Solution-Class-12-Physics-Chapter-15-Structure of Atoms and Nuclei-MSBSHSE

(B) singly ionized helium

Explanation :

According to Bhor's theory, the radius of an atomic number z atom's nth orbit is:

r ∝ \(\frac{n^2}{Z}\) , Where n is the orbit number.

(i) For hydrogen, n = 1 and Z = 1

Therefore, r ∝ \(\frac{n^2}{Z}=\frac{1}{1}\) = 1

(ii) For Helium, , n = 1 and Z = 2

Therefore, r ∝ \(\frac{n^2}{Z}=\frac{1}{2}\)

(iii) For Deuteron. n = 1 and Z = 1

Therefore, r ∝ 1

(iv) For Tritium, n = 1 and Z = 1

Therefore, r ∝ 1

Hence, the smallest is Helium. The correct option is Singly ionized helium.

(B) 4

Explanation :

Radius of the 4th orbit r₄ = 0.529 × 10^-8 × 4²

Radius of the 8th orbit r₈ = 0.529 × 10^-8 × 8²

∴ \(\frac{r_8}{r_4}=\frac{0.529×10^{-8}×8^2}{0.529×10^{-8}×4^2}=\frac{64}{16}\) = 4

(D) n = 8 to n = 7

(D) density

Explanation :

The formula used to calculate the density of a nucleus is,

Density = \(\frac{3m_N}{4πr_0^3}\)

Where m_N = mass of a nuclean

Therefore, the density of a nucleus does not depend on the mass number.

(B) \(\frac{N}{4}\)

Explanation :

The number of nuclei in a radioactive sample at a given time is N.

After the first half-life the remaining nuclei are, \(\frac{N}{2}\)

After the second half-life the remaining nuclei are, \(\frac{N/2}{2}\) = \(\frac{N}{4}\)

Bohr's atomic model postulates (for hydrogen atom) are as follows:

• The electron moves in a circular orbit around the nucleus. The centripetal force is caused by the positive nuclear charge's attraction to the negatively charged electron.
• The electron can only rotate without emitting energy in allowed or stable orbits, where its angular momentum is equal to an integral multiple of h/2π, where h is Planck's constant.
• Energy is only radiated when the electron jumps from one orbit to another with lower energy. The energy of the emitted quantum of electromagnetic radiation, or photon, equals the energy difference between the two states.

• Rutherford's atomic model has several challenges, including the assumption that electrons orbit the nucleus in circular motion.
• The circular motion is an accelerated one. According to classical electromagnetic theory, an accelerating charge constantly emits energy. As a result, an electron should continue to emit energy while moving around in its orbit.
• Due to the loss of energy, the radius of its orbit should continue to decrease. As a result, the electron should travel in a spiral pattern before falling into the nucleus in a very short period of time, on the order of 10^-16 seconds for a hydrogen atom. Therefore, the atom should be unstable. We exist because atoms are stable.
• If the electron moves along such a spiral path, the radius of its orbit would continuously decrease. As a result, the speed and frequency of revolution of the electron would go on increasing. The electron, therefore, would emit radiation of continuously changing frequency, and hence give rise to a continuous spectrum. However, atomic spectrum is a line spectrum.

Therefore, Rutherford's model is not suitable for understanding the atomic structure.

• ∝ -decay : A radioactive transformation in which an ∝ –particle is emitted is called ∝-decay.
• β-decay : A radioactive transformation in which a β –particle is emitted is called β-decay.
• γ-decay : In this type of decay, gamma rays are emitted by the parent nucleus. A nucleus in an excited state emits a γ-ray photon in a transition to the lower energy state. ∝- and β —particle emissions are often accompanied by γ -rays.

• Excitation energy of an electron in an atom : The energy required to transfer an electron from the ground state to an excited state (a state of higher energy) is called the excitation energy of the electron in that state.
• Binding energy : Binding energy of an electron in an atom is defined as the minimum energy that should be provided to an orbital electron to remove it from the atom such that its total energy is zero.
• Ionization energy : Ionization energy of an electron in an atom is defined as the minimum energy required to remove the least strongly bound electron from a neutral atom such that its total energy is zero.

