What is the respective number of and -particles emitted in the following radioactive decay?
1. 6 and 8
3. 6 and 6
3. 8 and 8
4. 8 and 6
A free neutron decays into a proton, an electron and:
1. A beta particle.
2. An alpha particle.
3. An antineutrino.
4. A neutrino.
| 1. | mass number reduced by \(2\) |
| 2. | mass number reduces by \(6\) |
| 3. | atomic number is reduced by \(2\) |
| 4. | atomic number remains unchanged |
A nuclear decay is expressed as:
\(_{6}^{11}\mathrm{C}\rightarrow _{5}^{11}\mathrm{B}+\beta^{+}+\mathrm{X}\)
Then the unknown particle \(X\) is:
1. neutron
2. antineutrino
3. proton
4. neutrino
| 1. | \({}_{7}^{13}\mathrm{N}\) | 2. | \({}_{5}^{10}\mathrm{B}\) |
| 3. | \({}_{4}^{9}\mathrm{Be}\) | 4. | \({}_{7}^{14}\mathrm{N}\) |
A nuclear reaction given by; \({ }_{Z}^{A} ~{X} \rightarrow{ }_{Z+1}^{A} {Y}+e^{-}+\bar{v}\) represents:
| 1. | fusion | 2. | fission |
| 3. | \(\beta^{-} \)decay | 4. | \(\gamma^{-}\)decay |
The nucleus absorbs an energetic neutron and emits a beta particle . The resulting nucleus is
1.
2.
3.
4.
A radioactive nucleus undergoes a series of decay according to the scheme
\(A \xrightarrow{\alpha} A_1 \xrightarrow{\beta} A_2 \xrightarrow{\alpha} A_3 \xrightarrow{\gamma} A_4\)
1. \(172~\text{and}~69\)
2. \(174~\text{and}~70\)
3. \(176~\text{and}~69\)
4. \(176~\text{and}~70\)
\(A\rightarrow B+_{2}^{4}\mathrm{He}\)
\(B\rightarrow C+2e^{-}\)
Then:
| 1. | \(A\) and \(C\) are isotopes |
| 2. | \(A\) and \(C\) are isobars |
| 3. | \(A\) and \(B\) are isotopes |
| 4. | \(A\) and \(B\) are isobars |
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