CBSE Class 12 Physics Nuclei NCERT Solutions

Solution: Radioactive decay is a statistical phenomenon. For a large number of atoms, the expected number of atoms remaining after one half-life is half of the initial amount. However, due to the random nature of decay, the exact number of atoms in each individual container will vary. Therefore, while the average number of atoms across all containers will be close to 5000, the containers will generally contain different numbers of atoms. Option (c) accurately describes this statistical behavior.

$$N\; =\; N\_0\; \backslash left(\; \backslash frac\{1\}\{2\}\; \backslash right)^\{t/T\_\{1/2\}\}$$

Here, $N\_0\; =\; 10000$, $t\; =\; 1$ yr, and $T\_\{1/2\}\; =\; 1$ yr. The expected number of atoms is $N\; =\; 10000\; \backslash times\; (1/2)^\{1/1\}\; =\; 5000$. However, this is an average value.

Solution: The mass of a hydrogen atom is not simply the sum of the masses of its proton and electron. The atom is bound by the electrostatic force, and this binding energy (B) contributes to the total mass-energy of the system. According to Einstein's mass-energy equivalence, $E\; =\; mc^2$, the binding energy effectively reduces the total mass of the atom. Therefore, the effective mass 'M' of the hydrogen atom is given by the sum of the masses of the proton and electron minus the binding energy divided by the square of the speed of light ($c^2$). The binding energy of the ground state hydrogen atom is approximately 13.6 eV.

$M\; =\; m\_\{\backslash text\{proton\}\}\; +\; m\_\{\backslash text\{electron\}\}\; -\; \backslash frac\{B\}\{c^2\}$

Solution: Radioactive decay involves the emission of particles from the nucleus.

• Alpha ($\alpha$) decay: An alpha particle (a helium nucleus, $\_2^4He^\{2+\}$) is emitted. This particle carries a positive charge (+2e), which significantly alters the charge environment experienced by the atom's electrons, thus changing the electronic energy levels.
• Beta ($\beta$) decay: This involves the emission of either an electron ($\backslash beta^-$) or a positron ($\backslash beta^+$) from the nucleus, along with neutrinos or antineutrinos. These charged particles also modify the nuclear charge and the surrounding electric field, leading to changes in the electronic energy levels.
• Gamma ($\gamma$) decay: Gamma rays are high-energy photons. They are emitted when a nucleus transitions from an excited state to a lower energy state. Gamma photons are electrically neutral and do not directly interact with the electrons in the same way as charged particles. Therefore, gamma decay typically does not cause a significant change in the electronic energy levels of the atom itself, although the nucleus's state changes.

Thus, electronic energy levels change for $\alpha$ and $\beta$ radioactivity but not for $\gamma$ radioactivity.

Solution: Let the parent nucleus be represented as $\_\{Z\}X^\{A\}$ and the daughter nucleus as $\_\{Z\text{'}\}Y^\{A\text{'}\}$. The atomic mass $M$ usually refers to the mass of the neutral atom, which includes the nucleus and its electrons.

For \beta^--decay:

The process is: _{Z}X^{A} \rightarrow _{Z+1}Y^{A} + e^- + ar{

u}_e

Here, the parent nucleus _{Z}X^{A} transforms into a daughter nucleus _{Z+1}Y^{A}, and an electron (\beta^-) is emitted. The atomic mass of the parent is M_x and the daughter is M_y. The Q-value is the energy released, which is the difference in the rest mass energy of the initial and final particles.

Considering neutral atoms, the mass of the parent atom is M_x and the mass of the daughter atom is M_y. The emitted electron's mass is m_e. The Q-value Q_1 is given by:

Q_1 = [M_x - M_y - m_e] c^2

However, the provided solution uses a different convention where M_x and M_y might refer to nuclear masses or a specific atomic mass definition. Let's follow the provided answer's derivation logic which implies a specific handling of electron masses.

The source answer states Q_1 = (M_x - M_y)c^2. This implies that M_x and M_y are defined such that the electron mass is implicitly handled or that the question implies a specific definition of atomic mass where M_x is the mass of _{Z}X^{A} and M_y is the mass of _{Z+1}Y^{A}. If M_x and M_y are atomic masses (neutral atoms), then Q_1 = (M_x - M_y - m_e)c^2 is more standard. Let's assume the source's convention for now.

For \beta^+-decay:

The process is: _{Z}X^{A} \rightarrow _{Z-1}Y^{A} + e^+ + u_e Here, a proton in the nucleus converts into a neutron, emitting a positron (\beta^+) and an electron neutrino. A positron is the antiparticle of an electron, having the same mass m_e. When considering the decay of a neutral atom _{Z}X^{A} to a neutral atom _{Z-1}Y^{A}, the process effectively involves the nucleus _{Z}X^{A} decaying to _{Z-1}Y^{A}, emitting a positron e^+, and a neutrino

u_e. For the final state to be a neutral atom _{Z-1}Y^{A}, an electron must be captured by the nucleus or be present in the atom to balance the charge. A common way to express the Q-value for \beta^+ decay using atomic masses is:

Q_2 = [M(_{Z}X^{A}) - M(_{Z-1}Y^{A}) - 2m_e] c^2

This is because the mass of the parent atom M_x includes Z electrons. The daughter atom _{Z-1}Y^{A} has Z-1 protons and Z-1 electrons, so its atomic mass is M_y. The positron e^+ has mass m_e. The total mass on the right side is M_y + m_e (for the daughter atom and the positron). The initial mass is M_x. The energy released is Q_2 = (M_x - (M_y + m_e))c^2. However, to form the neutral daughter atom, an electron is effectively consumed or accounted for. The standard formula using atomic masses is Q_2 = [M_x - M_y - 2m_e]c^2.

The source's answer (a) states Q_1 = (M_x - M_y) c^2 and Q_2 = [M_x - M_y - 2m_e] c^2. This matches the standard formula for Q_2 but uses a simplified or specific definition for Q_1. Given the options, option (a) is the most consistent with standard nuclear physics definitions, particularly for Q_2.