For the first line in the Lyman series,

\(\frac{1}{λ_{L1}}=R(\frac{1}{1^2}-\frac{1}{2^2})=R(1-\frac{1}{4})\) = \(\frac{3R}{4}\)

∴ v_L1 = \(\frac{c}{λ_{L1}}\) = \(\frac{3Rc}{4}\), where v denotes the frequency, c the speed of light in free space and R the Rydberg constant.

For the limit of the Lyman series,

\(\frac{1}{λ_{L∞}}=R(\frac{1}{1^2}-\frac{1}{∞})= R\)

∴ v_L∞  = \(\frac{c}{λ_{L∞}}\) = Rc

For the limit of the Balmer series,

\(\frac{1}{λ_{B∞}}=R(\frac{1}{2^2}-\frac{1}{∞})= \frac{R}{4}\)

∴ v_B∞  = \(\frac{c}{λ_{B∞}}\) = \(\frac{Rc}{4}\)

∴ v_L∞  - v_B∞ = Rc - \(\frac{Rc}{4}\) = \(\frac{3Rc}{4}\) = v_L1

Hence, the result.

Bohr's atomic model postulates are as follows:

• The electron moves in a circular orbit around the nucleus. The centripetal force is caused by the positive nuclear charge's attraction to the negatively charged electron.
• The electron can only rotate without emitting energy in allowed or stable orbits, where its angular momentum is equal to an integral multiple of h/2π, where h is Planck's constant.
• Energy is only radiated when the electron jumps from one orbit to another with lower energy. The energy of the emitted quantum of electromagnetic radiation, or photon, equals the energy difference between the two states.

Expression for the energy of an electron in the atom :

Consider the electron revolving in the nth orbit around the nucleus of an atom with the atomic number Z.

Let m and -e be the mass and the charge of the electron, r the radius of the orbit and v the linear speed of the electron.

According to Bohr’s first postulate, centripetal force on the electron = electrostatic force of attraction exerted on the electron by the nucleus

∴ \(\frac{mv^2}{r}=\frac{Ze^2}{4πε_0r^2}\)      …..(1)

where ε₀ is the permittivity of free space.

∴ Kinetic energy (KE) of the electron = \(\frac{1}{2}mv^2=\frac{Ze^2}{8πε_0r}\)     …..(2)

The electric potential due to the nucleus of charge +Ze at a point at a distance r from it is

V = \(\frac{1}{4πε_0}.\frac{Ze}{r}\)

∴  Potential energy (PE) of the electron = charge on the electron x electric potential

= \(-e×\frac{1}{4πε_0}.\frac{Ze}{r}\) = \(-\frac{Ze^2}{4πε_0r}\)   ……(3)

Hence, the total energy of the electron in the n^th orbit is

E = KE + PE = \(-\frac{Ze^2}{4πε_0r}\) + \(\frac{Ze^2}{8πε_0r}\)

∴ E = \(-\frac{Ze^2}{8πε_0r}\)       …..(4)

This shows that the total energy of the electron in the n^th orbit of the atom is inversely proportional to the radius of the orbit as Z, ε₀ and e are constants.

The radius of the nth orbit of the electron is

r = \(\frac{ε_0h^2n^2}{πmZe^2}\)    ….(5)

where h is Planck's constant.

From Eqs. (4) and (5), we get,

E_n = \(-\frac{Ze^2}{8πε_0}(\frac{πmZe^2}{ε_0h^2n^2})\) = \(-\frac{mZ^2e^4}{8ε_0^2h^2n^2}\)     ….(6)

This gives the expression for the energy of the electron in the n^th Bohr orbit. The minus sign in the expression shows that the electron is bound to the nucleus by the electrostatic force of attraction.

As m, Z, e, ε₀ and h are constant, we get

E_n ∝ 1/n²

i.e., the energy of the electron in a stationary energy state is discrete and is inversely proportional to the square of the principal quantum number.

According to Bohr’s third postulate for the model of the hydrogen atom, an atom radiates energy only when an electron jumps from a higher energy state to a lower energy state and the energy of the quantum of electromagnetic radiation emitted in this process is equal to the energy difference between the two states of the electron. This emission of radiation gives rise to a spectral line.

