## Structure of Atoms and Nuclei

### Maharashtra Board-Class-12th-Physics-Chapter-15

### Solutions

In solving problems, use m_{e} = 0.00055 u = 0.511 MeV/c, ^{2}m_{p} = 1.00728 u = 938.272 MeV/c, ^{2}m_{n} = 1.00866 u = 939.565 MeV/cand ^{2} m_{H} = 1.007825 u. |

**Question 1. Choose the correct option.**

**(i) In which of the following systems will the radius of the first orbit of the electron be smallest?**

**(A) hydrogen **

**(B) singly ionized helium**

**(C) deuteron **

**(D) tritium**

(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.

**(ii) The radius of the 4th orbit of the electron will be smaller than its 8th orbit by a factor of**

**(A) 2 **

**(B) 4**

**(C) 8 **

**(D) 16**

(B) 4

**Explanation :**

Radius of the 4th orbit r_{4} = 0.529 × 10^{-8} × 4^{2}

Radius of the 8th orbit r_{8} = 0.529 × 10^{-8} × 8^{2}

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

**(iii) In the spectrum of hydrogen atom which transition will yield longest wavelength?**

**(A) n = 2 to n = 1 **

**(B) n = 5 to n = 4**

**(C) n = 7 to n = 6 **

**(D) n = 8 to n = 7**

(D) *n *= 8 to *n *= 7

**(iv) Which of the following properties of a nucleus does not depend on its mass number?**

**(A) radius **

**(B) mass**

**(C) volume **

**(D) density**

(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.

**(v) If the number of nuclei in a radioactive sample at a given time is N, what will be the number at the end of two half-lives?**

**(A) \(\frac{N}{2}\)**

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

**(C) \(\frac{3N}{2}\)**

**(D) \(\frac{N}{8}\)**

(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, \(\frac{N/2}{2}\) = \(\frac{N}{4}\)

**Question 2. Answer in brief.**

**(i) State the postulates of Bohr’s atomic model.**

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.

**(ii) State the difficulties faced by Rutherford’s atomic model.**

- 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.

**(iii) What are alpha, beta and gamma decays?**

**∝****-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.

**(iv) Define excitation energy, binding energy and ionization energy of an electron in an atom.**

**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.

**(v) Show that the frequency of the first line in Lyman series is equal to the difference between the limiting frequencies of Lyman and Balmer series.**

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.

**Question 3. State the postulates of Bohr’s atomic model and derive the expression for the energy of an electron in the atom.**

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 ε_{0} 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, ε_{0} 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, ε_{0} and h are constant, we get

E_{n} ∝ 1/n^{2}

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.

**Question 4. Starting from the formula for energy of an electron in the nth orbit of hydrogen atom, derive the formula for the wavelengths of Lyman and Balmer series spectral lines and determine the shortest wavelengths of lines in both these series.**

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 ε_{0} = 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 = ∞

**Question 5. Determine the maximum angular speed of an electron moving in a stable orbit around the nucleus of hydrogen atom.**

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 ε_{0} = 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 10 ^{16} rad/s**

This is the required quantity.

**Question 6. Determine the series limit of Balmer, Paschen and Pfund series, given the limit for Lyman series is 912 Å.**

**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.

**Question 7. Describe alpha, beta and gamma decays and write down the formulae for the energies generated in each of these decays.**

**(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)

**Question 8. Explain what are nuclear fission and fusion giving an example of each. Write down the formulae for energy generated in each of these processes.**

**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 ^{235}U 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 10^{7} K to 10^{8} 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 10^{8} K.

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

(deuteron + triton → Helium nucleus + neutron)

**Question 9. Describe the principles of a nuclear reactor. What is the difference between a nuclear reactor and a nuclear bomb?**

**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/c^{2}

The binding energy of an alpha particle = (Zm_{p} +Nm_{n} - M)c^{2}

= (2m_{p} + 2m_{n} - M)c^{2}

= [(2)(1.00728u) +2(1.00866 u) - 4.00151 u]c^{2}

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

= 28.289655 MeV

= 28.289655 × 10^{6} eV × 1.602 × 10^{-19} J

= 4.532002731 x 10^{-12} J

**Question 11. An electron in hydrogen atom stays in its second orbit for 10 ^{-8} s. How many revolutions will it make around the nucleus in that time?**

**Given** : z = 1, m = 9.1 x 10^{-31} kg, e = 1.6 × 10^{-19} C, ε_{0} = 8.85 x 10^{-12} C^{2}/N.m^{2}, 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 × 10 ^{7}**

**Question 12. Determine the binding energy per nucleon of the americium isotope ** ** \(_{95}^{244}Am\)****, given the mass of ** ** ****to be 244.06428 u.**

