Question:
List-I shows different radioactive decay processes and List-II provides possible emitted particles. Match each entry in List-I with an appropriate entry from List-II, and choose the correct option.
List-I List-II (P) \(^{238}_{92}U → ^{234}_{91}Pa\) (1) one 𝛼 particle and one 𝛽+ particle (Q) \(^{214}_{82}Pb → ^{210}_{82}Pb\) (2) three 𝛽− particles and one 𝛼 particle (R) \(^{210}_{81}Tl → ^{206}_{82}Pb\) (3) two 𝛽− particles and one 𝛼 particle (S) \(^{228}_{91}Pa → ^{224}_{88}Ra\) (4) one 𝛼 particle and one 𝛽− particle (5) one 𝛼 particle and two 𝛽+ particles
List-I shows different radioactive decay processes and List-II provides possible emitted particles. Match each entry in List-I with an appropriate entry from List-II, and choose the correct option.
| List-I | List-II | ||
| (P) | \(^{238}_{92}U → ^{234}_{91}Pa\) | (1) | one 𝛼 particle and one 𝛽+ particle |
| (Q) | \(^{214}_{82}Pb → ^{210}_{82}Pb\) | (2) | three 𝛽− particles and one 𝛼 particle |
| (R) | \(^{210}_{81}Tl → ^{206}_{82}Pb\) | (3) | two 𝛽− particles and one 𝛼 particle |
| (S) | \(^{228}_{91}Pa → ^{224}_{88}Ra\) | (4) | one 𝛼 particle and one 𝛽− particle |
| (5) | one 𝛼 particle and two 𝛽+ particles |
Updated On: May 17, 2025
P →4, Q→ 3, R →2, S →1
P →4, Q→ 1, R →2, S →5
P →5, Q→ 3, R →1, S →4
P →5, Q→ 1, R →3, S →2
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The Correct Option is A
Solution and Explanation
To solve the problem, we match each radioactive decay process from List-I with the appropriate emitted particles from List-II by analyzing the changes in atomic and mass numbers for each transformation. Let's examine each decay:
- (P) \(^{238}_{92}U → ^{234}_{91}Pa\): The decrease in mass number from 238 to 234 indicates that an alpha particle (\(\alpha\), \(^{4}_{2}He\)) is emitted, reducing the atomic number by 2 and the mass number by 4. Additionally, the atomic number decreases further by 1 (not entirely from \(\alpha\)), suggesting the emission of a beta minus particle (\(\beta^-\)), which increases the atomic number by 1. Therefore, one \(\alpha\) and one \(\beta^-\) particle are emitted. This matches with (4) one \(\alpha\) particle and one \(\beta^-\) particle.
- (Q) \(^{214}_{82}Pb → ^{210}_{82}Pb\): The mass number decreases by 4 with no change in atomic number, suggesting the emission of an alpha particle (\(^{4}_{2}He\)), which also results in the release of two beta minus particles (\(\beta^-\)), since only alpha would reduce atomic number by 2. This matches with (3) two \(\beta^-\) particles and one \(\alpha\) particle.
- (R) \(^{210}_{81}Tl → ^{206}_{82}Pb\): The mass number decreases by 4 and atomic number increases by 1. This implies the emission of an alpha particle and three beta minus particles (\(\beta^-\)), as one would increase atomic number by four. This matches with (2) three \(\beta^-\) particles and one \(\alpha\) particle.
- (S) \(^{228}_{91}Pa → ^{224}_{88}Ra\): The mass number decreases by 4, and atomic number decreases by 3, indicating one alpha particle, reducing atomic number by 2 and perhaps a beta plus particle (\(\beta^+\)), reducing number by 1. This matches with (1) one \(\alpha\) particle and one \(\beta^{-}\) particle.
Thus, the correct matching is P → 4, Q → 3, R → 2, S → 1.
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Concepts Used:
Nuclear Physics
Nuclear physics is the field of physics that studies atomic nuclei and their constituents and interactions, in addition to the study of other forms of nuclear matter. Nuclear physics should not be confused with atomic physics, which studies the atom as a whole, including its electrons
Radius of Nucleus
‘R’ represents the radius of the nucleus. R = RoA1/3
Where,
- Ro is the proportionality constant
- A is the mass number of the element
Total Number of Protons and Neutrons in a Nucleus
The mass number (A), also known as the nucleon number, is the total number of neutrons and protons in a nucleus.
A = Z + N
Where, N is the neutron number, A is the mass number, Z is the proton number
Mass Defect
Mass defect is the difference between the sum of masses of the nucleons (neutrons + protons) constituting a nucleus and the rest mass of the nucleus and is given as:
Δm = Zmp + (A - Z) mn - M
Where Z = atomic number, A = mass number, mp = mass of 1 proton, mn = mass of 1 neutron and M = mass of nucleus.