Maharashtra State Board 12th Physics Solutions Chapter 15 Structure of Atoms and Nuclei
In solving problems, use me= 0.00055 u = 0.5110 MeV/c2, mp= 1.00728 u, mn= 1.00866u, mH= 1.007825 u, u = 931.5 MeV, e = 1.602 × 10-19C, h = 6.626 × 10-34Js, ε0= 8.854 × 10-12SI units and me = 9.109 × 10-31kg.
1. Choose the correct option.
i) In which of the following systems will the radius of the first orbit of the electron be the smallest?
(A) hydrogen
(B) singly ionized helium
(C) deuteron
(D) tritium
Answer:
(D) tritium
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
Answer:
(B) 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
Answer:
(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
Answer:
(D) density
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)
(B)
(C)
(D)
Answer:
(B)
2. Answer in brief.
i) State the postulates of Bohr’s atomic model.
Answer:
The postulates of Bohr’s atomic model (for the hydrogen atom) :
ii) State the difficulties faced by Rutherford’s atomic model.
Answer:
(1) According to Rutherford, the electrons revolve in circular orbits around the atomic nucleus. The circular motion is an accelerated motion. According to the classical electromagnetic theory, an accelerated charge continuously radiates energy. Therefore, an electron during its orbital motion, should go on radiating energy. Due to the loss of energy, the radius of its orbit should go on decreasing. Therefore, the electron should move along a spiral path and finally fall into the nucleus in a very short time, of the order of 10-16s in the case of a hydrogen atom. Thus, the atom should be unstable. We exist because atoms are stable.
(2) 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 con-tinuous spectrum. However, atomic spectrum is a line spectrum.
iii) What are alpha, beta and gamma decays?
Answer:
(a) 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.
Example :
Q = [mu– mTh– mα]c2
(b) 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.
Example :
where is the neutrino emitted to conserve the momentum, energy and spin.
Q = [mu– mTh– mα]c2
In a β+-decay, the atomic number of the nucleus decreases by 1 and the mass number remains unchanged.
Example : {aligned}
&30 \\
&15
{aligned} P →{ }_{14}^{30} Si+{ }_{+1}^{0} e+v_{e}
where veis the neutrino emitted to conserve the momentum, energy and spin.
Q = [mP– mSi– me]c2
[Note : The term fi particle refers to the electron (or positron) emitted by a nucleus.]
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.
iv) Define excitation energy, binding energy and ionization energy of an electron in an atom.
Answer:
(1) 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.
(2) 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.
(3) 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.
Answer:
For the first line in the Lyman series,
∴ vL1= , 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,
Hence, the result.
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
∴ ……………. (1)
where ε0is the permittivity of free space.
∴ Kinetic energy (KE) of the electron
= ………….. (2)
The electric potential due to the nucleus of charge +Ze at a point at a distance r from it is
V =
∴ Potential energy (PE) of the electron
= charge on the electron × electric potential
= – e × …………….. (3)
Hence, the total energy of the electron in the nth orbit is
E = KE + PE =
∴ E = ………….. (4)
This shows that the total energy of the electron in the nth orbit of the atom is inversely proportional to the radius of the orbit as Z, ε0and e are constants. The radius of the nth orbit of the electron is
r = …………….. (5)
where h is Planck’s constant.
From Eqs. (4) and (5), we get,
En= ……………… (6)
This gives the expression for the energy of the electron in the nth 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, ε0and h are constant, we get
En∝
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.
[ Note : Energy levels are most conveniently expressed in electronvolt. Hence, substituting the values of m, e, £0 and h, and dividing by the conversion factor 1.6 × 10-19J/eV,
En≅ (in eV)
For hydrogen, Z = 1
∴ En≅ (in eV).
The energy of the electron in a hydrogen atom,
when it is in an orbit with the principal quantum
number n, is
En=
where m = mass of electron, e = electronic charge, h = Planck’s constant and = permittivity of free space.
