Solutions-Class-11-Science-Chemistry-Chapter-13-Nuclear Chemistry and Radioactivity-Maharashtra Board

(b) \(_1^2\)H + \(_1^1\)H → \(_2^3\)He

(a) \(_{15}^{30}\)P

(a) 0.132 y⁻¹

(c) ³²P

(c) nuclear fusion

• Nuclides with even protons and even neutrons are most stable, as they form proton−proton and neutron−neutron pairs, ensuring the stability of the nucleus.
• There is one−third number of stable nuclides with either Z or N odd compared to both are even. But these nuclides are less stable compared to even number of protons and neutrons.
• There are only four stable nuclides with odd Z and odd N. Generally they are unstable due to the presence of two unpaired nucleous and about 2% of the earth's crust consists of such nuclides.

Nuclear fission :

• This involves the splitting of a heavy nucleus of an atom into two nearly equal fragments along with the release of large amount of energy.
• Consider the nuclear fission of ²³⁵U isotope by neutron.

\(_{92}^{232}\)U + \(_1^0\)n  →     \(_{56}^{142}\)Ba  + ₃₆Kr  + 3 \(_1^0\)n + energy

• Each fission may lead to different products. The mass of fission products is less than the mass of parent nucleus and energy equivalent to loss in mass or mass defect is emitted in the reaction.
• The reaction indicates that fission of one ²³⁵U nucleus emits 3 neutrons which further bring about the fission of three ²³⁵U nuclei and this process continues resulting a chain reaction releasing a large amount of energy in a short time.
• Fission of ²³⁵U takes place by 400 ways yielding 800 fission products many of which are radioactive producing new elements on disintegration.
• The energy released per fission is about 200 MeV. This self sustained chain reaction is highly explosive.
• This energy may be utilised for destructive purposes, defence purpose as well as for peaceful purposes (nuclear reactor−power plant), etc.

The nuclides with odd number of both protons and neutrons are least stable due to absence of a partner nucleon for pairing with one odd nucleon. The presence of unpaired nucleons imparts instability to the nuclides.

For example, \(_{93}^{237}\)Np, \(_{97}^{247}\)Bk.

• Beta decay occurs when an unstable nucleus emits a beta particle and energy. A beta particle is either an electron or a positron. An electron is a negatively charged particle, and a positron is a positively charged electron (or anti−electron).
• When the beta particle is an electron, the decay is called beta−minus (ß−) decay. In beta−minus decay, a neutron breaks down to a proton and an electron, and the electron is emitted from the nucleus.
• When the beta particle is a positron, the decay is called beta−plus (ß+) decay. In beta−plus decay, a proton breaks down to a neutron and a positron, and the positron is emitted from the nucleus.
• Thus, beta−minus decay occurs when a nucleus has too many neutrons relative to protons (i.e., N/Z >1) and beta−plus decay occurs when a nucleus has too few neutrons relative to protons (i.e., N/Z <1).
• The nuclides having N/Z > 1, have a tendency to lose ß⁻ ( e) reducing the ratio N/Z and attaining stability.

\(_{Z}^{A}\)X → \(_{Z+1}^{A}\)Y + \(_{-1}^{0}\)e (ß⁻).

• The nuclides having N/Z < 1 have a tendency to lose β⁺ e or position to attain stability).

\(_{Z}^{A}\)X → \(_{Z-1}^{A}\)Y + \(_{+1}^{0}\)e (ß⁺).

• Nuclear transmutation is transformation of a stable nucleus into another stable or unstable nucleus.

(_{7}^{14}\)N +  (_{2}^{4}\)He →  (_{8}^{17}\)O + (_{1}^{1}\)H

• Artificial radioactivity is nuclear transmutation in which a stable nucleus is converted into radioactive nucleus by high velocity projectiles.

(_{5}^{10}\)B      +     (_{2}^{4}\)He    →         (_{7}^{13}\)N      + (_{0}^{1}\)n

Stable              +           Projectile    →             Radioactive

• Thus, the new isotope formed in artificial radioactivity is unstable and radioactive whereas in nuclear transmutation the new isotope formed is stable.

The energy released equivalent to the mass lost (defect) during the formation of an atom (or nuclide) is called nuclear binding energy.

Binding energy per nucleon = \(\bar{B}\) = \(\frac{Δm×931.4}{A}\)

The nuclear stability can be explained by plotting binding energy per nucleon as a function of mass number A of various nuclides. The parabolic curve obtained is called binding energy curve.

