Atomic Nucleus Short Question And Answers
Question 1. If there is no electron in the nucleus then how does -emission take place from the nucleus?
Answer:
Inside a nucleus when a neutron is converted to a proton, an electron is produced. Hence, it comes out as a β -particle. The nuclear force does not influence this electron
Question 2. The nucleus of a radioactive element emits an α -particle and then emits 2 β -particles subsequently. Prove that the product (daughter nucleus) is an isotope of the original elements
Answer:
According to the law of radioactivity, if an α -particle is emitted from an element atomic number decreases by 2. Again, if β -particle is emitted it increases by 1. So if a particular nucleus of an element emits α – particle and then 2 β-particles, the change in its atomic number =(-2 +1 + 1) = 0.
Again isotopes are elements having equal atomic numbers but different mass numbers. So the daughter nucleus is an isotope of the parent one.
Question 3. What is meant by the statement that the half-life of radium is 1622 years
Answer:
This statement means that the amount of radium present in a radioactive sample will be halved after 1622 years due to radioactive disintegration. After another 1622 years, i.e., (1622 × 2) years from the present, the amount of radium will become \(\frac{1}{4}\) the present amount. In the past, the radioactive disintegration of radium has occurred in the same way.
Question 4. What will be the change in the ratio of neutrons to pro¬ tons of a nucleus if
- β -particle is emitted
- α – positron is emitted and
- ϒ -A ray photon is a melt
Answer:
- a β -particle i.e., an electron is emitted when a neutron changes into a proton and the ratio will decrease.
- Again a positron is emitted from the nucleus when a proton converts to a neutron and the ratio will increase.
- However, if a γ -ray photon is emitted the ratio of neutrons to protons will remain unchanged
Question 5. Why is a neutron used as an ideal particle for bombarding the nucleus of elements in a nuclear reaction?
Answer:
Since a neutron has no charge, it is not repelled by the positively charged nucleus. Therefore, even a very weak neutron can bombard a nucleus and initiate a nuclear reaction. Hence neutron is used for bombarding the nucleus in a nuclear reaction.
WBBSE Class 12 Atomic Nucleus Short Q&A
Question 6. What is the difference between a chemical reaction and a nuclear reaction?
Answer:
The valence electron of an atom takes part in a chemical reaction and no new element is formed in the process. In a nuclear reaction, the nucleus of the atom changes and results in the formation of an atom of another element.
The energy involved in a chemical reaction is less and ofthe order of eV. Whereas, in a nuclear reaction, the energy involved is very high and of the order of MeV
Question 7. What effect will be noticed when a source of α -particles is introduced in a charged gold leaf electroscope?
Answer:
The leaves will collapse very fast, α -particle ionizes the dry air in the electroscope, making it a good conductor, and the charge from the leaves flows to earth through this conducting air.
Question 8. Is mass defect always positive or negative?
Answer:
Mass defect \(\Delta m=\frac{\Delta E}{c^2}\) The binding energy AE of every nucleus is positive; consequently, the mass defect is positive without any exception.
Question 9. State radioactive decay law. Write down the relation between, the radius of the nucleus and the mass number of an atom
Answer:
Radioactive law states that the number of nuclei undergoing decay per unit time at any instant is proportional to the total number of nuclei in the sample at that instant
Question 10. Draw the variation of binding energy per nucleon with the mass number of atoms and indicate the stable and unstable
Answer:
The almost horizontal region in the middle portion of the graph indicates the stable region. The region to the extreme left and the sloping region to the extreme right indicate unstable regions. regions on the diagram
Question 11. Write down the equation of 0 -decay. Why is the detection of neutrinos difficult?
Answer:
X (parent) → Y (daughter) + β(β- electron) + ν– (antineutrino)
The charge and mass of both neutrinos and anti-neutrinos are zero. Hence it is experimentally difficult to detect them.