The energy of the electron in a hydrogen atom, when it is in an orbit with the principal quantum number n, is

E_n = \(-\frac{me^4}{8ε_0^2h^2n^2}\) 

where m =mass of electron, e = electronic charge, h = Planck's constant and ε₀ = permittivity of free space.

Let E_m be the energy of the electron in a hydrogen atom when it is in an orbit with the principal quantum number m and E_n, its energy in an orbit with the principal quantum number n, n < m. Then

E_m = \(-\frac{me^4}{8ε_0^2h^2m^2}\)   and E_n = \(-\frac{me^4}{8ε_0^2h^2n^2}\)

Therefore, the energy radiated when the electron jumps from the higher energy state to the lower energy state is

E_m − E_n = \(-\frac{me^4}{8ε_0^2h^2m^2}-(-\frac{me^4}{8ε_0^2h^2n^2})\)

              = \(\frac{me^4}{8ε_0^2h^2}(\frac{1}{n^2}-\frac{1}{m^2})\)

This energy is emitted in the form of a quantum of radiation (photon) with energy hv, where v is the frequency of the radiation.

∴ E_m − E_n = hv

∴ v = \(\frac{E_m-E_n}{h}\) = \(\frac{me^4}{8ε_0^2h^3}(\frac{1}{n^2}-\frac{1}{m^2})\)

The wavelength of the radiation is λ = c/v, where c is the speed of radiation in free space.

The wave number, \(\bar{v}\) = 1/λ = v/c

∴ \(\bar{v}\) = 1/λ = \(\frac{me^4}{8ε_0^2h^3c}(\frac{1}{n^2}-\frac{1}{m^2})\)= \(R(\frac{1}{n^2}-\frac{1}{m^2})\)

where R = \(\frac{me^4}{8ε_0^2h^3c}\) is a constant called the Rydberg constant.

This expression gives the wave number of the radiation emitted and hence that of a line in hydrogen spectrum.

[Note : Z =1 for hydrogen atom]

For the Laymen series : n= 1, m =2,3,4,..., ∞

1/λ_L  = \(R(\frac{1}{1^2}-\frac{1}{m^2})\)

and for the shortest wavelength line in this series,

1/λ_Ls  = \(R(\frac{1}{1^2})\)  as m = ∞

For the Balmer series : n= 2, m = 3,4,5, ..., ∞

1/λ_B  = \(R(\frac{1}{2^2}-\frac{1}{m^2})\)

and for the shortest wavelength line in this series,

1/λ_Bs  = \(R(\frac{1}{4})\) as m = ∞

The radius of the nth Bohr orbit is

r = \(\frac{ε_0h^2n^2}{πmZe^2}\)    ……(1)

and the linear speed of an electron in this orbit is

v = \(\frac{Ze^2}{2ε_0nh}\)     ……(2)

where ε₀ = permittivity of free space, h= Planck's constant, n = principal quantum number, m= electron mass, e = electronic charge and Z = atomic number of the atom.

Since angular speed ω = \(\frac{v}{r}\), then from Eqs. (1) and (2) we get,

ω = \(\frac{v}{r}\) = \(\frac{Ze^2}{2ε_0nh}.\frac{πmZe^2}{ε_0h^2n^2}\) = \(\frac{πmZ^2e^4}{2ε_0^2n^3h^3}\)

∴ ω_max = \(\frac{πme^4}{2ε_0^2h^3}\)  …(for Z=1 and n=1)

∴ ω_max = \(\frac{(3.142)(9.1×10^{-31})(1.6×10^{-19})}{2×(8.85×10^{-12})(6.63×10^{-34})}\) = 4.105 x 10¹⁶ rad/s

This is the required quantity.