Inferences from binding energy curve :

• Light nuclides (A < 30) : For light nuclides, the peaks in curve with A values in multiples of 4 are very stable. They are \(_{2}^{4}\)He, \(_{6}^{12}\)C, \(_{8}^{16}\)O
• Medium mass nuclides : (30 < A < 90) : In these nuclides, the binding energy per nucleon, \(\bar{B}\) increases from 8 MeV for A = 6 to 8.3 MeV for A having values 28 to 32 and further remains nearly constant 8.5 MeV. The nuclides falling on the maximum are most stable. ⁵⁶Fe having \(\bar{B}\) values 8.79 MeV is extremely stable.
• Heavy nuclides (A > 90) : For heavy nuclides the \(\bar{B}\) values decrease from 8.79 to 7.7 MeV. ²⁰⁹Bi is the most stable nuclide. Beyond this all nuclides are unstable and radioactive.

Alpha decay :

• The emission of α −particle from the radioactive nuclei is called α −decay.
• α −particle has +2 unit charge and 4 u mass.
• Hence on−particle is identified with helium nucleus and designated with
• On emission of α −particle, the parent element loses 2 protons and 2 neutrons, thus losing 4 u mass.
• The daughter element thus formed has mass 4 units less and charge or atomic number less by 2 units than its parent element.

\(_{Z}^{A}\)X    →     \(_{Z-2}^{A-4}\)Y   +   \(_{2}^{4}\)He

Parent   →    Daughter   +  α-particle

\(_{88}^{226}\)Ra    →     \(_{88}^{226}\)Rn   +    \(_{2}^{4}\)He

Parent   →     Daughter  +   α-particle

• The daughter element formed occupies two places (groups) to the left of the parent element in the periodic table.

Nuclear fusion offers greater energy than nuclear fission but requires extremely high temperatures, surpassing 10⁸K which is achieved by nuclear fission. Hence, nuclear fusion cannot be used to produce energy.

The nuclear stability can be explained with the help of neutrons to proton (N/Z) ratio.

• When for stable nuclides the number of neutrons (N) is plotted as a function of number of protons (Z), a curve is obtained. This curved belt is called nuclear stability zone.

• The nuclides which exist outside the stability zone are unstable and radioactive.
• A straight line below the belt has equal number of neutrons and protons or the slope N/Z =1.
• Nuclides lighter than \(_{20}^{40}\)Ca are stable and have N = Z.
• For stable nuclides heavier than calcium, the ratio N/Z is greater than one.
• In heavier nuclides with the increasing number of protons there arises large coulombic repulsion among them. With increased number of neutrons the protons in the nuclei get separated decreasing the repulsion and increasing stability. Therefore, heavier nuclides require more number of neutrons than protons to attain a stability.

The age of the wood sample can be determined by radiocarbon dating as ¹⁴C becomes a part of a plant due to the photosynthesis reaction (i.e., absorption of [¹⁴CO₂ + ¹²CO₂].

The activity (N) of a given wood sample and that of a fresh sample of the live plant (N_o) is measured, where, N_o denotes the activity of the given sample at the time of death.

The age of the given wood sample, can be determined by applying following Formulae:

t = \(\frac{2.303}{λ}\)log10\(\frac{N_0}{N}\)

where λ = \(\frac{0.693}{5730y}\) = 1.21 × 10⁻⁴ y⁻¹

\(_{6}^{13}\)C and \(_{7}^{13}\)N are mirror nuclei in which the number of protons and neutrons differ by 1.

\(_{54}^{118}\)Xe → γ + \(_{54}^{118}\)I

The most stable nuclide known is \(_{26}^{56}\)Fe.

• It has two magic numbers namely 26 protons and 56 − 26 = 30 neutrons.
• They have even Z and N. This gives extra stability to the nuclide.

The relation between decay constant (λ) and half life (t_1/2) of a radioelement is,

t_1/2  = \(\frac{0.693}{λ}\)

• The helium atom is composed of 2 protons and 2 neutrons (or 1 neutron) along with 2 electrons in the outer shell.
• On the other hand, α−particle constitutes 2 protons and 2 neutrons bound together to form a particle that is similar to helium (except the presence of electrons).
• Helium is one of the inert gas which is stable (duplet complete) whereas α−particle is unstable and highly reactive.

Pairs of Isotones :

(i) \(_{5}^{10}\)B, and \(_{6}^{11}\)C,  (ii) \(_{13}^{27}\)Al, and \(_{14}^{28}\)S

Mirror nuclei: Since there are no isobars the given set of nuclides does not contain a pair of mirror nuclei.

