Q1:
In the nuclear emission stated above, the mass number and atomic number of the product Q respectively, are (NEET 2024)
(a) 280, 81
(b) 286, 80
(c) 288, 82
(d) 286, 79
Ans: (d) 286, 79
To determine the mass number and atomic number of the final product Q in the provided nuclear reactions, we need to analyze the impact of each type of decay (alpha, positron emission, beta-minus decay, and electron capture) on the mass number and atomic number of the initial element
1. Alpha decay (α decay): In alpha decay, an alpha particle (which is a nucleus) is emitted. This reduces the mass number by 4 units and the atomic number by 2 units.
2. Beta plus decay (positron emission, e+): During positron emission, a proton in the nucleus is transformed into a neutron, and a positron is emitted. This decreases the atomic number by 1 but does not change the mass number.
3. Beta-minus decay (β− decay): In a beta-minus decay, a neutron in the nucleus converts into a proton and an electron (beta particle) and an antineutrino are emitted. This results in an increase in the atomic number by 1, while the mass number remains unchanged.
4. Electron capture (e− capture): During electron capture, an atomic electron is absorbed by the nucleus, causing a proton to convert into a neutron. This process decreases the atomic number by 1 without altering the mass number.
From the calculations above, the mass number of Q is 286, and its atomic number is 79.
D: Mass number = 286, Atomic number = 79
Q2: The ratio of nuclear densities and nuclear volumes of and
are, respectively: (NEET 2024)
(a) 13 : 1 and 14 : 1
(b) 14 : 1 and 1 : 1
(c) 1 : 1 and 14 : 1
(d) 1 : 1 and 13 : 1
Ans: (c)
Nuclear density is nearly constant for all nuclei because it depends on mass and volume, and both increase proportionally with the mass number.
So, the ratio of nuclear densities of Fe-56 to He-4 is 1:1.
The volume of a nucleus is proportional to its mass number.
Fe has mass number 56 and He has 4.
Therefore, the volume ratio of Fe to He is 56:4, which simplifies to 14:1.
Q3: An excited heavy nucleus emits radiations in the following sequence: (NEET 2024)
Where Z and A are the atomic and mass number of element X, respectively. The possible emitted particles or radiations in the sequence, respectively are:
(a) e⁺, α, e⁻, α, γ
(b) e⁻, α, e⁺, α, γ
(c) α, e⁻, α, e⁺, γ
(d) α, e⁺, α, e⁻, γ
Ans: (c)
Q4: Water is used as a coolant in a nuclear reactor because of its: (NEET 2024)
(a) High thermal expansion coefficient
(b) High specific heat capacity
(c) Low density
(d) Low boiling point
Ans: (b)
Water is an excellent coolant because it has a high specific heat capacity, meaning it can absorb and carry away a large amount of heat without undergoing a significant increase in temperature. This makes it very effective for removing the heat produced during nuclear fission in reactors. Options like low density or low boiling point would not be ideal for a coolant.
Q5: Select the correct statements among the following: (NEET 2024)
A. Slow neutrons can cause fission in more effectively than fast neutrons.
B. α-rays are helium nuclei.
C. β-rays are fast-moving electrons or positrons.
D. γ-rays are electromagnetic radiations of wavelengths larger than X-rays.
Choose the most appropriate answer from the options given below:
(a) A, B, and C only
(b) A, B, and D only
(c) A and B only
(d) C and D only
Ans: (a)
Statement A is correct: Slow (thermal) neutrons are more effective in inducing fission in Uranium-235 because they are more likely to be captured by the nucleus.
Statement B is correct: Alpha particles (α-rays) are indeed nuclei of helium atoms (2 protons and 2 neutrons).
Statement C is correct: Beta particles (β-rays) are either electrons (β⁻) or positrons (β⁺), moving at high speeds.
Statement D is incorrect: γ-rays have shorter wavelengths (higher energy) than X-rays, not longer.
