Atoms, Nuclei & Radioactivity Test - 1

Concept:

There are three known isotope of hydrogen 1H (protium), 2H (deuterium), and 3H (tritium). 

Explanation

From the above explanation, we can see that the hydrogen atom has three isotopes  Protium, Deuterium & Tritium.

Extra points:

Protium:

The most common is the ordinary hydrogen usually called protium. It has one proton in the nucleus and an electron revolving around it.

Protium is a selective “proton pump inhibitor”, a medicine which reduces the amount of acid produced in your stomach. It is used for treating acid-related diseases of the stomach and intestine.

Deuterium:

The second isotope of hydrogen is called heavy hydrogen or deuterium. It consists of one proton and one neutron in the nucleus and electron revolving around it.

Deuterium is not radioactive and does not represent a significant toxicity hazard. Water enriched in molecules that include deuterium instead of protium is called Heavy water. Deuterium and its compounds are used as a non-radioactive label in chemical experiments and insolvent.

It is widely used in prototype fusion reactors and has their application in military, industrial and scientific fields.

In nuclear fusion reactors, it is used as a tracer and it is responsible to slow down neutrons in heavy water moderated fission reactors.

Tritium:

The third isotope of hydrogen is called tritium. It consists of one proton and two neutrons in the nucleus and an electron revolving around it.

The trace amount of tritium occurs naturally because of the interaction of cosmic rays with atmospheric gases. Tritium has also been released during nuclear weapon tests. The most common methods of producing tritium are by bombarding a natural isotope of lithium, lithium-6 with neutrons in a nuclear reactor.

CONCEPT:

• Nuclear fusion is a reaction in which two or more atomic nuclei come closer enough to form one or more different atomic nuclei and subatomic particles (neutrons or protons).
• A hydrogen bomb is an immensely powerful bomb whose destructive power comes from the rapid release of energy during the nuclear fusion of isotopes of hydrogen (deuterium and tritium), using an atom bomb as a trigger.
• Sun is the best example of nuclear fusion in which smaller nuclei of atoms fused at very high temperature and pressure into a larger nucleus.
   

\(\frac{2}{1}H+\frac{2}{1}H \xrightarrow{at\ high \Delta T}{} \frac{4}{2}He\)

EXPLANATION:

• Nuclear fusion in the sun is a type of nuclear reaction in which there is a collision of two or more atomic nuclei at high energy levels. There is a formation of a new nucleus.
• In the Sun, the temperature is around 15 Million kelvins and pressure are millions of times higher than the Earth's surface.
• At this high temperature and pressure, the Coulomb barrier is overcome and nuclear fusion takes place.
• This fusion happens inside the core of the sun and the energy later escapes to the surface of the sun.
• It is responsible for the generation of solar radiation.

• Rutherford Experiment: Rutherford with his co-workers initiated a series of groundbreaking experiments that completely changed the accepted model of the atom.
• Very thin sheets of gold foil were bombarded with fast-moving alpha particles.
  • Bombardment of alpha particles on the gold foil which showed that a very small percentage of alpha particles were deflected.
  • This experiment showed that 'The nuclear model of the atom consists of a small and dense positively charged interior surrounded by a cloud of electrons.

Figure (A) The experimental setup for Rutherford's gold foil

Figure (B) The plum pudding model (top) says that all of the alpha particles should have been passed through the gold foil with no deflection or very little deflection. Rutherford found that a small percentage of alpha particles were deflected at large angles, which could be explained by an atom with a very small, dense, positively-charged nucleus at its center (bottom).

EXPLANATION:

• In the Rutherford atomic model, alpha particles are bombarded with Very thin sheets of gold foil.
• Rutherford’s atomic model accounts for the positively charged central core of an atom.
• The correct option is 1.

Key Points

Bohr model of the atom was proposed by Neils Bohr.

• Bohr explained that a nucleus (positively charged) is surrounded by negatively charged electrons.
• According to Bhor, Electron which is moving in an orbital does not lose or radiate energy.
• Bohr was awarded the Nobel Prize in Physics in 1922.
• The Bohr model of an atom came as a modification to Rutherford’s model of an atom.
• Postulates of Bohr’s model of an atom are:
  • In an atom, electrons move in a fixed circular path around a positively charged nucleus. This path is known as orbits or shells
  • These orbits or shells have fixed energy.
  • The energy levels are represented by an integer (n = 1, 2, 3, 4, …) known as a quantum number.
  • If an electron moves from lower energy level to higher energy level, then it will do so by gaining energy and if it moves from higher energy level to lower energy level then it will do so by losing energy.