The decay constant A of a radioelement is represented as,

λ = \(\frac{2.303}{t}\)log₁₀\(\frac{N_0}{N}\)

where No is initial number of radioactive atoms at t − 0 and N is after time t.

∴ t = \(\frac{2.303}{λ}\)log₁₀\(\frac{N_0}{N}\)

At half life period, t = t_1/2 the number of atoms becomes N = N₀/2

∴ t_1/2  = \(\frac{2.303}{λ}\)log₁₀\(\frac{N_0}{N_0/2}\)

= \(\frac{2.303}{λ}\)log₁₀ 2

= \(\frac{2.303×0.3010}{λ}\)

= \(\frac{0.693}{λ}\)

If a is activity of a radioelement then,

log₁₀ a = \(-\frac{λt}{2.303}\) + log₁₀ a₀

where a₀ is initial activity.

When log₁₀ a is plotted versus t a straight line with negative slope is obtained,

∴ slope = \(-\frac{λt}{2.303}\)

Units of radioactivity : Rate of radioactive decay is expressed in dps.

(i) The unit for the radioactivity is curie (Ci).

1 Ci = 3.7 x 10¹⁰ dps (disintegrations per second)

(ii) Another unit of radioactivity is Becquerel (Bq)

1 Bq = 1 dps. Thus, 1Ci = 3.7 x 10¹⁰ dps = 3.7 x 10¹⁰ Bq

Given : t_1/2 = 900 min. λ = ?

λ = \(\frac{0.693}{t_{1/2}}\) = \(\frac{0.693}{900}\) = 7.7 x 10⁻⁴ min⁻¹

Given : λ = 0.017 h⁻¹, t_1/2 = ?

t_1/2 = \(\frac{0.693}{λ}\) = \(\frac{0.693}{0.017}\) = 40.76 h

Given : Total B.E. = 508 MeV; A = 58, B.E. per nucleon = ?

B.E. per nucleon = \(\frac{\text{Total B.E.}}{\text{Number of nucleons}}\)

= \(\frac{508}{A}\)

= \(\frac{508}{58}\) = 8.759 MeV

Binding energy per nucleon = 8.759 MeV.

Given : Z = 16, A = 32; m = 31.97 u; m_H = 1.0078 u; m_n = 1.0087 u

Mass defect = Δm

= Z × m_H + (A − Z)m_n − m

= 16 × 1.0078 + (32 − 16)×1.0087 − 31.97

=16.1248 + 16.1392 − 31.97

= 0.294 u

Mass defect = Δm =0.294 u.

\(_{1}^{1}\)H + \(_{1}^{1}\)H  →  \(_{1}^{2}\)H +    \(_{+1}^{0}\)e

\(_{1}^{1}\)H + \(_{1}^{2}\)H  →  \(_{2}^{3}\)He

\(_{2}^{3}\)He + \(_{2}^{3}\)He  →   \(_{2}^{4}\)He + 2\(_{1}^{1}\)H

\(_{2}^{3}\)He + \(_{1}^{1}\)H  →   \(_{2}^{4}\)He + \(_{+1}^{0}\)e

• During nuclear fusion enormous energy is released. High temperature is required for nuclear fusion reaction which is attained by carrying out nuclear reaction prior to it. These reactions occur in the Sun and stars continuously.

Decrease in mass number = 232 − 208 = 24.

Since a emission decreases mass number by 4, there must be emission of 6 α particles.

Since each α particle emission decreases atomic number by 2,

final atomic number should be 90 − 12 = 78.

But since each β particle emission increases atomic number by 1, there must be emission of 4 β particles.

∴ 6 α and 4 β particles are emitted.

A      →         B     +  \(_{2}^{4}\)He (∝)

Group 18     Group 16

A is in group 18 of periodic table.

• The emission of one α−particle decreases the mass number by 4 whereas the emission of ß−particles has no effect on the mass number.
• Net decrease in mass number = 222 − 214 = 8. This decrease is only due to α−particle. Hence, number of α −particle emitted = 8/4 = 2
• Now, the emission of one α −particle decreases the atomic number by 2 and one ß−particle emission increases it by 1.
• The net decrease in atomic number = 86 − 84 = 2
• The emission of 2 α−particles causes a decrease in the atomic number by 4. However, the actual decrease is only 2. It means atomic number increases by 2. This increase is due to the emission of 2 ß−particles.

Thus, 2 α and 2 B−particles are emitted.

Given : t_1/2 = 110 min; t = 20 min. Fraction of 19F sample decays = ?