Given : λ_L∞ = 912 Å

For hydrogen spectrum, \(\frac{1}{λ}\) = R_H\((\frac{1}{n^2}-\frac{1}{m^2}\)

∴ \(\frac{1}{λ_{L∞}}\) = R_H\((\frac{1}{1^2}-\frac{1}{∞})\) = R_H     …….(1)

as n = 1, m = ∞

\(\frac{1}{λ_{B∞}}\) = R_H\((\frac{1}{4}-\frac{1}{∞})\) = \(\frac{R_H}{4}\)  …….(2)

as n = 2, m = ∞

\(\frac{1}{λ_{Pa∞}}\) = R_H\((\frac{1}{9}-\frac{1}{∞})\) = \(\frac{R_H}{9}\) …….(3)

as n = 3, m = ∞

\(\frac{1}{λ_{Pf∞}}\) = R_H\((\frac{1}{25}-\frac{1}{∞})\) = \(\frac{R_H}{25}\) …….(4)

as n = 5, m = ∞

From Eqs. (1) and (2), we get,

\(\frac{λ_{B∞}}{λ_{L∞}}\) = \(\frac{R_H}{R_H/4}\) = 4

∴ λ_B∞ = 4 λ_L∞ = 4(912) = 3648 Å

This is the series limit of the Balmer series.

From Eqs. (1) and (3), we get,

\(\frac{λ_{Pa∞}}{λ_{L∞}}\) = \(\frac{R_H}{R_H/9}\) = 9

∴ λ_Pa∞ = 9 λ_L∞ = 9(912) = 8208 Å

This is the series limit of the Paschen series.

From Eqs. (1) and (4), we get,

\(\frac{λ_{Pf∞}}{λ_{L∞}}\) = \(\frac{R_H}{R_H/25}\) = 25

∴ λ_Pa∞ = 25 λ_L∞ = 25(912) = 22800 Å

This is the series limit of the Pfünd series.

(i) α -decay : A radioactive transformation in which an α –particle is emitted is called α-decay. In an α—decay, the atomic number of the nucleus decreases by 2 and the mass number decreases by 4.

Formula for the energies generated :

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

(ii) β-decay : A radioactive transformation in which a β –particle is emitted is called β-decay. In a β-decay, the atomic number of the nucleus increases by 1 and the mass number remains unchanged.

Formula for the energies generated :

\(_{Z}^{A}X→_{-1}^{\,0}β+_{Z+1}^{A}Y\) + energy released

(iii) γ-decay : A given nucleus does not emit α- and β –particles simultaneously. However, on emission of α or β -particles, most nuclei are left in an excited state.

A nucleus in an excited state emits a γ-ray photon in a transition to the lower energy state. Hence, α- and β —particle emissions are often accompanied by γ -rays.

Formula for the energies generated in γ-decay :

\(_{Z}^{A}X^*→_{0}^{0}γ+_{Z}^{A}X\)(energy released is carried by the γ –ray photon)

Nuclear fission : It is a nuclear reaction in which a heavy nucleus of an atom, such as that of uranium, splits into two or more fragments of comparable size, either spontaneously or as a result of bombardment of a neutron on the nucleus (induced fission).

• It is followed by emission of two or three neutrons.
• The mass of the original nucleus is more than the sum of the masses of the fragments. This mass difference is released as energy, which can be enormous as in the fission of ²³⁵
• Nuclear fission was discovered by Lise Meitner, Otto Frisch, Otto Hahn and Fritz Strassmann in 1938.

Fission of uranium-235 :

The products of the fission of ²³⁵U by thermal neutrons are not unique. A variety of fission fragments are produced with mass number A ranging from about 72 to about 138, subject to the conservation of mass-energy, momentum, number of protons (Z) and number of neutrons (N).

A few typical fission equations are :

(1) \(_{92}^{235}U+_{0}^{1}n→_{92}^{236}U→_{54}^{140}Xe+_{38}^{94}Sr+2_{0}^{1}n\) + 200 MeV

(235 + 1 = 236 = 140 + 94 + 2)

(2) \(_{92}^{235}U+_{0}^{1}n→_{92}^{236}U→_{56}^{144}Ba+_{36}^{90}Kr+2_{0}^{1}n\) +200 MeV

(235 + 1 = 236 = 144 + 90 + 2)

(3) \(_{92}^{235}U+_{0}^{1}n→_{92}^{236}U→_{57}^{148}La+_{35}^{85}Br+3_{0}^{1}n\) +200 MeV

(235 + 1 = 236 = 148 + 85 + 3) .

