Ncert Solution Class 12th Physics Chapter – 12 Atom
| Textbook | NCERT |
| class | Class – 12th |
| Subject | Physics |
| Chapter | Chapter – 12 |
| Chapter Name | Atom |
| Category | Class 12th Physics Question & Answer |
| Medium | English |
| Source | last doubt |
Ncert Solution Class 12th Physics Chapter – 12 Atom
?Chapter – 12?
✍Atom✍
?Question & Answer?
Q1. Choose the correct alternative from the clues given at the end of each statement:
(a) The size of the atom in Thomson’s model is ………. the atomic size in Rutherford’s model. (much greater than/no different from/much less than.)
(b) ) In the ground state of ………. electrons are in stable equilibrium, while in ………. electrons always experience a net force. (Thomson’s model/ Rutherford’s model.)
(c)A classical atom based on ………. is doomed to collapse. (Thomson’s model/ Rutherford’s model.)
(d) An atom has a nearly continuous mass distribution in a ……….but has a highly non-uniform mass distribution in ……….(Thomson’s model/ Rutherford’s model.)
(e)The positively charged part of the atom possesses most of the mass in ………. (Rutherford’s model/both the models.)
Answer:
(a) The size of the atom in Thomson’s model is no different from the atomic size in Rutherford’s model.
(b) In the ground state of Thomson’s model electrons are in stable equilibrium, while in Rutherford’s model electrons always experience a net force.
(c) A classical atom based on Rutherford’s model is doomed to collapse.
(d) An atom has a nearly continuous mass distribution in a Thomson’s model but has a highly non-uniform mass distribution in Rutherford’s model.
(e) The positively charged part of the atom possesses most of the mass in both the models.
Q2. Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?
Answer: We know that the mass of the incident alpha particle (6.64 × 10-27kg) is more than the mass of hydrogen (1.67 × 10-27Kg). Hence, the target nucleus is lighter, from which we can conclude that the alpha particle would not rebound. Implying to the fact that solid hydrogen isn’t a suitable replacement to gold foil for the alpha particle scattering experiment.
Q3. What is the shortest wavelength present in the Paschen series of spectral lines?
Answer:
Q4. A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?
Answer:
Separation of two energy levels in an atom, E = 2.3 eV = 2.3 × 1.6 × 10 −19 = 3.68 × 10 −19 J
Let ν be the frequency of radiation emitted when the atom transits from the upper level to the lower level.
We have the relation for energy as: E = hv
Where, h = Planck’s constant = 6.62 x 10-34 Js
∴ V = E/h = 3.68 x 10-19 / 6.62 x 10-32 = 5.55 x 1014 Hz
Hence, the frequency of the radiation is 5.6 × 10 14 Hz.
Q5. The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?
Answer:
Ground state energy of hydrogen atom, E = − 13.6 eV
The total energy of hydrogen atom is -13.6 eV. The kinetic energy is equal to the negative of the total energy.
Kinetic energy = − E = − (− 13.6) = 13.6 eV
Potential energy = negative of two times of kinetic energy.
Potential energy = − 2 × ( 13.6 ) = − 27.2 eV
Q6. A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.
Answer:
For ground level, n 1 = 1
Let E1 be the energy of this level. It is known that E1 is related with n1 as:
E1 = -13.6/n12 eV
= -13.6/12 = -13.6 eV
The atom is excited to a higher level, n2 = 4.
Let E2 be the energy of this level.
∴ E2 = -13.6/n22 eV
= -13.6/42 = -13.6/16 eV
The amount of energy absorbed by the photon is given as:
E = E2 – E1
= (-13.6 /16) – (-13.6/1)
= 13.6 X 15/16 eV
= (13.6 X 15/16) X 1.6 X 10-19 = 2.04 X 10-18 J
For a photon of wavelengthλ, the expression of energy is written as:
E = hc/λ
Where,
h = Planck’s constant = 6.6 × 10−34 Js
c = Speed of light = 3 × 108 m/s
∴ λ = hc/E
= (6.6×10-34x3x108)/(2.04×10-18)
= 9.7×10-8 m = 97 nm
And, frequency of a photon is given by the relation,
v = c/λ
= (3×108)/(9.7×10-8) ≈ 3.1 x 1015 Hz
Hence, the wavelength of the photon is 97 nm while the frequency is 3.1 × 1015 Hz.
Q7. (a) Using the Bohr’s model, calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels.
