Beta Decay MCQ Quiz in मल्याळम - Objective Question with Answer for Beta Decay - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Mar 24, 2025
Latest Beta Decay MCQ Objective Questions
Top Beta Decay MCQ Objective Questions
Beta Decay Question 1:
During \(\beta\)-decay (beta minus), the emission of antineutrino particle is supported by which of the following statement(s)?
Answer (Detailed Solution Below)
Beta Decay Question 1 Detailed Solution
For (b) Through experiments it has been observed that direction of emitted electron and recoiling nuclei are almost never exactly opposite as required for linear momentum to be conserved, so there must be another particle balancing the momentum.
For (c) During \(\beta\)- decay, the energy of electron is found to vary continuously from \(0\) to a maximum value (this maximum value is a characteristic of nuclide). To explain this experimental observation, we also need some other particle.
Beta Decay Question 2:
The \(\beta\)-decay process, discovered around 1900, is basically the decay of a neutron \(n\). In the laboratory, a proton \(p\) and an electron \(e^{-}\) are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process, i.e. \(n\rightarrow p+e^{-}+\overline{v}_{e}\), around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino \(\overline{v}_{e}\) to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is \(0.8\times 10^{6}eV\). The kinetic energy carried by the proton is only the recoil energy.
What is the maximum energy of the anti-neutrino?
Answer (Detailed Solution Below)
Beta Decay Question 2 Detailed Solution
Beta Decay Question 3:
The \(\beta\)-decay process, discovered around 1900, is basically the decay of a neutron \(n\). In the laboratory, a proton \(p\) and an electron \(e^{-}\) are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process, i.e. \(n\rightarrow p+e^{-}+\overline{v}_{e}\), around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino \(\overline{v}_{e}\) to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is \(0.8\times 10^{6}eV\). The kinetic energy carried by the proton is only the recoil energy.
If the anti-neutrino had a mass of 3 \(eV/c^{2}\) (where \(c\) is the speed of light) instead of zero mass, what should be the range of the kinetic energy, \(K\), of the electron?
Answer (Detailed Solution Below)
Beta Decay Question 3 Detailed Solution
Beta Decay Question 4:
In the following equation representing β- decay, the number of neutrons in the nucleus X is
\({ }_{83}^{210} \mathrm{Bi} \rightarrow \mathrm{X}+\mathrm{e}^{-1}+\overline{\mathrm{v}}\)
Answer (Detailed Solution Below)
Beta Decay Question 4 Detailed Solution
Given:
The β-decay equation:
21083Bi → X + e-1 + v
Concept Used:
β-decay:
- In β-decay, a neutron transforms into a proton, emitting an electron and an antineutrino.
- The atomic number increases by 1 while the mass number remains the same.
Definitions:
- Neutron: A subatomic particle with no charge and a mass slightly greater than that of a proton.
- Proton: A positively charged subatomic particle found in the nucleus of an atom.
- Electron: A negatively charged subatomic particle.
Formula Used:
- Number of Neutrons = Mass Number - Atomic Number
Calculation:
The new element X after β-decay has:
Mass Number = 210
Atomic Number = 84
Number of Neutrons = 210 - 84
⇒ Number of Neutrons = 126
∴ The number of neutrons in the nucleus X is 126.
Beta Decay Question 5:
Mx and My denote the atomic masses of the parent and the daughter nuclei respectively in a radioactive decay. The Q-value for a β– decay is Q1 and that for a β+ decay is Q2. If me denotes the mass of an electron, then which of the following statements is correct?
Answer (Detailed Solution Below)
Beta Decay Question 5 Detailed Solution
Concept:
The energy released or absorbed during a reaction is known as Q value.
Q value = [ (sum of reactants) - (sum of products)] c2
Solution:
Suppose Mx and My denote atomic masses and daughter nuclei respectively. The Q values for β- and β+ decay are Q1 and Q2 respectively.
The nuclear reaction of β- decay is given by
⇒ AXZ → AYZ+1 + 0e-1 + ν¯ + Q1
Q1 = [(Mx - Zme) - (My -(Z+1)me) - me] c2
⇒ Q1 = [Mx -My]c2
Similarly, the nuclear reaction of β+ decay is given by
⇒ AXZ → AYZ-1 + 0e1 + ν¯ + Q2
The correct answer is option (1).
Beta Decay Question 6:
Which of the following particle is emitted by a radioactive element during the beta decay?
Answer (Detailed Solution Below)
Beta Decay Question 6 Detailed Solution
CONCEPT:
- Radioactivity: Radioactivity is a process by which the nucleus of an unstable atom loses energy by emitting radiation.
- Two forces, namely the force of repulsion that is electrostatic and the powerful forces of attraction of the nucleus acts in the nucleus. These two forces are considered extremely strong in nature. The instability of the nucleus increases as the size of the nucleus increases because the mass of the nucleus becomes a lot to concentrate. That’s the reason why atoms of Plutonium, Uranium are extremely unstable and show radioactivity.
- The three types of radioactive decay are classified as follows:
- Alpha Decay: The radioactive decay in which the ejected particle is an alpha particle.
- Beta Decay: Beta decay occurs in one of the two ways:
- Beta Minus Decay: When the nucleus emits an electron and an antineutrino in a process that changes a neutron to a proton
- Beta Plus Decay: When the nucleus emits a positron and a neutrino in a process that changes a proton to a neutron.