Nuclear fusion : A type of nuclear reaction in which lighter atomic nuclei (of low atomic number) fuse to form a heavier nucleus (of higher atomic number) with the release of enormous amount of energy is called nuclear fusion.

Very high temperatures, of about 10⁷ K to 10⁸ K, are required to carry out nuclear fusion. Hence, such a reaction is also called a thermonuclear reaction.

Example ;The D-T reaction, being used in experimental fusion reactors, fuses a deuteron and a triton nuclei at temperatures of about 10⁸ K.

\(_{1}^{2}D+_{1}^{3}T→_{2}^{4}He+_{0}^{1}n\)+ 17.6 MeV

(deuteron + triton → Helium nucleus + neutron)

Principles of a nuclear reactor : In a nuclear reactor fuel rods are used to provide a suitable fissionable material such as . Control rods are used to start or stop the reactor. Moderators are used to slow down the fast neutrons ejected in a nuclear fission to the appropriate lower speeds. Material used as a coolant removes the energy released in the nuclear reaction by converting it into thermal energy for production of electricity.

Difference between a nuclear reactor and a nuclear bomb :

In a nuclear reactor, a nuclear fission chain reaction is used in a controlled manner, while in a nuclear bomb, the nuclear fission chain reaction is not controlled, releasing tremendous energy in a very short time interval.

Question 10. Calculate the binding energy of an alpha particle given its mass to be 4.00151 u.

Answer :

Given : M =4.00151 u, m_p = 1.00728 u, m_n = 1.00866u, 1 u = 931.5 MeV/c²

The binding energy of an alpha particle = (Zm_p +Nm_n - M)c²

= (2m_p + 2m_n - M)c²

= [(2)(1.00728u) +2(1.00866 u) - 4.00151 u]c²

= (2.01456 + 2.01732 - 4.00151)(931.5) MeV

= 28.289655 MeV

= 28.289655 × 10⁶ eV × 1.602 × 10^-19 J

= 4.532002731 x 10^-12 J

Given : z = 1, m = 9.1 x 10^-31 kg, e = 1.6 × 10^-19 C, ε₀ = 8.85 x 10^-12 C²/N.m², h = 6.63 x 10^-34 J.s, n = 2, t = 10^-8 s

The periodic time of the electron in a hydrogen atom,

T = \(\frac{4ε_0^2h^3n^3}{πme^4}\)

= \(\frac{4(8.85×10^{-12})^2(6.63×10^{-34})^3(2)^3}{3.142(9.1×10^{-31})(1.6×10^{-19})^4}\) = 3.898 × 10^-16 s

Let N be the number of revolutions made by the electron in time t. Then, t = NT.

∴ N = \(\frac{t}{T}\) = \(\frac{10^{-8}}{3.898×10^{-16}}\) = 2.565 × 10⁷

Given : Z = 95, N = 244 - 95 = 149, m_p =1.00728 u, m_n = 1.00866 u,

M = 244.06428 u, 1u = 931.5MeV/c²

The binding energy per nucleon,

\(\frac{E_B}{A}\) = \(\frac{(Zm_p+Nm_n-M)c^2}{A}\)

= \(\frac{[95(1.00728)+149(1.00866)-244.06428]uc^2}{244}\)

= \((\frac{95.6916+150.29034-244.06428}{244})\)(931.5) MeV nucleon

= 7.3209 MeV/nucleon

Given : M₁ (\(_{3}^{7}Li\) atom) =7.016 u, M₂ (He atom) = 4.0026u, m_p = 1.00728 u, 1u = 931.5MeV/c²

ΔM = M₁ + m_p - 2M₂

= [7.016 + 1.00728 - 2(4.0026)] u

= 0.01808 u = (0.01808)(931.5) MeV/c²

= 16.84152 MeV/c

Therefore, the energy released in the nuclear reaction = (ΔM)c² =16.84152 MeV

(a) \(_{88}^{226}Ra\) → \(_{2}^{4}∝\) + \(_{86}^{222}Em\)

Em (Emanation) ≡ Rn (Radon)

Here, ∝ particle is emitted and radon is formed.