Answer:
(a) Let ν1 be the orbital speed of the electron in a hydrogen atom in the ground state level, n1 = 1. For charge (e) of an electron, ν1 is given by the relation,
ν 1 = e2/n14πϵ0(h/2π) = e2/2ϵ0h
Where, e = 1.6 × 10−19 C
ϵ0 = Permittivity of free space = 8.85 × 10-12 N−1 C2 m−2
h = Planck’s constant = 6.62 × 10−34 Js
∴ ν1 = (1.6×10-19)2/2×8.85×10-12x6.62×10-34 = 0.0218 x 108 = 2.18 x 106 m/s
For level n2 = 2, we can write the relation for the corresponding orbital speed as:
ν2 = e2/n22ϵ0h = (1.6×10-19)2/2x2x8.85×10-12x6.62×10-34 = 1.09 x 106 m/s
And, for n3 = 3, we can write the relation for the corresponding orbital speed as:
ν3 = e2/n32ϵ0h = (1.6×10-19)2/3x2x8.85×10-12x6.62×10-34 = 7.27 x 105 m/s
Hence, the speed of the electron in a hydrogen atom in n = 1, n=2, and n=3 is 2.18 × 10 6 m/s, 1.09 × 10 6 m/s, 7.27 × 10 5 m/s respectively.
(b) Let T 1 be the orbital period of the electron when it is in level n1 = 1.
Orbital period is related to orbital speed as:
T1 = 2πr1/ν 1
Where, r1 = Radius of the orbit
= n12h2ϵ0/πme2
h = Planck’s constant = 6.62 × 10−34 Js
e = Charge on an electron = 1.6 × 10−19 C
ϵ0 = Permittivity of free space = 8.85 × 10−12 N−1 C2 m−2
m = Mass of an electron = 9.1 × 10−31 kg
∴ T1 = 2πr1/ν 1
= (2πx(1)2x(6.62×10-34)2×8.85×10-12)/2.18×106xπx9.1×10-31x(1.6×10-19)2
= 15.27×10-17 = 1.527×10-16 s
For level n 2 = 2, we can write the period as:
T2 = 2πr2/ν 2
Where, r2 = Radius of the electron in n2 = 2
= (n2)2h2ϵ0/πme2
∴ T2 = 2πr2/ν2
= (2πx(2)2x(6.62×10-34)2x8.85×10-12)/1.09 x 106 x π x 9.1 x 10-31 x (1.6 x 10-19)2
= 1.22 x 10-15 s
And, for level n 3 = 3, we can write the period as:
T3 = 2πr3/ν 3
Where, r 3 = Radius of the electron in n 3 = 3
= (n3)2h2ϵ0/πme2
∴ T3 = 2πr3/ν 3
= (2πx(3)2x(6.62×10-34)2×8.85×10-12)/7.27 x 105 x π x 9.1 x 10-31 x (1.6 x 10-19)2
= 4.12 x 10-15 s
Hence, the orbital period in each of these levels is 1.52 × 10 −16 s, 1.22 × 10 −15 s, and 4.12 × 10 −15 s respectively.
Q8. The radius of the innermost electron orbit of a hydrogen atom is 5.3×10–11 m. What are the radii of the n = 2 and n =3 orbits?
Answer:
The radius of the innermost orbit of a hydrogen atom, r1 = 5.3 × 10−11 m.
Let r2 be the radius of the orbit at n = 2. It is related to the radius of the inner most orbit as:
r_2 = ( n ) ^ { 2 } r_ 1 = 4 × 5.3 × 10-11 = 2.12 × 10-10 m
For n = 3, we can write the corresponding electron radius as:
r_3 = ( n ) ^ { 2 } r_ 1 = 9 × 5.3 x 10-11 = 4.77 × 10-10 m
Therefore, 2.12 × 10−10 m and 4.77 × 10−10 m are the radii of an electron for n = 2 and n = 3 orbits respectively.
Q9. A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
Answer:
It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV.
Also, the energy of the gaseous hydrogen in its ground state at room temperature is −13.6 eV.
When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes −13.6 + 12.5 eV i.e., −1.1 eV.
Orbital energy is related to orbit level (n) as:
E = -13.6/(n)2 eV
For n = 3, E = -13.6/(9)2 = -1.5 eV
This energy is approximately equal to the energy of gaseous hydrogen. It can be concluded that the electron has jumped from n = 1 to n = 3 level.