Q.1. The half life of a radioactive substance is 20 minutes. In how much time, the activity of substance drops to (1/16)th of its initial value?
(a) 20 minutes
(b) 40 minutes
(c) 60 minutes
(d) 80 minutes
Ans: (d) 80 minutes
Solution:
(a)
(b)
(c)
(d)
Ans: (b)
Solution:
Q.3. A nucleus with mass number 240 breaks into fragments each of mass number 120, the binding energy per nucleon of unfragmented nuclei is 7.6 MeV while that of fragments is 8.5 MeV. The total gain in the Binding Energy in the process is:
(a) 804 MeV
(b) 216 MeV
(c) 0.9 MeV
(d) 9.4 MeV
Ans: (b) 216 MeV
Solution:
Given binding energy per nucleon of X, Y & Z are 7.6 MeV, 8.5 MeV & 8.5 MeV respectively.
Gain in binding energy is :-
Q = Binding Energy of products – Binding energy of reactants
= (120 × 8.5 × 2) – (240 × 7.6) MeV
= 216 MeV
Q.4. A radioactive nucleus undergoes spontaneous decay in the sequence
Where Z is the atomic number of element X. The possible decay particles in the sequence are :
(a) β+,α,β-
(b) β-,α,β+
(c) α,β-,β+
(d) α,β+,β-
Ans: (a) β+,α,β-
Solution:β+ decreases atomic number by 1
α decreases atomic number by 2
β- decreases atomic number by 1
Q.5. The half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be:
(a) 2/3
(b) 2/3√2
(c) 1/2
(d) 1/2√2
Ans: (d) 1/2√2
Solution:
Q.6. The energy equivalent of 0.5 g of a substance is :
(a) 1.5×1013 J
(b) 0.5×1013J
(c) 4.5×1016J
(d) 4.5×1013J
Ans: (d) 4.5×1013J
Solution:
E = Δmc2 = 0.5 × 10-3 × (3 × 108)2
= 0.5 ×10-3 × 9 ×1016
E = 4.5 ×1013 J
Q.7. When a uranium isotopeis bombarded with a neutron, it generates
three neutrons and :
(a)
(b)
(c)
(d)
Ans: (c)
Solution:
92 = 36 + Z; ← A + 89 + 3 = 235 ⇒ A = 144
Z = 56
Q.8. For a radioactive material, half-life is 10 minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is:
(a) 20
(b) 10
(c) 30
(d) 15
Ans: (a) 20
Solution:
Q.9. Radioactive material 'A' has decay constant '8 λ' and material 'B' has decay constant 'λ'. Initially they have same number of nuclei. After what time, the ratio of number of nuclei of material 'B' to that 'A' will be 1/e?
(a) 1/7λ
(b) 1/8λ
(c) 1/9λ
(d) 1/10λ
Ans: (a) 1/7λ
Solution:
Q.10. The Binding energy per nucleon of nuclei are 5.60 MeV and 7.06 MeV, respectively. In the nuclear reaction
the value of energy Q released is :
(a) 8.4 MeV
(b) 17.3 MeV
(c) 19.6 MeV
(d) −2.4 MeV
Ans: (b) 17.3 MeV
Solution:
Q.11. A radio isotope ‘X’ with a half life 1.4 × 109 years decays to ‘Y’ which is stable. A sample of the rock from a cave was found to contain ‘X’ and ‘Y’ in the ratio 1 : 7. The age of the rock is
(a) 4.20 × 109 years
(b) 8.40 × 109 years
(c) 1.96 × 109 years
(d) 3.92 × 109 years
Ans: (a) 4.20 × 109 years
Solution:
Ratio of X:Y is given = 1:7
That is,
⇒ 7mx = my
Let, the initial total mass is m.
Therefore, time taken to become 1/8 unstable part
= 3 x T1/2
= 3 x 1.4 x 109
= 4.2 x 109 y
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