Radioactive Decay MCQ Quiz in मल्याळम - Objective Question with Answer for Radioactive Decay - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

CONCEPT:

• The half-life of a radioactive element (T1/2): The time interval in which mass of a radioactive substance or the number of atoms reduced to half of its initial value.
• The expression for the half-life is

\({T_{\frac{1}{2}}} = \frac{{0.693}}{\lambda }\)
Where λ = is the decay rate constant

EXPLANATION:

• The half-life is given by the formula:

\(\Rightarrow {{\rm{T}}_{1/2}} = \frac{{{\rm{ln\;}}2}}{{\rm{\lambda }}}= \frac{{0.693}}{\lambda }\)

• The half-life of a radionuclide is equal to natural log of 2 / disintegration constant λ. Therefore option 1 is correct.

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 on.
  • That’s the reason why atoms of Plutonium, Uranium are extremely unstable and show radioactivity.
• Half-life: Half-life is the time period after which half of the quantity of the radioactive element gets decayed.

CALCULATION:

Given Half-life = 2 years, total time = 6 years, and let the initial amount of radioactive substance (N_o) = 1

Since half-life is 2 years so 6 years is equal to 3 half-lives.

• We know that the fraction of substance remains after n half-lives,

\(\Rightarrow N'=\frac{N_{o}}{2^{n}}\)

Where N = amount remaining after n half-lives and N_o = initial amount

Let 

\(\Rightarrow N'=\frac{N_{o}}{2^{3}}\)

\(\Rightarrow N'=\frac{N_{o}}{8}=\frac{1}{8}\)

• Hence, option 2 is correct.

CONCEPT:​

• Half-life: The time required for a quantity to reduce to half of its initial value is called the half-life.

\(T_{1/2}=\frac{ln\space2} {λ }\)

where \(T_{1/2}\) is half-life and λ is radioactive decay constant.​

CALCULATION:

Given: \(\lambda=6.93\times 10^{-3}\)

• Half-life,

​ \(\Rightarrow T_{1/2}=\frac{ln\space2} {λ }\)

\(\Rightarrow T_{1/2}=\frac{ln\space2} {6.93\times 10^{-3} }=\frac{0.693} {6.93\times 10^{-3} }=10^2year\)

So the correct answer is option 2.

The correct answer is option 2) i.e. an isotope of X

CONCEPT:

Radioactive decay: Radioactive decay is the spontaneous breakdown of the nucleus of an atom along with the release of energy and matter from the nucleus.

Alpha Decay: The radioactive decay in which the parent nucleus breakdown into a daughter nucleus and an alpha particle. 

The equation for α-decay is given by:

 \(_{Z}^{A}\textrm{X}\rightarrow _{Z-4}^{A-4}\textrm{Y}+_{2}^{4}{α}\)

Beta Decay: The radioactive decay in which the parent nucleus emits an electron and an antineutrino and converts a neutron to a proton.

The equation for β-decay is given by:

 \(_{Z}^{A}\textrm{X}\rightarrow _{Z+1}^{A}\textrm{Y}+e^{-} +\bar{\nu }\)

Where X is the parent nucleus, Y is the daughter nucleus, A is the atomic mass, Z is the atomic number, e- is the electron, \(\bar\nu\) is the antineutrino.

EXPLANATION:

Let \(_{Z}^{A}X\) be the nucleus.

After α decay, the resultant nucleus will be \(_{Z-2}^{A-4}X\)

Further, \(_{Z-2}^{A-4}X\) undergoes β decay two times. Hence, the resultant nucleus will be \(_{Z-2+1+1}^{A-4}X = _{Z}^{A-4}X\)

Therefore, the atomic number of the resultant nucleus remains unchanged. Hence, the resultant nucleus is an isotope of X.

Additional Information

Concept:

Radioactivity:

• It is the emission of energy in the form of ionizing radiation.
• Radioactivity is the phenomenon exhibited by an atom’s nuclei due to nuclear instability.
• In 1896, Henry Becquerel discovered this phenomenon.
• Radioactivity is a process by which the nucleus of an unstable atom loses energy by emitting radiation
• There are three types of decay: Alpha, Beta, and Gamma.
• ​Radioactive decay law: N = N₀ e^{-λ t} where λ = disintegration constant or decay constant, N₀ = number of radioactive nuclei in the sample at time t= 0, N = number of radioactive nuclei in the sample at time t
• The half-life of the sample is, \(t_{1/2}=\frac{ln2}{λ}\) where λ = decay constant

Calculation:

Given,

The radioactive decay constant, λ = 0.00693 per year

The half-life of the substance, \(t_{1/2}=\frac{ln2}{\lambda}\)

\(t_{1/2}=\frac{0.693}{0.00693} = 100 ~years\)

Hence, the half-life of the substance is 100 years.

Concept:

Radioactive decay: 

When the n/p ratio of an atomic nucleus is high enough to make it unstable, it decays. Radioactive decay mainly emits α- particles, β- particles, and gamma rays.

