When a nucleus reverts from an excited state to the ground state, the energy difference between the — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Nuclear Energy Levels Just like electrons occupy discrete quantized energy levels around the nucleus, nucleons (protons and neutrons) inside the nucleus also reside in discrete nuclear energy levels. After a nuclear reaction or a radioactive decay ($\alpha$ or $\beta$), the daughter nucleus is often left in a high-energy excited state, denoted $\ce{^A_Z X^*}$. Step 2 - De-excitation Releases γ-rays To reach stability (the ground state), the excited nucleus releases the excess energy as a photon: $$\ce{^A_Z X^* -> ^A_Z X + \gamma}$$ The energy difference: $\Delta E = h\nu$ Nuclear energy gaps are in the range $10^4$ eV to several MeV — far larger than atomic electron transitions (a few eV). This enormous energy difference means the emitted photons have: - Very high frequency $\nu$ - Very short wavelength $\lambda$ This high-energy electromagnetic radiation is called **γ-rays (gamma rays)**. Crucially, $A$ and $Z$ do not change during γ emission. Step 3 - Why Each Option is Correct or Incorrect - **(A) α-particle**: A helium-4 nucleus emitted by heavy nuclei to reduce their $n/p$ ratio. Emission reduces both $A$ and $Z$. Not emitted during simple nuclear de-excitation. - **(B) β-particle**: An electron or positron emitted when a neutron converts to a proton (or vice versa) to stabilize the $n/p$ ratio. Changes $Z$ but not $A$. Not emitted during de-excitation. - **(C) γ-rays**: **Correct.** Pure electromagnetic energy released when an excited nucleus transitions to a lower energy state. No change in $A$ or $Z$. - **(D) neutrino**: A neutral lepton emitted alongside a $\beta$ particle to conserve energy and lepton number. Not emitted during simple de-excitation. $$\boxed{\text{Answer: C}}$$