**Given :** Z = 95, N = 244 - 95 = 149, m_{p} =1.00728 u, m_{n} = 1.00866 u,

M = 244.06428 u, 1u = 931.5MeV/c^{2}

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**

**Question 13. Calculate the energy released in the nuclear reaction \(_{3}^{7}Li\) ****+ p ****→** **2**∝ **given mass of \(_{3}^{7}Li\) ****atom and of helium atom to be 7.016 u and 4.0026 u respectively.**

**Given** : M_{1} (\(_{3}^{7}Li\) atom) =7.016 u, M_{2} (He atom) = 4.0026u, m_{p} = 1.00728 u, 1u = 931.5MeV/c^{2}

ΔM = M_{1} + m_{p} - 2M_{2}

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

= 0.01808 u = (0.01808)(931.5) MeV/c^{2}

= 16.84152 MeV/c

Therefore, the energy released in the nuclear reaction = (ΔM)c^{2} =16.84152 MeV

**Question 14. Complete the following equations describing nuclear decays.**

**(a) \(_{88}^{226}Ra\)** ** ****→ ****∝ +**

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

**(c) ** ** \(_{90}^{228}Th\) ****→ ****∝ +**

**(d) \(_{7}^{12}N\) ** ** ****→ ** ** \(_{6}^{12}C\) +**

(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.

**Question 15. Calculate the energy released in the following reactions, given the masses to be \(_{88}^{223}Ra\)** **: 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. Ignore neutrino energy.**

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

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

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

**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/c^{2}

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

The energy released in this reaction = (ΔM)c^{2}

= [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)c^{2}

= [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)c^{2}

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

= 1.47177 MeV

**Question 16. Sample of carbon obtained from any living organism has a decay rate of 15.3 decays per gram per minute. A sample of c arbon obtained from very old charcoal shows a disintegration rate of 12.3 disintegrations per gram per minute. Determine the age of the old sample given the decay constant of carbon to be 3.839 × 10 ^{-12} per second.**

**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) = A_{0}e^{-}^{λ}^{t}

∴ e^{λ}^{t} = \(\frac{A_0}{A}\), ∴ λt = log_{e}\((\frac{A_0}{A})\)

∴ t = \(\frac{2.303}{λ}\)log_{10 }\((\frac{A_0}{A})\)

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

= \(\frac{2.303}{3.839×10^{-12}}\)(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 10^{12} s = **5.687 x 10 ^{10} s**

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

**Question 17. The half-life of \(_{38}^{90}Sr\)** **is 28 years. Determine the disintegration rate of its 5 mg sample.**

**Given** : T_{1/2} = 28 years = 28 x 3.156 × 10^{7} s = 8.837 × 10^{8} s, M = 5mg = 5 ×10^{-3} g

90 grams of \(_{38}^{90}Sr\) contain 6.02 x 10^{23} atoms

Hence, here, N = \(\frac{(6.02×10^{23})(5×10^{-3})}{90}\) = 3.344 x 10^{19} atoms

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

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

**= 2.622 ×10 ^{10} disintegrations per second**

**Question 18. What is the amount of \(_{27}^{60}Co\)**** ****necessary to provide a radioactive source of strength 10.0 mCi, its half-life being 5.3 years?**

**Given** : Activity = 10.0 mCi = 10.0 × 10^{-3} Ci = (10.0 × 10^{-3})(3.7 × 10^{10}) dis/s = 3.7 × 10^{8} dis/s

T_{1/2} = 5.3 years = (5.3)(3.156 × 10^{7}) s = 1.673 × 10^{8} 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 × 10^{16} atoms

= 60 grams of contain 6.02 x 10^{23} atoms

∴ Mass of 8.933 x 10^{16} 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.

**Question 19. Disintegration rate of a sample is 10 ^{10} per hour at 20 hrs from the start. It reduces to 6.3 **

**×**

**10**

^{9}per hour after 30 hours. Calculate its half life and the initial number of radioactive atoms in the sample.**Given** : A(t_{1}) = 10^{10} per hour, where t_{1} = 20 h,

A(t_{2}) = 6.3 × 10^{9} per hour, where t_{2} = 30 h

A(t) = A_{0}e^{-}^{λ}^{t} , ∴ A(t_{1}) = A_{0}e^{-}^{λ}^{t}^{1} and A(t_{2}) = A_{0}e^{-}^{λ}^{t}^{2}

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

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

∴ 1.587 = e^{10}^{λ}, ∴ 10λ = 2.303 log_{10}(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, A_{0} = A(t_{1})e^{λ}^{t}^{1} = 10^{10}e^{(0.04622)(20)} = 10^{10}e^{0.9244}

Let x = e^{0.9244 }, ∴ 2.303 log_{10}x = 0.9244

∴ log_{10}x = \(\frac{0.9244}{2.303}\) = 0.4014

∴ x = antilog 0.4014 = 2.52

∴ A_{0} = 2.52 x 10^{10} per hour

Now A_{0} = N_{0}λ, ∴ N_{0 }= \(\frac{A_0}{λ}\) = \(\frac{2.52×10^{10}}{0.04622}\) = **5.452 x 10 ^{11}**