What happens when an α- particle is emitted:

ZXA → Z-2XA-4 + 2α4

Mass number A to (A - 4) and atomic number Z to (Z - 2).

What happens when a β- particle is emitted:

ZXA —→ Z+1XA + -1β0

The mass number will remain the same A and atomic number Z to (Z + 1).

Calculation:

Given:

α-decay:

\(_Z^AX \rightarrow _{Z-2}^{A-4}X\;+\; _{2}^{4}α\)

1st β-decay:

\(_{Z-2}^{A-4}X \rightarrow_{Z-1}^{A-4}X\;+\;^{0}_{-1}β\)

2nd β-decay:

\(_{Z-2}^{A-4}X \rightarrow_{Z}^{A-4}X\;+\;^{0}_{-1}β\)

Additional Information

What happens when γ- rays are emitted:

• Gamma decay is the emission of energy in the form of photons from the nucleus of the atom.
• Gamma rays are electromagnetic waves.
• They are pure energy. They have no mass or atomic number.

CONCEPT:

• Radioactive decay: When the n/p ratio of an atomic nucleus is high enough to make it unstable, it decays.
• Radioactive decay emits mainly alpha α, beta β, and gamma γ rays.
• Alpha Emission:
  • An alpha particle is identical to a helium nucleus, is made up of two protons and two neutrons bound together.
  • Its atomic number decreases by two and mass number or atomic mass decreases by four.
  • Atomic number Z to Z-2 and mass number A to A-4.
• Beta Emission: 
  • Nucleon becomes proton and electron, the electron leaves the atom as a beta particle but the proton stays in the nucleus.
  • Atomic number increases by 1 but the mass number or atomic mass stays the same. 
  • Atomic number Z to Z+1 and Mass number will remain the same A. 
• Gamma emission:
  • The emission of energy from the nucleus of the atom in the form of photons.
  • Gamma rays are electromagnetic waves.
  • They are pure energy. They have no mass or atomic number.

EXPLANATION:

• When a nucleus is decayed due to radioactivity, it emits α- particles, β- particles, and gamma rays.

• Gamma decay will have no effect on mass or atomic number.
• Gamma rays are pure energy waves.
• It emits in the form of packets of energy.
• So the correct answer is option 3.

CONCEPT: 

• Radioactive decay is the spontaneous disintegration of the unstable nuclei into stable nuclei by emitting spontaneous radiations like alpha, beta
• Gamma decay is a type of radioactive decay in which the nucleus drops from a higher energy level to a lower energy level by emitting a photon   
• No new elements are created in gamma decay 

EXPLANATION: 

•  Gamma decay is the way through which the nucleus drops from a high energy level to a lower energy level through the emission of high energy photon. Hence, option 3 is the answer

Concept:

• Alpha rays (α): Alpha ray, also called alpha particles alpha ray, contains two protons and two neutrons bound together into a particle just like a helium-4 nucleus.
• Beta rays (β): Beta rays, also called beta particles or beta radiation, maybe a high-energy, high-speed electron or positron emitted by the radioactive decay of an atomic nucleus during the process of beta decay.
• Gamma rays (γ): A gamma-ray, or gamma radiation, is penetrating electromagnetic radiation arising from the radioactive decay of atomic nuclei.

Radioactivity:

• The phenomenon of spontaneous emission of radiations by heavy elements is called radioactivity.
• The elements which show this phenomenon is called radioactive elements.

Nuclear Radiations:

• According to Rutherford's experiment when a sample of a radioactive substance is put in a lead box and allows the emission of radiation through a small hole only.
• When the radiation enters into the external electric field, it split into three parts (α – rays, β – rays, and γ – gamma rays).

Explanation:

• α rays have high ionization power → Correct.
• Isotopes are always radioactive → Incorrect.
• β rays are deflected towards the cathode when the electric field is applied → Incorrect.
• Gamma rays are negatively charged particles → Incorrect.

 Important Points

• The ability of radiation to damage molecules is known as ionizing power.
• The ability of each type of radiation to pass through matter is expressed in terms of penetration power.
• we know radiation is caused because of radioactivity and the table below represents the penetration power and ionization power of different radiation

• Penetrating powers γ – rays > β – rays > α -rays

CONCEPT:

• Half-life: The time required for a quantity to reduce to half of its initial value is called the half-life.

\(T_{1/2}=\frac{0.693} {λ }\)

where \(T_{1/2}\) is half-life and λ is radioactive decay constant.

• Mean life: The average lifetime of all the nuclei of a particular unstable atomic species is called mean life.

\(τ=\frac{1}{λ}\)

\(τ=\frac{T_{1/2}}{0.693}\)

where τ is mean life, λ is radioactive decay constant, and \(T_{1/2}\) is the half-life of the element.

CALCULATION:

Given that 100 elements emit in 2 sec and 50 elements and next 2 sec,

50 elements are just half of 100 elements. So,

Half-life \(T_{1/2}\) = 2 sec

Mean life \(τ=\frac{T_{1/2}}{0.693}\)

\(τ=\frac{2}{0.693}\) sec

So the correct answer is option 3.