(b) \(_{8}^{19}O\) → e⁻ + \(_{9}^{19}F\)

Here, e⁻ ≡ \(_{-1}^{0}β\)  is emitted and fluorine is formed.

(c) \(_{90}^{228}Th\) → \(_{2}^{4}∝\) + \(_{88}^{224}Ra\)

Here, ∝ particle is emitted and radium is formed.

(d) \(_{7}^{12}N\) → \(_{6}^{12}C\) + \(_{1}^{0}β\)

\(_{1}^{0}β\) is e⁺ (positron)

Here, β⁺ is emitted and carbon is formed.

Given :  \(_{88}^{223}Ra\) : 223.0185 u, \(_{82}^{209}Pb\) : 208.9811 u,

\(_{6}^{14}C\) : 14.00324 u, \(_{92}^{236}U\) : 236.0456,

\(_{56}^{140}Ba\) : 139.9106 u, \(_{36}^{94}Kr\) : 93.9341 u,

\(_{6}^{11}C\) : 11.01143 u, \(_{5}^{11}B\) : 11.0093 u.

m_n = 1.00866 u, m(e⁺) = 0.00055 u,1u = 931.5 MeV/c²

(a) \(_{88}^{223}Ra\)  → \(_{82}^{209}Pb\) + \(_{6}^{14}C\)

The energy released in this reaction = (ΔM)c²

= [223.0185 - (208.9811 + 14.00324)](931.5)MeV

= 31.820004 MeV

(b) \(_{92}^{236}U\)  → \(_{56}^{140}Ba\)  + \(_{36}^{94}Kr\) + 2n

The energy released in this reaction = (ΔM)c²

= [236.0456 - (139.9106 + 93.9341 + (2)(1.00866)](931.5)MeV

= 171.00477 MeV

(c) \(_{6}^{11}C\) → \(_{5}^{11}B\) + e⁺ + neutrino

The energy released in this reaction = (ΔM)c²

= [11.01143 - (11.0093 + 0.00055)] (931.5) MeV

= 1.47177 MeV

Given : 15.3 decays per gram per minute (living organism), 12.3 disintegrations per gram per minute (very old charcoal). Hence, we have,

\(\frac{A(t)}{A_0}\) = \(\frac{12.3}{15.3}\), λ = 3.839 x 10^-12 per second

A(t) = A₀e^-λt

∴ e^λt = \(\frac{A_0}{A}\), ∴ λt = log_e\((\frac{A_0}{A})\)

∴ t = \(\frac{2.303}{λ}\)log₁₀ \((\frac{A_0}{A})\)

= \(\frac{2.303}{3.839×10^{-12}}\)log₁₀ \((\frac{15.3}{12.3})\)

= \(\frac{2.303}{3.839×10^{-12}}\)(log 15.3 – log 12.3)

= \(\frac{2.303}{3.839×10^{-12}}\)(1.1847 – 1.0899)

= \(\frac{2.303×0.0948}{3.839}\)x 10¹² s = 5.687 x 10¹⁰ s

= \(\frac{5.687×10^{10}}{3.156×10^7}\) = 1802 years

Given : T_1/2 = 28 years = 28 x 3.156 × 10⁷ s = 8.837 × 10⁸ s, M = 5mg = 5 ×10^-3 g

90 grams of \(_{38}^{90}Sr\) contain 6.02 x 10²³ atoms

Hence, here, N = \(\frac{(6.02×10^{23})(5×10^{-3})}{90}\) = 3.344 x 10¹⁹ atoms

∴ The disintegration rate = N_λ = N\(\frac{0.693}{T_{1/2}}\)

=\(\frac{(3.344×10^{19})(0.693)}{90}\)

= 2.622 ×10¹⁰ disintegrations per second

Given : Activity = 10.0 mCi = 10.0 × 10^-3 Ci = (10.0 × 10^-3)(3.7 × 10¹⁰) dis/s = 3.7 × 10⁸ dis/s

T_1/2 = 5.3 years = (5.3)(3.156 × 10⁷) s = 1.673 × 10⁸ s

Decay constant, λ = \(\frac{0.693}{T_{1/2}}\) = \(\frac{0.693}{1.673×10^8}\)  s^-1

= 4.142 × 10^-9 s^-1

Activity = N_λ

∴ N = \(\frac{activity}{λ}\) = \(\frac{3.7×10^8}{4.1423×10^{-9}}\) = 8.933 × 10¹⁶ atoms

= 60 grams of  contain 6.02 x 10²³ atoms

∴ Mass of 8.933 x 10¹⁶ atoms of  = \(\frac{8.933×10^{16}}{6.02×10^{23}}\) × 60 = 8.903 × 10^-6 g = 8.903 μg

This is the required amount.

