21.01\; -\; x)\backslash \%\; =\; (21.01\; -\; 9.30)\backslash \%\; =\; 11.71\backslash \%$$

Answer: The abundance of $^{25}_{12}$Mg is approximately 9.30%, and the abundance of $^{26}_{12}$Mg is approximately 11.71%.

Solution:

The neutron separation energy is the energy required to remove a neutron from a nucleus. This energy is equivalent to the mass defect of the process, converted into energy. The conversion factor is $1$ u $= 931.5$ MeV.

For $^{41}_{20}$Ca:

The process of removing a neutron from $^{41}_{20}$Ca can be represented as:

$$^\{41\}\_\{20\}\backslash text\{Ca\}\; \backslash longrightarrow\; ^\{40\}\_\{20\}\backslash text\{Ca\}\; +\; ^\{1\}\_\{0\}\backslash text\{n\}$$

The masses involved are:

• Mass of $^{41}_{20}$Ca nucleus, $m(^{41}_{20}\text{Ca}) = 40.962278$ u
• Mass of $^{40}_{20}$Ca nucleus, $m(^{40}_{20}\text{Ca}) = 39.962591$ u
• Mass of neutron, $m_n = 1.008665$ u

The mass defect ($\Delta m$) for this reaction is the difference between the mass of the reactants and the mass of the products:

$$\backslash Delta\; m\; =\; m(^\{41\}\_\{20\}\backslash text\{Ca\})\; -\; [m(^\{40\}\_\{20\}\backslash text\{Ca\})\; +\; m\_n]$$

$$\backslash Delta\; m\; =\; 40.962278\; \backslash text\{\; u\}\; -\; (39.962591\; \backslash text\{\; u\}\; +\; 1.008665\; \backslash text\{\; u\})$$

$$\backslash Delta\; m\; =\; 40.962278\; \backslash text\{\; u\}\; -\; 40.971256\; \backslash text\{\; u\}$$

$$\backslash Delta\; m\; =\; -0.008978\; \backslash text\{\; u\}$$

The neutron separation energy ($S_n$) is the energy equivalent of this mass defect. Note that the calculation above represents the mass of the initial nucleus minus the mass of the final nucleus and neutron. The energy required to *separate* the neutron is the positive value of this difference if the initial nucleus is heavier, or the energy released if the final products are lighter. In this context, we calculate the energy required to break the nucleus apart.

The energy required to remove the neutron is:

$$S\_n\; =\; [m(^\{40\}\_\{20\}\backslash text\{Ca\})\; +\; m\_n\; -\; m(^\{41\}\_\{20\}\backslash text\{Ca\})]\; \backslash times\; 931.5\; \backslash text\{\; MeV/u\}$$

$$S\_n\; =\; [39.962591\; \backslash text\{\; u\}\; +\; 1.008665\; \backslash text\{\; u\}\; -\; 40.962278\; \backslash text\{\; u\}]\; \backslash times\; 931.5\; \backslash text\{\; MeV/u\}$$

$$S\_n\; =\; [40.971256\; \backslash text\{\; u\}\; -\; 40.962278\; \backslash text\{\; u\}]\; \backslash times\; 931.5\; \backslash text\{\; MeV/u\}$$

$$S\_n\; =\; 0.008978\; \backslash text\{\; u\}\; \backslash times\; 931.5\; \backslash text\{\; MeV/u\}$$

$$S\_n\; \backslash approx\; 8.363007\; \backslash text\{\; MeV\}$$

For $^{27}_{13}$Al:

The process of removing a neutron from $^{27}_{13}$Al can be represented as:

$$^\{27\}\_\{13\}\backslash text\{Al\}\; \backslash longrightarrow\; ^\{26\}\_\{13\}\backslash text\{Al\}\; +\; ^\{1\}\_\{0\}\backslash text\{n\}$$

The masses involved are:

• Mass of $^{27}_{13}$Al nucleus, $m(^{27}_{13}\text{Al}) = 26.981541$ u
• Mass of $^{26}_{13}$Al nucleus, $m(^{26}_{13}\text{Al}) = 25.986895$ u
• Mass of neutron, $m_n = 1.008665$ u

The mass defect ($\Delta m$) for this reaction is:

$$\backslash Delta\; m\; =\; m(^\{27\}\_\{13\}\backslash text\{Al\})\; -\; [m(^\{26\}\_\{13\}\backslash text\{Al\})\; +\; m\_n]$$

$$\backslash Delta\; m\; =\; 26.981541\; \backslash text\{\; u\}\; -\; (25.986895\; \backslash text\{\; u\}\; +\; 1.008665\; \backslash text\{\; u\})$$

$$\backslash Delta\; m\; =\; 26.981541\; \backslash text\{\; u\}\; -\; 26.995560\; \backslash text\{\; u\}$$

$$\backslash Delta\; m\; =\; -0.014019\; \backslash text\{\; u\}$$

The neutron separation energy ($S_n$) is the energy equivalent of the positive mass difference:

$$S\_n\; =\; [m(^\{26\}\_\{13\}\backslash text\{Al\})\; +\; m\_n\; -\; m(^\{27\}\_\{13\}\backslash text\{Al\})]\; \backslash times\; 931.5\; \backslash text\{\; MeV/u\}$$

$$S\_n\; =\; [25.986895\; \backslash text\{\; u\}\; +\; 1.008665\; \backslash text\{\; u\}\; -\; 26.981541\; \backslash text\{\; u\}]\; \backslash times\; 931.5\; \backslash text\{\; MeV/u\}$$

$$S\_n\; =\; [26.995560\; \backslash text\{\; u\}\; -\; 26.981541\; \backslash text\{\; u\}]\; \backslash times\; 931.5\; \backslash text\{\; MeV/u\}$$

$$S\_n\; =\; 0.014019\; \backslash text\{\; u\}\; \backslash times\; 931.5\; \backslash text\{\; MeV/u\}$$

$$S\_n\; \backslash approx\; 13.059\; \backslash text\{\; MeV\}$$

Answer: The neutron separation energy for $^{41}_{20}$Ca is approximately 8.363007 MeV, and for $^{27}_{13}$Al, it is approximately 13.059 MeV.