Let Embe the energy of the electron in a hydrogen atom when it is in an orbit with the principal quantum number m and E, its energy in an orbit with the principal quantum number n, n < m. Then
Em= and En=
Therefore, the energy radiated when the electron jumps from the higher energy state to the lower energy state is
Em– En=
=
This energy is emitted in the form of a quantum of radiation (photon) with energy hv, where V is the frequency of the radiation.
∴ Em– En= hv
∴ v =
The wavelength of the radiation is λ =
where c is the speed of radiation in free space.
The wave number,
where is a constant called the Ryd berg constant.
This expression gives the wave number of the radiation emitted and hence that of a line in hydrogen spectrum.
For the Lyman series, n = 1,m = 2, 3, 4, ………… ∞
∴ and for the shortest wavelength line m this series, as m = ∞.
For the Balmer series, n = 2, m = 3, 4, 5, … ∞.
∴ and for the shortest wavelength line in this series, as m = ∞
[Note: Johannes Rydberg (1854—1919), Swedish spectroscopist, studied atomic emission spectra and introduced the idea of wave number. The empirical formula where m and n are simple integers, is due to Rydberg. When we consider the finite mass of the nucleus, we find that R varies slightly from element to element.]
Obtain the formula for ω and continue as follows :
This is required quantity.
(b) 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.
Example :
where is the neutrino emitted to conserve the momentum, energy and spin.
Q = [mu– mTh– mα]c2
In a β+-decay, the atomic number of the nucleus decreases by 1 and the mass number remains unchanged.
Example : {aligned}
&30 \\
&15
{aligned} P →{ }_{14}^{30} Si+{ }_{+1}^{0} e+v_{e}
where veis the neutrino emitted to conserve the momentum, energy and spin.
Q = [mP– mSi– me]c2
[Note : The term fi particle refers to the electron (or positron) emitted by a nucleus.]
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.
The products of the fission of235U 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
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 108K.
(2) The value of the energy released in the fusion of two deuteron nuclei and the temperature at which the reaction occurs mentioned in the textbook are probably misprints.]
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.
[Note : The first nuclear bomb (atomic bomb) was dropped on Hiroshima in Japan on 06 August 1945. The second bomb was dropped on Nagasaki in Japan on 9 August 1945.]
(b)
Here, e–≡ is emitted and fluorine is formed.
(c)
Here, α particle is emitted and radium is formed.
(d)
is e+(positron)
Here, β+is emItted and carbon is formed.
(b)
The energy released in this reaction =
(∆M) c2= [236.0456 – (139.9106 + 93.9341 + (2)(1.00866)1(93 1.5)MeV
= 171.00477 MeV
(c) + neutrino
The energy released in this reaction = (∆M) c2
= [11.01143 – (11.0093 + O.00055)](931.5) MeV
= 1.47177 MeV
(b)A0= N0A ∴ N0= = A0τ
= (7.4 × 104)(3.391 × 107)
= 2.509 × 1012nuclei
This is the required number.
(c) A(t) = A0e-λt= 2e-(2.949 × 10-8)(3.156 × 107)
= 2e-0.9307= 2 / e0.9307
Let x = e0.9307∴ Iogex = 0.9307
∴ 2.303log10x = 0.9307
∴ log10x = = 0.4041
∴ x = antilog 0.4041 = 2.536
∴ A (t) = μCi = 0.7886 μCi
(a) A = Nλ ∴ N =
= 6.654 × 1010
Number of atoms in 1 g of carbon =
=5.017 × 1022
= 0.7539 × 1012
∴ 114C atom per 0.7539 × 1012atoms of carbon
∴ 414C atoms per 3 × 1012atoms of carbon
(b) Present activity per gram =
= 0.09666 dis/s per gram
A0= 0.255 dis/s per gram
Now, A(t) = A0e-λt
This is the required quantity.
12th Physics Digest Chapter 15 Structure of Atoms and Nuclei Intext Questions and Answers
Use your brain power (Textbook Page No. 336)