CONCEPT:

In the Bohr model, spectrum series are given below:

• The Lyman series: It includes the lines emitted by transitions of the electron from an outer orbit of quantum number n2 > 1 to the 1st orbit of quantum number n1 = 1.
  • All the energy wavelengths in the Lyman series lie in the ultraviolet band.
• The Balmer series: It includes the lines due to transitions from an outer orbit n₂ > 2 to the orbit n₁ = 2.
  • Four of the Balmer lines lie in the "visible" part of the spectrum.

Paschen series (Bohr series, n1 = 3)

Brackett series (n1 = 4)

Pfund series (n1 = 5)​

EXPLANATION:

• All the energy wavelengths in the Lyman series lie in the ultraviolet band.
• So the correct answer is option 4.​​

Additional Information

• When atoms are excited they emit light of certain wavelengths that correspond to different colors Due to the electron making transitions between two energy levels in an atom.
• The light emission can be seen as a series of colored lines, known as atomic spectra.
  • These atomic spectra are divided into a number of spectral series. The wavelength related to these series is given by the Rydberg formula.
  • The energy differences between levels in the Bohr model is given by the Rydberg formula. Hence the wavelengths of emitted/absorbed photons

\({\displaystyle {1 \over λ }=Z^{2}R_{∞ }\left({1 \over {n_{1}}^{2}}-{1 \over {n_{2}}^{2}}\right)}\)

CONCEPT:

• Binding Energy: The energy needed to break a nucleus into its constituent protons and neutrons.
• Binding energy per nucleon (BEN): Energy needs to remove a nucleon from a nucleus.
• Mass defect: Difference between the mass of a nucleus and the total mass of its constituent nucleons.

Binding Energy (B) = Δm × 931 MeV 

​Where Δm = mass defect.

CALCULATION:

Given - Binding Energy (B) = 2.23 MeV

The mass defect in amu is

⇒ B = Δm × 931 MeV 

\(\Rightarrow \Delta m = \frac{2.23}{931}=0.0024\, \, amu\)

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

The correct answer is α - Particle.

Key Points

• Polonium-215 → Lead-211 + alpha decay.
  • Polonium-215 has a half-life of 1.78 milliseconds and Lead-211 has a half-life of 36.1 minutes.
• A nuclear reaction is a process in which two nuclei or nuclear particles collide, to produce different products than the initial particles.

Important Points

• Plutonium decays mainly by emitting alpha radiation.
  • The emission of an alpha particle by a plutonium atom begins a series of radioactive decays, called a decay series.
• The half-life of Pu-238 is 90 years and 24,000 years for Pu-239.
• Plutonium-238 and plutonium-239 initially decay by alpha radiation, both are also associated with gamma radiation release.
• Plutonium-241 is initially associated with beta radiation and then later gamma radiation.

Additional Information

• Plutonium (Pu) is an element with atomic number 94. An atom of plutonium has 94 protons in its nucleus and 94 electrons orbiting the nucleus. Plutonium is a transuranic element.
• There are 18 different isotopes of plutonium.
• An important example of nuclear fission is the splitting of the uranium-235 nucleus when it is bombarded with neutrons.
• The two general kinds of nuclear reactions are nuclear decay reactions and nuclear transmutation reactions.

CONCEPT:

• Radioactive material: The material which can decay spontaneously is called as radioactive material.
• Half-life: The time interval in which half of the material gets decayed is called the half-life of the material.
• The decay constant: The probability of decay per unit time of radioactive material is called the decay constant.

\({\bf{Decay}}\;{\bf{constant}}\;\left( {\bf{λ }} \right) = \frac{{ln2}}{{{t_{\frac{1}{2}}}}}\)

Half-life (t1/2) = ln2/λ

Note: ln2 = 0.693

EXPLANATION:

Given: Half-life = T

Decay constant = T₀

The relation between half-life and decay constant for a radioactive element is given:

T = 0.693/T₀

So option 1 is correct.

CONCEPT:

As per radioactive decay law, the total number of nuclei of radioactive compounds after radioactive decay in the sample is given by given equation

\(N=N_0 e^{-λ t}\)

where N is the number of nuclei of radioactive compounds after radioactive decay, N0 is the number of nuclei of radioactive compounds initially, λ is the decay constant and t is the time of radioactive decay.

• The half-life of a radioactive element (T1/2): The time interval in which the mass of a radioactive substance or the number of atoms is 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

CALCULATION:

Given - T = 9 hours, t = 3 hours

Number of half-lives (n) in 9 hours 

\(\Rightarrow n = \frac{T}{t}=\frac{9}{3}=3 \,hours\)

As we know,

\(\Rightarrow N=N_o(\frac{1}{2})^n\)

\(\Rightarrow N=N_o(\frac{1}{2})^3=\frac{N_o}{8}\)