λ = \(\frac{0.693}{t_{1/2}}\) = \(\frac{0.693}{110}\) = 6.3 x 10⁻³ min⁻¹

λ = \(\frac{2.303}{t}\)log₁₀\(\frac{N_0}{N}\)

∴ log₁₀\(\frac{N_0}{N}\) = \(\frac{λt}{2.303}\) = \(\frac{6.3×10^{-3}×20}{2.303}\) = 0.05471

∴ \(\frac{N_0}{N}\) = Antilog 0.05471 = 1.134

∴ Fraction of sample left = \(\frac{N_t}{N_0}\) = \(\frac{1}{1.134}\) = 0.8818

∴ Fraction of sample decayed = 1 − 0.8818 = 0.1182

Given : t_1/2 = 87.8 d; t = 180 d; Percentage of sample left = ?

λ = \(\frac{0.693}{t_{1/2}}\) = \(\frac{0.693}{87.8}\) = 7.893 x 10⁻³ d⁻¹

λ = \(\frac{2.303}{t}\)log₁₀\(\frac{N_0}{N}\)

∴ log₁₀\(\frac{N_0}{N}\) = \(\frac{λt}{2.303}\) = \(\frac{7.893×10^{-3}×180}{2.303}\) = 0.6169

∴ \(\frac{N_0}{N}\) = Antilog 0.6169 = 4.139

∴ Fraction left = \(\frac{N_t}{N_0}\) = \(\frac{1}{4.139}\) = 0.2416

∴ Percentage of sample remained = 0.2416 × 100 = 24.16%

Percentage of sample ³⁵S left = 24.16.

Given : t_1/2 = 78 h; t = ? for decay of 12% of sample of Ga

λ = \(\frac{0.693}{t_{1/2}}\) = \(\frac{0.693}{78}\)  = 8.885 x 10⁻³ h⁻¹

λ = \(\frac{2.303}{t}\)log₁₀\(\frac{N_0}{N}\)

∴ t = \(\frac{2.303}{λ}\)log₁₀\(\frac{N_0}{N}\)

= \(\frac{2.303}{8.885×10^{-3}}\)log₁₀\(\frac{100}{(100-12)}\)

= \(\frac{2.303}{8.885×10^{-3}}\)log₁₀ 1.136

= \(\frac{2.303×0.05538}{8.885×10^{-3}}\) = 14.35 h

Time to decay of 12% of sample of Ga = 14.35 h

Given : m₀ = 0.5 g; m = 0.0788 J; t = 8 days; t_1/2 = ?

λ = \(\frac{2.303}{t}\)log₁₀\(\frac{m_0}{m}\)

= \(\frac{2.303}{8}\)log₁₀\(\frac{0.5}{0.0788}\)

= \(\frac{2.303}{8}\)log₁₀ 6.345

= \(\frac{2.303×0.8024}{8}\) = 0.231 d⁻¹

t_1/2 = \(\frac{0.693}{λ}\) = \(\frac{0.693}{0.231}\) = 3.0 d

Half life, t_1/2 = 3.0 d.

Given : Percentage decay = 65%; t = 4.2 d; t_1/2 = ?

Percentage of element left = 100 − 65 = 35%

λ = \(\frac{2.303}{t}\)log₁₀\(\frac{N_0}{N}\)

= \(\frac{2.303}{4.2}\)log₁₀\(\frac{100}{35}\)

= \(\frac{2.303}{4.2}\)log₁₀ 2.857

= \(\frac{2.303×0.4559}{4.2}\) = 0.25 d⁻¹

t_1/2 = \(\frac{0.693}{λ}\) = \(\frac{0.693}{0.25}\)  = 2.772 d

Half life, t_1/2 = 2.772 d.

Given : For \(_{36}^{84}\)Kr, Z = 36, A = 84, m_H = 1.0078 u, m_n = 1.0087 u, m = 83.913 u

B.E. per nucleon = ?

Δm = Z x m_H + (A − Z) x m_n − m

= 36 × 1.0078 + (84 − 36) × 1.0087 − 83.913

= 36.2808 + 48 × 1.0086 − 83.913

= 36.2808 + 48.4176 − 83.913

= 0.7854 u

B.E. per nucleon = \(\frac{Δm×931.4}{A}\) = \(\frac{0.7854×931.4}{84}\) = 8.708 MeV

B.E. per nucleon = 8.708 MeV.

Given : Atomic masses :

m_Ir =173.97 u; m_Re = 169.96 u; m_He = 4.0026 u

\(_{77}^{174}\)Ir  → \(_{75}^{170}\)Re + \(_{2}^{4}\)He

Δm  = (Total mass of reactants) – (Total mass of products)

= m_Ir− (m_Re + m_He)

= 173.97 − (169.96 + 4.0026)

= 173.97 − 173.9626

= 7.4 × 10⁻³ u

Energy released, E = Δm x 931.4 = 7.4 × 10⁻³ × 931.4 = 6.892 MeV

Energy released = 6.892 MeV.