Given : A(t₁) = 10¹⁰ per hour, where t₁ = 20 h,

A(t₂) = 6.3 × 10⁹ per hour, where t₂ = 30 h

A(t) = A₀e^-λt , ∴ A(t₁) = A₀e^-λt1 and A(t₂) = A₀e^-λt2

∴ \(\frac{A(t_1)}{A(t_2)}\) = \(\frac{e^{-λt_1}}{e^{-λt_2}}=e^{λ(t_2-t_1}\)

∴ \(\frac{10^{10}}{6.3×10^9}= e^{λ(30-20)}\) = e^10λ

∴ 1.587 = e^10λ, ∴ 10λ = 2.303 log₁₀(1.587)

∴ λ = (0.2303)(0.2007) = 0.04622 per hour

The half life of the material, T_1/2 = \(\frac{0.693}{λ}\) = \(\frac{0.693}{0.04622}\) = 14.99 hours

Now, A₀ = A(t₁)e^λt1 = 10¹⁰e^{(0.04622)(20)} = 10¹⁰e^0.9244

Let x = e^0.9244 , ∴ 2.303 log₁₀x = 0.9244

∴ log₁₀x = \(\frac{0.9244}{2.303}\) = 0.4014

∴ x = antilog 0.4014 = 2.52

∴ A₀ = 2.52 x 10¹⁰ per hour

Now A₀ = N₀λ, ∴ N₀ = \(\frac{A_0}{λ}\) = \(\frac{2.52×10^{10}}{0.04622}\) = 5.452 x 10¹¹

This is the initial number of radioactive atoms in the sample.

Given : T_1/2 = 272 d = 272 x 24 x 60 × 60 s = 2.35 × 10⁷ s,

A₀ = 2.0 μCi = 2.0 x 10^-6 x 3.7 × 10¹⁰ = 7.4 × 10⁴ dis/s. t = 1 year = 3.156 × 10⁷ s

(a) T_1/2 = \(\frac{0.693}{λ}\)  = 0.693 τ

∴ The mean lifetime for ⁵⁷Co = τ = \(\frac{T_{1/2}}{0.693}\) = \(\frac{2.35×10^7}{0.693}\) = 3.391 × 10⁷ s

The decay constant for ⁵⁷Co = λ = \(\frac{1}{τ}\) = \(\frac{1}{3.391×10^7}\) = 2.949 × 10^-8 s^-1

(b) A₀ = N₀λ, ∴ N₀ = \(\frac{A_0}{λ}\) = A₀τ = (7.4 × 10⁴)(3.391 × 10⁷) = 2.509 × 10¹² nuclei

This is the required number.

(c) A(t) = A₀e^-λt =  = 2e^-0.9307= 2/e^0.9307

Let x = e^0.9307

∴ log x = 0.9307

∴ 2.303 log₁₀ x = 0.9307

∴ log₁₀ x = \(\frac{0.9307}{2.303}\) = 0.4041

∴ x = antilog 0.4041 = 2.536

∴ A(t) = \(\frac{2}{2.536}\)μCi = 0.7886 μCi

Given : \(_{15}^{32}P\) : T_1/2 = 14.3 d

∴ λ₁ = \(\frac{0.693}{14.3d}\) = 0.04846 d^-1

\(_{15}^{33}P\) : T_1/2 = 25.3 d, ∴ λ₂ = \(\frac{0.693}{25.3d}\) = 0.02739 d^-1

At time t = 0, \(\frac{N_{O1}λ_1}{N_{O2}λ_2}\) = \(\frac{90%}{10%}\) = 9     ….(1) and

