Question

Match the LIST-I with LIST-II

LIST-I (Type of decay in Radioactivity)		LIST-II (Reason for stability)
A.	Alpha decay	III.	Nucleus is mostly heavier than Pb (Z=82)
B.	Beta negative decay	IV.	Nucleus has too many neutrons relative to the number of protons
C.	Gamma decay	I.	Nucleus has excess energy in an excited state
D.	Positron Emission	II.	Nucleus has too many protons relative to the number of neutrons

Choose the correct answer from the options given below:

• A. A - I, B - II, C - III, D - IV
• B. A - I, B - III, C - II, D - IV
• C. A - I, B - II, C - IV, D - III
• D. A - III, B - IV, C - I, D - II

Hint:

Think of nuclear decay as a nucleus's way of adjusting its proton-to-neutron ratio to reach the "valley of stability". - Too heavy? \(\rightarrow\) Alpha decay. - Too many neutrons? \(\rightarrow\) Beta-minus decay. - Too many protons? \(\rightarrow\) Positron emission or electron capture. - Too much energy? \(\rightarrow\) Gamma decay.

Answer 1

The Correct Option is D

Step 1: Analyze each type of radioactive decay. - A. Alpha decay: The emission of an alpha particle (\(^{4}_{2}\text{He}\)) reduces the mass number by 4 and the atomic number by 2. This is the most common decay mode for very heavy nuclei (typically with Z>82, i.e., heavier than Lead) to become more stable. This matches III. - B. Beta negative decay: A neutron transforms into a proton, emitting an electron (\(e^-\)) and an antineutrino. This process (\(n \to p + e^- + \bar{\nu}_e\)) increases the proton number and decreases the neutron number. It occurs in nuclei that have an excess of neutrons compared to protons (a high N/Z ratio). This matches IV. - C. Gamma decay: An excited nucleus transitions to a lower energy state by emitting a high-energy photon (gamma ray). The number of protons and neutrons does not change. This occurs when a nucleus has excess energy, often after a previous alpha or beta decay. This matches I. - D. Positron Emission (\(\beta^+\) decay): A proton transforms into a neutron, emitting a positron (\(e^+\)) and a neutrino. This process (\(p \to n + e^+ + \nu_e\)) decreases the proton number and increases the neutron number. It occurs in nuclei that have an excess of protons compared to neutrons (a low N/Z ratio). This matches II.
Step 2: Combine the matches to find the correct sequence. - A \(\rightarrow\) III - B \(\rightarrow\) IV - C \(\rightarrow\) I - D \(\rightarrow\) II This sequence is A - III, B - IV, C - I, D - II, which corresponds to option (4).