If N₀ = 1 then N = 1 − 3/4 = 1/4 ; t = 60 min. t_1/2 = ?

λ = \(\frac{2.303}{t}\)log₁₀\(\frac{N_0}{N}\)

= \(\frac{2.303}{60}\)log₁₀\(\frac{1}{1/4}\)

= \(\frac{2.303}{60}\)log₁₀ 4

= \(\frac{2.303×0.6021}{60}\) = 0.02311 min⁻¹

t_1/2 =  \(\frac{0.693}{λ}\) = \(\frac{0.693}{0.02311}\) = 30 min

Half life, t_1/2 = 30 min.

Given : Mass of Radium = 0.1 g, t = 1 ∴ y = 1 x 365 x 24 x 60 x 60 s; Number of particles emitted = ?

Disintegrations per year = 0.1 x 3.7 × 10¹⁰ × 1 x 365 × 24 × 60 × 60 = 1.167 × 10¹⁷

Number of ∝−particles emitted = 1.167 x 10¹⁷.

Given : N_o = 1Ci = 3.7 x 10¹⁰ dps; N_t = 1.5 ×10⁴ dps; t = 303 d; t_1/2 =?

λ = \(\frac{2.303}{t}\)log₁₀\(\frac{N_0}{N}\)

= \(\frac{2.303}{3.3}\)log₁₀\(\frac{3.7×10^{10}}{1.5×10^4}\)

= \(\frac{2.303}{3.3}\)log₁₀ 2.467 × 10⁶

= \(\frac{2.303×6.3992}{3.3}\) = 0.04858

t_1/2 = \(\frac{0.693}{λ}\) = \(\frac{0.693}{0.04858}\) = 14.27 d

Half life, t_1/2 = 14.27 d.

Given : t_1/2 = 3.82 d; Decay = 99.9% ∴ N = 100 − 99.9 = 0.1; t =?

λ = \(\frac{0.693}{t_{1/2}}\) = \(\frac{0.693}{3.82}\)  = 0.1814 d⁻¹

λ = \(\frac{2.303}{t}\)log₁₀\(\frac{N_0}{N}\)

∴ t = \(\frac{2.303}{λ}\)log₁₀\(\frac{N_0}{N}\)

= \(\frac{2.303}{0.1814}\)log₁₀\(\frac{100}{0.1}\)

= \(\frac{2.303×3}{0.1814}\) = 38.08 d

Time required = 38.08 d

Given :

Loss in mass Δm = 4.34 x 10⁹ kg s⁻¹ = 4.34 × 10¹² g s⁻¹ = \(\frac{4.34×10^{12}}{1.66×10^{-24}}\) u s⁻¹

Energy radiated  = Δm x 931.4

= \(\frac{4.34×10^{12}}{1.66×10^{-24}}\) x 931.4 MeV s⁻¹

= \(\frac{4.34×10^{12}}{1.66×10^{-24}}\) x 931.4 x 10⁶ eV s⁻¹

= \(\frac{4.34×10^{12}}{1.66×10^{-24}}\) x 931.4 x 10⁶ × 1.6 × 10⁻¹⁹ J s⁻¹

= \(\frac{4.34×10^{12}}{1.66×10^{-24}}\) x 931.4 x 10⁶ × 1.6 × 10⁻²² kJ s⁻¹

= 3.896 x 10²³ kJ s⁻¹

Energy released = 3.896 x 10²³ kJ s⁻¹

Given : Initial rate of disintegration = R₀ = 16.0 dps g⁻¹ , Final rate of disintegration = R = 7.0 dps g⁻¹ , t_1/2 = 5730 y; t = ?

λ = \(\frac{0.693}{t_{1/2}}\) = \(\frac{0.693}{5730}\) = 1.209 x 10⁻⁴ y⁻¹

λ = \(\frac{2.303}{t}\)log₁₀\(\frac{R_0}{R}\)

∴ t = \(\frac{2.303}{λ}\)log₁₀\(\frac{R_0}{R}\)

= \(\frac{2.303}{1.209×10^{-4}}\)log₁₀\(\frac{16}{7}\)

= \(\frac{2.303}{1.209×10^{-4}}\)log₁₀ 2.286

= \(\frac{2.303×0.3591}{1.209×10^{-4}}\) = 6840 y

Age of wood sample = 6840 y.