- Gamma Decay: A radioactive nucleus first decays by the emission of an α or β particle. The daughter nucleus that results is usually left in an excited state and it can decay to a lower energy state by emitting a gamma-ray photon.
EXPLANATION:
- Beta Decay: Beta decay occurs in one of the two ways:
- Beta Minus Decay: When the nucleus emits an electron and an antineutrino in a process that changes a neutron to a proton
- Beta Plus Decay: When the nucleus emits a positron and a neutrino in a process that changes a proton to a neutron.
- Hence, option 1 is correct.
Beta Decay Question 7:
The loss of one neutron is equivalent to:
Answer (Detailed Solution Below)
Beta Decay Question 7 Detailed Solution
Concept:
Explanation:
-
The loss of one neutron from an atom is equivalent to both the loss of one β-particle and the increase of one proton.
-
This is because a neutron can decay into a proton and an electron, with the electron being emitted as a β-particle.
-
The resulting atom will therefore have one less neutron and one more proton than the original atom. This process is known as beta decay.
- When an isotope has too many neutrons, it can attain greater nuclear stability if one of the neutrons decays to protons
- Such ejection leads to emission of a beta ray as follows:
0n1 → 1H1 + -1e0
- Beta ray emission occurs when the ratio \(n\over p\) is higher than that expected for stability.
Thus, the loss of one proton is equivalent to the ejection of one β particle.
Important Points
- Decay of one neutron will also mean generation of one proton.
Beta Decay Question 8:
Which of the following particle is emitted by a radioactive element during the beta decay?
Answer (Detailed Solution Below)
Beta Decay Question 8 Detailed Solution
CONCEPT:
- Radioactivity: Radioactivity is a process by which the nucleus of an unstable atom loses energy by emitting radiation.
- Two forces, namely the force of repulsion that is electrostatic and the powerful forces of attraction of the nucleus acts in the nucleus. These two forces are considered extremely strong in nature. The instability of the nucleus increases as the size of the nucleus increases because the mass of the nucleus becomes a lot to concentrate. That’s the reason why atoms of Plutonium, Uranium are extremely unstable and show radioactivity.
- The three types of radioactive decay are classified as follows:
- Alpha Decay: The radioactive decay in which the ejected particle is an alpha particle.
- Beta Decay: Beta decay occurs in one of the two ways:
- Beta Minus Decay: When the nucleus emits an electron and an antineutrino in a process that changes a neutron to a proton
- Beta Plus Decay: When the nucleus emits a positron and a neutrino in a process that changes a proton to a neutron.
- Gamma Decay: A radioactive nucleus first decays by the emission of an α or β particle. The daughter nucleus that results is usually left in an excited state and it can decay to a lower energy state by emitting a gamma-ray photon.
EXPLANATION:
- Beta Decay: Beta decay occurs in one of the two ways:
- Beta Minus Decay: When the nucleus emits an electron and an antineutrino in a process that changes a neutron to a proton
- Beta Plus Decay: When the nucleus emits a positron and a neutrino in a process that changes a proton to a neutron.
- Hence, option 1 is correct.
Beta Decay Question 9:
Neutrino is a particle emitted in ?
Answer (Detailed Solution Below)
Beta Decay Question 9 Detailed Solution
The Correct answer is β-decay.
Key Points
Beta Decay:
Neutrino is a subatomic particle with no electrical charge and a very small mass. They are difficult to detect because they have minimal interaction with matter. To detect neutrinos, very large and sensitive detectors are required.
Beta decay is a process in which either a neutron is converted into a proton (β− decay) or a proton is converted into a neutron (β+ decay).
During β− decay, an electron and a new particle named antineutrino are created and emitted from the nucleus,
\(n \rightarrow p+e+\bar \nu\)
During β+ decay, a positron and a neutrino are created and emitted from the nucleus,
\(p \rightarrow n + e^+ + \nu\)
β− decay: \(_Z X^A \rightarrow _{Z+1}Y^A +e +\bar \nu\)
β+ decay: \(_Z X^A \rightarrow _{Z-1}Y^A +e^+ + \nu\)
Additional Information
- α-decay (Alpha Decay):
- In alpha decay, the nucleus emits an alpha particle, which consists of 2 protons and 2 neutrons (essentially a helium nucleus).
- Alpha decay is typically seen in heavy elements like uranium or radium. No neutrino is involved in this process.
- γ-decay (Gamma Decay):
- In gamma decay, a nucleus releases excess energy in the form of gamma rays (high-energy photons).
- Gamma decay usually follows other types of decay (like α or β decay) when the nucleus is in an excited state. Neutrinos are not involved in γ-decay.
Beta Decay Question 10:
What happens when a beta particle is emitted from a nucleus?
Answer (Detailed Solution Below)
Beta Decay Question 10 Detailed Solution
CONCEPT:
- What happens when a β- particle is emitted:
- The original atom is changed into a different element when a beta particle is emitted.
- Neutron in the nucleus becomes electron and neutron, the electron leaves the atom as a beta particle but the proton stays in the nucleus.
- The atomic number increases by 1 and the mass number or atomic mass stay the same.
- The mass number will remain the same A and atomic number Z to Z+1.
ZXA —→ Z+1XA + -1β0
EXPLANATION:
- When a beta particle is emitted from a nucleus, the atom mass remains the same.
- When a beta particle is emitted from a nucleus, the atomic number increases by 1.
- So the correct answer is option 1.