At time t, \(\frac{N_{O1}λ_1e^{-λ_1t}}{N_{O2}λ_2e^{-λ_2t}}\) = \(\frac{15%}{85%}\)     ….(2)

Dividing Eq. (1) by Eq. (2), we get,

\(\frac{N_{O1}λ_1}{N_{O2}λ_2}\)×\(\frac{N_{O2}λ_2e^{-λ_2t}}{N_{O1}λ_1e^{-λ_1t}}\) = \(\frac{9}{3/17}\) = \(\frac{153}{3}\)

∴ \(e^{(λ_1-λ_2)t}=\frac{153}{3}\)

∴ (λ₁ - λ₂)t = 2.303 log₁₀\(\frac{153}{3}\)= 2.303 (log₁₀ 153 - log₁₀ 3)

∴ (0.04846 - 0.02739) t = 2.303 (2.1847 - 0.4771)

∴ t = \(\frac{(2.303)(1.7076)}{0.02107}\) = 186.6 days

This is the required time.

Given : T_1/2 = 5730 y ∴ λ = \(\frac{0.693}{5730×3.156×10^7}\) = 3.832 × 10^-12 s^-1,

A = 0.255 Bq per gram of carbon in part (a); M = 500mg = 500 × 10^-3 g,

174 decays in one hour = \(\frac{174}{3600}\)  = 0.04833 dis/s in part (b) [per 500 mg]

(a) A = Nλ,  ∴ N = \(\frac{A}{λ}\) = \(\frac{0.255}{3.832×10^{-12}}\) = 6.654 × 10¹⁰

Number of atoms in 1 g of carbon = \(\frac{6.02×10^{23}}{12}\) = 5.017 × 10²²

\(\frac{5.017×10^{22}}{6.654×10^{10}}\) = 0.7539 × 10¹²

∴ 1 ¹⁴C atom per 0.7539 x 10¹² atoms of carbon

∴ 4 ¹⁴C atoms per 3 x 10¹² atoms of carbon

(b) Present activity per gram = \(\frac{0.04833}{500×10^{-3}}\) = 0.09666 dis/s per gram 

A₀ = 0.255 dis/s per gram

Now, A(t) = A₀e^-λt

∴ λt = 2.303 log₁₀ \((\frac{A_0}{A})\) = 2.303 log₁₀ \((\frac{0.255}{0.09666})\) 

∴ t = \(\frac{2.303log2.683}{3.832×10^{-12}}\) = 25.32 x 10¹⁰ s = \(\frac{25.32×10^{10}}{3.156×10^7}\) = 8023 years

This is the required quantity.

Given : Power=3000 MW = 3 × 10⁹ J/s

∴ Energy to be produced each day = 3 × 10⁹ × 86400 J each day = 2.592 × 10¹⁴ J each day

Energy per fission = 202.79 MeV = 202.79 × 10⁶ × 1.6 × 10^-19 J = 3.245 x 10^-11 J

∴ Number of fissions each day = \(\frac{2.592×10^{14}}{3.245×10^{-11}}\) = 7.988 x 10²⁴ each day

0.235 kg of ²³⁵U contains 6.02 x 10²³ atoms

∴ M = \((\frac{7.988×10^{24}}{6.02×10^{23}})\)(0.235) = 3.118 kg =

This is the required quantity.

Given : Average atomic mass of magnesium 24.312 u, \(_{12}^{24}Mg\) : (23.98504 u), \(_{12}^{25}Mg\) : (24.98584 u), \(_{12}^{26}Mg\) : (25.98259 u), \(_{12}^{24}Mg\): 78.99% by mass

∴ 24.312 = \(\frac{(2.98504)(78.99)+(24.98584)x+(25.98259)(100-78.99-x)}{100}\)

∴ 24.312 = 18.9457831 + \(\frac{24.98584x}{100}\) +  25.98259 - 20.52364784 - \(\frac{25.98259x}{100}\)

∴ \(\frac{0.99675}{100}\) x = 0.09272526,  ∴ x = 9.30%

100 - 78.99 - 9.30 = 11.71

∴ \(_{12}^{25}Mg\) : 9.30% by mass and

∴ \(_{12}^{26}Mg\): 11.71% by mass