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

**Question 20. The isotope ^{57}Co decays by electron capture to ^{57}Fe with a half-life of 272 d. The ^{57}Fe nucleus is produced in an excited state, and it almost instantaneously emits gamma rays. **

**(a) Find the mean lifetime and decay constant for ^{57}Co. **

**(b) If the activity of a radiation source ^{57}Co is 2.0 μCi now, how many ^{57}Co nuclei does the source contain? **

**(c) What will be the activity after one year?**

**Given **: T_{1/2} = 272 d = 272 x 24 x 60 × 60 s = 2.35 × 10^{7} s,

A_{0} = 2.0 μCi = 2.0 x 10^{-6} x 3.7 × 10^{10} = 7.4 × 10^{4} dis/s. t = 1 year = 3.156 × 10^{7} s

**(a)** T_{1/2} = \(\frac{0.693}{λ}\) = 0.693 τ

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

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

**(b)** A_{0} = N_{0}λ, ∴ N_{0 }= \(\frac{A_0}{λ}\) = A_{0}τ = (7.4 × 10^{4})(3.391 × 10^{7}) = **2.509 × 10 ^{12} nuclei**

This is the required number.

**(c)** A(t) = A_{0}e^{-}^{λ}^{t} = = 2e^{-0.9307}= 2/e^{0.9307}

Let x = e^{0.9307}

∴ log x = 0.9307

∴ 2.303 log_{10} x = 0.9307

∴ log_{10} x = \(\frac{0.9307}{2.303}\) = 0.4041

∴ x = antilog 0.4041 = 2.536

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

**Question 21. A source contains two species of phosphorous nuclei, \(_{15}^{32}P\) ****( T_{1/2} = 14.3 d) and \(_{15}^{33}P\)**

**(**

*T*_{1/2}= 25.3 d). At time*t*= 0, 90% of the decays are from \(_{15}^{32}P\)**. How much time has to elapse for only 15% of the decays to be from \(_{15}^{32}P\)**

**?**

**Given** : \(_{15}^{32}P\) : *T*_{1/2} = 14.3 d

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

\(_{15}^{33}P\) : *T*_{1/2} = 25.3 d, ∴ λ_{2} = \(\frac{0.693}{25.3d}\) = 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,

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

∴ \(e^{(λ_1-λ_2)t}=\frac{153}{3}\)

∴ (λ_{1} - λ_{2})t = 2.303 log_{10}\(\frac{153}{3}\)= 2.303 (log_{10} 153 - log_{10} 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.

**Question 22. Before the year 1900 the activity per unit mass of atmospheric carbon due to the presence of 14C averaged about 0.255 Bq per gram of carbon. (a) What fraction of carbon atoms were ^{14}C? (b) An archaeological specimen containing 500 mg of carbon, shows 174 decays in one hour. What is the age of the specimen, assuming that its activity per unit mass of carbon when the specimen died was equal to the average value of the air? Half-life of ^{14}C is 5730 years?**

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 × 10^{10}

Number of atoms in 1 g of carbon = \(\frac{6.02×10^{23}}{12}\) = 5.017 × 10^{22}

\(\frac{5.017×10^{22}}{6.654×10^{10}}\) = 0.7539 × 10^{12}

∴ 1 ^{14}C atom per 0.7539 x 10^{12} atoms of carbon

∴ 4 ^{14}C atoms per 3 x 10^{12} atoms of carbon

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

A_{0} = 0.255 dis/s per gram

Now, A(t) = A_{0}e^{-}^{λ}^{t}

∴ λt = 2.303 log_{10} \((\frac{A_0}{A})\) = 2.303 log_{10 }\((\frac{0.255}{0.09666})\)

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

This is the required quantity.

**Question 23. How much mass of ^{235}U is required to undergo fission each day to provide 3000 MW of thermal power? Average energy per fission is 202.79 MeV**

**Given** : Power=3000 MW = 3 × 10^{9} J/s

∴ Energy to be produced each day = 3 × 10^{9} × 86400 J each day = 2.592 × 10^{14} J each day

Energy per fission = 202.79 MeV = 202.79 × 10^{6} × 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 10^{24} each day

0.235 kg of ^{235}U contains 6.02 x 10^{23} atoms

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

This is the required quantity.

**Question 24. In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are \(_{12}^{24}Mg\) (23.98504 u), \(_{12}^{25}Mg\)(24.98584 u) and \(_{12}^{26}Mg\) (25.98259 u). The natural abundance of \(_{12}^{24}Mg\) is 78.99% by mass. Calculate the abundances of other two isotopes.**

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**

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