CBSE Class 12 Physics Nuclei NCERT Solutions - Additional Exercises

Solution:

We are given the following information:

• Average atomic mass of magnesium, $m\; =\; 24.312\; \backslash text\{\; u\}$
• Mass of $^{24}_{12}$Mg isotope, $m\_1\; =\; 23.98504\; \backslash text\{\; u\}$
• Mass of $^{25}_{12}$Mg isotope, $m\_2\; =\; 24.98584\; \backslash text\{\; u\}$
• Mass of $^{26}_{12}$Mg isotope, $m\_3\; =\; 25.98259\; \backslash text\{\; u\}$
• Abundance of $^{24}_{12}$Mg, $\backslash eta\_1\; =\; 78.99\backslash \%$

Let the abundance of $^{25}_{12}$Mg be $\backslash eta\_2\; =\; x\backslash \%$. Since the sum of abundances of all isotopes must be 100%, the abundance of $^{26}_{12}$Mg will be $\backslash eta\_3\; =\; (100\; -\; 78.99\; -\; x)\backslash \%\; =\; (21.01\; -\; x)\backslash \%$.

The formula for the average atomic mass is the weighted average of the masses of the isotopes:

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

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:

$$2431.2\; =\; 1894.5783096\; +\; 24.98584x\; +\; (25.98259\; \backslash times\; 21.01)\; -\; (25.98259\; \backslash times\; x)$$

Calculate the term $25.98259\; \backslash times\; 21.01$:

$25.98259\; \backslash times\; 21.01\; =\; 545.8942159$

Now, substitute this back into the equation:

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

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 to solve for x:

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

$0.99675x\; =\; 9.2725255$

$x\; =\; \backslash frac\{9.2725255\}\{0.99675\}$

$x\; \backslash approx\; 9.300\backslash \%$

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

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

$\backslash eta\_3\; =\; 21.01\; -\; x\; =\; 21.01\; -\; 9.300\; =\; 11.71\backslash \%$

Therefore, the abundances of the other two isotopes are:

• $^{25}_{12}$Mg: $9.30\backslash \%$
• $^{26}_{12}$Mg: $11.71\backslash \%$

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 reaction where a neutron is ejected from the nucleus.

We will use the conversion factor: $1\; \backslash text\{\; u\}\; =\; 931.5\; \backslash text\{\; MeV\}$.

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

The process of removing a neutron from $^{41}_{20}$Ca can be represented by the nuclear reaction:

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

The mass defect ($\backslash Delta\; m$) for this reaction is the difference between the mass of the initial nucleus and the sum of the masses of the resulting nucleus and the neutron:

$$\backslash Delta\; m\; =\; m(^\{41\}\_\{20\}\backslash text\{Ca\})\; -\; [m(^\{40\}\_\{20\}\backslash text\{Ca\})\; +\; m(^\{1\}\_\{0\}\backslash text\{n\})]$$

Substitute the given masses:

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

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

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

The neutron separation energy ($E\_\{sep\}$) is the energy equivalent of this mass defect. Note that the separation energy is a positive value, representing the energy *required*. The negative mass defect here indicates that the initial nucleus is slightly less massive than the sum of its parts, which is unusual for stable nuclei and might indicate an error in the provided data or a misunderstanding of the definition in this context. However, following the standard procedure for separation energy calculation, we consider the energy required to break apart the nucleus.

Let's re-evaluate the calculation assuming the question implies the energy to *add* a neutron to $^{40}$Ca to form $^{41}$Ca, or the energy released when $^{41}$Ca *decays* by neutron emission. The definition states 'energy required to remove a neutron', which implies the mass of the products must be greater than the reactant for energy to be *required*. If the mass defect is negative, it means energy is released, not required. Let's assume the question intends to calculate the binding energy related to neutron removal, which is often positive.

A more common approach is to calculate the energy released when a neutron is *added* to form the heavier nucleus, or the energy required to *break* the heavier nucleus. The energy required to remove a neutron from $^{41}$Ca is the energy difference: $E\_\{sep\}\; =\; [m(^\{40\}\_\{20\}\backslash text\{Ca\})\; +\; m(^\{1\}\_\{0\}\backslash text\{n\})]\; -\; m(^\{41\}\_\{20\}\backslash text\{Ca\})$.

$$E\_\{sep\}\; =\; [39.962591\; \backslash text\{\; u\}\; +\; 1.008665\; \backslash text\{\; u\}]\; -\; 40.962278\; \backslash text\{\; u\}$$

$$E\_\{sep\}\; =\; 40.971256\; \backslash text\{\; u\}\; -\; 40.962278\; \backslash text\{\; u\}$$

$$E\_\{sep\}\; =\; 0.008978\; \backslash text\{\; u\}$$

Now, convert this mass defect to energy in MeV:

$$E\_\{sep\}\; =\; 0.008978\; \backslash text\{\; u\}\; \backslash times\; 931.5\; \backslash text\{\; MeV/u\}$$

$$E\_\{sep\}\; \backslash approx\; 8.363007\; \backslash text\{\; MeV\}$$

Thus, the neutron separation energy for $^{41}_{20}$Ca is approximately 8.363007 MeV.

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

The process of removing a neutron from $^{27}_{13}$Al can be represented by the nuclear reaction:

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

The neutron separation energy ($E\_\{sep\}$) is calculated as:

$$E\_\{sep\}\; =\; [m(^\{26\}\_\{13\}\backslash text\{Al\})\; +\; m(^\{1\}\_\{0\}\backslash text\{n\})]\; -\; m(^\{27\}\_\{13\}\backslash text\{Al\})$$

Substitute the given masses:

$$E\_\{sep\}\; =\; [25.986895\; \backslash text\{\; u\}\; +\; 1.008665\; \backslash text\{\; u\}]\; -\; 26.981541\; \backslash text\{\; u\}$$

$$E\_\{sep\}\; =\; 26.995560\; \backslash text\{\; u\}\; -\; 26.981541\; \backslash text\{\; u\}$$

$$E\_\{sep\}\; =\; 0.014019\; \backslash text\{\; u\}$$

Now, convert this mass defect to energy in MeV:

$$E\_\{sep\}\; =\; 0.014019\; \backslash text\{\; u\}\; \backslash times\; 931.5\; \backslash text\{\; MeV/u\}$$

$$E\_\{sep\}\; \backslash approx\; 13.0590085\; \backslash text\{\; MeV\}$$

Thus, the neutron separation energy for $^{27}_{13}$Al is approximately 13.059 MeV.