During its de-excitation, the electrons can jump from n = 3 to n = 1 directly, which forms a line of the Lyman series of the hydrogen spectrum.
We have the relation for wave number for Lyman series as:
1/λ = Ry (1/12 – 1/n2)
Where,
Ry = Rydberg constant = 1.097 × 107 m−1
λ= Wavelength of radiation emitted by the transition of the electron
For n = 3, we can obtain λ as:
1/λ = 1.097 x 107 (1/12 – 1/32)
= 1.097×107(1-1/9) = 1.097×107x8/9
λ = 9/(8 x 1.097 x 107) = 102.55 nm
If the electron jumps from n = 2 to n = 1, then the wavelength of the radiation is given as:
1/λ = 1.097 x 107 (1/12 – 1/22)
= 1.097×107(1-1/4) = 1.097×107x3/4
λ = 4/(1.097×107x3) = 121.54nm
If the transition takes place from n = 3 to n = 2, then the wavelength of the radiation is given as:
1/λ = 1.097 x 107 (1/22 – 1/32)
= 1.097×107(1/4-1/9) = 1.097×107x5/36
λ = 36/(5×1.097×107) = 656.33 nm
This radiation corresponds to the Balmer series of the hydrogen spectrum.
Hence, in Lyman series, two wavelengths i.e., 102.5 nm and 121.5 nm are emitted. And in the Balmer series, one wavelength i.e., 656.33 nm is emitted.
Q10. In accordance with the Bohr’s model, find the quantum number that characterizes the earth’s revolution around the sun in an orbit of radius 3 × 1011 m with orbital speed 3 × 104 m/s. (Mass of earth = 6.0 × 1024 kg.)
Answer:
Radius of the orbit of the Earth around the Sun, r = 1.5 × 1011 m
Orbital speed of the Earth, ν = 3 × 104 m/s
Mass of the Earth, m = 6.0 × 1024 kg
According to Bohr’s model, angular momentum is quantized and given as:
mvr = nh/2π
Where,
h = Planck’s constant = 6.62 × 10−34 Js
n = Quantum number
∴ n = mvr2π/h
= (2πx6x1024x3x104x1.5×1011)/(6.62×10-34)
= 25.61×1073 = 2.6 x 1074
Hence, the quanta number that characterizes the Earth’ revolution is 2.6 × 1074 .
Q11. Choose a suitable solution to the given statements which justify the difference between Thomson’s model and Rutherford’s model
(a) In the case of scattering of alpha particles by a gold foil, average angle of deflection of alpha particles stated by Rutherford’s model is (less than, almost the same as, much greater than) stated by Thomson’s model.
(b) Is the likelihood of reverse scattering (i.e., dispersing of α-particles at points more prominent than 90°) anticipated by Thomson’s model ( considerably less, about the same, or much more prominent ) than that anticipated by Rutherford’s model?
(c) For a small thickness T, keeping other factors constant, it has been found that amount of alpha particles scattered at direct angles is proportional to T. This linear dependence implies?
(d) To calculate average angle of scattering of alpha particles by thin gold foil, which model states its wrong to skip multiple scattering?
Answer:
(a) almost the same
The normal point of diversion of alpha particles by a thin gold film anticipated by Thomson’s model is about the same as from anticipated by Rutherford’s model. This is on the grounds that the average angle was taken in both models.
(b) much less
The likelihood of scattering of alpha particles at points more than 90° anticipated by Thomson’s model is considerably less than that anticipated by Rutherford’s model.
(c) Dispersing is predominantly because of single collisions. The odds of a single collision increment linearly with the amount of target molecules. Since the number of target particles increment with an expansion in thickness, the impact likelihood depends straightly on the thickness of the objective.
(d) Thomson’s model
It isn’t right to disregard multiple scattering in Thomson’s model for figuring out the average angle of scattering of alpha particles by a thin gold film. This is on the grounds that a solitary collision causes almost no deflection in this model. Subsequently, the watched normal scattering edge can be clarified just by considering multiple scattering.
Q12. The gravitational attraction amongst proton and electron in a hydrogen atom is weaker than the coulomb attraction by a component of around 10−40. Another option method for taking a gander at this case is to assess the span of the first Bohr circle of a hydrogen particle if the electron and proton were bound by gravitational attraction. You will discover the appropriate response fascinating.