Question 12. In a nuclear decay, a nucleus emits one α -particle and then two β -particles one after another. Show that the final nucleus is an isotope ofthe formed nucleus.
Answer:
The atomic number decreases by 2 during or α – decay and it increases by 1 for each β -decay. So the atomic numbers of both P and S are equal to Z and hence they are isotopes of each other.
Short Answer Questions on Nuclear Physics
Question 13. R = R0A1/3 RQ = constant A = mass number), R = The option is the correct radius of the nucleus. Taking the relation shows that the nuclear density does not depend on mass number A
Answer:
The volume of a nucleus = \(\frac{4}{3} \pi R^3=\frac{4}{3} \pi R_0^3 A\)
Mass = Au
Density = \(\frac{A}{\frac{4}{3} \pi R_0^3 A}=\frac{3}{4 \pi R_0^3}\) it does not depend on A
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Atomic Nucleus Conclusion
1. The Law of conservation of mass-energy:
The total amount of mass and energy remains constant in the universe. Different types of transformation among them are possible but the creation or destruction of mass energy is not possible.
2. Nuclear force:
Inside the nucleus protons and neutrons are held together tightly due to a tremendous force of attraction between them. This force is called nuclear force.
3. Unified atomic mass unit:
The mass of part of one carbon- 1/12 atom is called 1 unified atomic mass unit (u). 4 Nuclear mass: Subtracting the mass of electrons present in an atom from the mass ofthe atom, the nuclear mass of that atom can be obtained.
4. Mass number:
The nearest whole number ofthe mass of an atom in the unified atomic mass unit is called the mass number of that atom. The sum of the number of protons and neutrons present in the nucleus of an atom is equal to its mass number
5. Isotopes:
The atoms of the same element having different mass numbers are called isotopes of that element
6. Isobars:
The atoms of the different elements having equal mass numbers are called isobars.
7. Isotones:
The atoms of the different elements having an equal number of neutrons in them are called isotones.
Conceptual Questions on Fission and Fusion
8. Atomic number:
The number of protons present inside the nucleus of an atom of an element is called the atomic number of that element
9. The magnitude of nuclear density is nearly 2 × 1014 g. cm-3
10. Radioactivity is a nuclear phenomenon of an element Due to radioactivity one element is converted into another ele¬ element
11. From a radioactive element α,β, and γ-rays are emitted.
12. No radioactive isotope can emit α,β and γ -rays simulta¬ neously
13. Half-life:
The time after which radioactive atoms present in a radioactive sample become half of its initial amount due to disintegration is called half-life.
14. Average life:
The time required for the number of a radioactive sample to fall to its initial number of atoms is called the average life of the radioactive element.
15. Artificial radioactivity:
When a naturally stable element is transformed into its unstable isotope by artificial means and the radioactive disintegration of that isotope in a natural way is possible then this phenomenon is known as artificial radioactivity.
16. Nuclear fission:
The phenomenon of splitting a heavy nucleus into two relatively lighter nuclei of comparable masses is known as nuclear fission.
17. Nuclear fusion:
The phenomenon of a combination of two or more lighter nuclei to form a heavy nucleus is called nuclear fusion.
18. E = mc² [c = velocity of light in vacuum]
19. 1 eV = 1.6 × 10-12 erg = 1.6 × 10-19 J
20. lu = 1.66 × 10-24 g = 931.2 MeV
21.
- ZXA → Z-2 Y A-4 + 2He4
- ZXA → Z+1 Y A + -1e0
22. N = N0 e-λt [N0 = number of radioactive atoms at t = 0, λ = number of radioactive atoms after time t = disintegration constant]
23. T \(\frac{0.693}{\lambda}\) T = half life]
24. N = \(\frac{N_0}{2^{t / T}}\)
25. τ = \(\frac{1}{\lambda}=1.443 T\)
τ = Average life
26. N = \(\frac{N_0}{e^{t / \tau}}\)