Answer:
Radius of the first Bohr orbit is given by the relation,
Where,
∈0 = Permittivity of free space
h = Planck’s constant = 6.63 × 10−34 Js
me = Mass of an electron = 9.1 × 10−31 kg
e = Charge of an electron = 1.9 × 10−19 C
mp = Mass of a proton = 1.67 × 10−27 kg
r = Distance between the electron and the proton
Coulomb attraction between an electron and a proton is given as:
Gravitational force of attraction between an electron and a proton is given as:
Where,
G = Gravitational constant = 6.67 × 10−11 N m2/kg2
If the electrostatic (Coulomb) force and the gravitational force between an electron and a proton are equal, then we can write:
Putting the value of equation (4) in equation (1), we get:
It is known that the universe is 156 billion light years wide or 1.5 × 1027 m wide. Hence, we can conclude that the radius of the first Bohr orbit is much greater than the estimated size of the whole universe.
Q13. Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n–1). For large n, show that this frequency equals the classical frequency of revolution of the electron in the orbit.
Answer:
It is given that a hydrogen atom de-excites from an upper level (n) to a lower level (n − 1).
We have the relation for energy (E1) of radiation at level n as:
Where
v1 = Frequency of radiation at level n
h = Planck’s constant
m = Mass of hydrogen atom
e = Charge on an electron
∈0 = Permittivityof free space
Now, the relation for energy (E2) of radiation at level (n − 1) is given as:
Where
v2 = Frequency of radiation at level (n − 1)
Energy (E) released as a result of de-excitation:
E = E2 − E1
hv = E2 − E1 ……………(iii)
Where,
v = Frequency of radiation emitted
Putting values from equations (i) and (ii) in equation (iii), we get:
For large n, we can write (2n − 1) ≃ 2n and (n − 1) ≃ n.
Classical relation of frequency of revolution of an electron is given as:
Where,
Velocity of the electron in the nth orbit is given as:
And, radius of the nth orbit is given as:
Putting the values of equations (vi) and (vii) in equation (v), we get:
Hence, the frequency of radiation emitted by the hydrogen atom is equal to its classical orbital frequency.
Q 14: Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (~ 10–10m).
(a) Construct a quantity with the dimensions of length from the fundamental constants e, me , and c. Determine its numerical value.
(b) You will find that the length obtained in (a) is many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in a non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr discard c and look for ‘something else’ to get the right atomic size. Now, the Planck’s constant h had already made its appearance elsewhere. Bohr’s great insight lay in recognising that h, me, and e will yield the right atomic size. Construct a quantity with the dimension of length from h, me, and e and confirm that its numerical value has indeed the correct order of magnitude.
Answer:
According to Coulomb’s law, the force between hydrogen nucleus and electron is giben as:
Q 15: The total energy of an electron in the first excited state of the hydrogen atom is about –3.4 eV.
(a) What is the kinetic energy of the electron in this state?
(b) What is the potential energy of the electron in this state?
(c) Which of the answers above would change if the choice of the zero of potential energy is changed?
Answer:
(a) Total energy of the electron, E = – 3.4 eV
The kinetic energy of the electron is equal to the negative of the total energy.
K.E = – E
= – (- 3.4 ) = + 3.4 eV
Kinetic energy = + 3.4 eV
(b) Potential energy (U) of the electron is equal to the negative of twice of the kinetic energy,
P.E = – 2 (K.E)
= – 2 x 3.4 = – 6.8 eV
Potential energy = – 6.8 eV
(c) The potential energy of the system depends on the reference point. If the reference point is changed from zero then the potential energy will change. The total energy is given by the sum of the potential energy and kinetic energy. Therefore, the total energy will also change.
Q16: If Bohr’s quantization postulate (angular momentum = nh/2π) is a basic law of nature, it should be equally valid for the case of planetary motion as well. Why then do we never speak of quantization of orbits of planets around the sun?
Answer: The quantum level for a planetary motion is considered to be continuous. This is because the angular momentum associated with planetary motion is largely relative to the value of Planck’s constant (h). 1070h is the order of the angular momentum of the Earth in its orbit. As the values of n increase, the angular momenta decreases. So, the planetary motion is considered to be continuous.
Q17: Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom [i.e., an atom in which a negatively charged muon (μ−) of mass about 207me orbits around a proton].
Answer:
The charge of the muon ,
So, at equilibrium,
The Rydberg’s formula for the hydrogen atom is
Where R is Rydberg constant for the Hydrogen atom and equals to 1.1
For shortest wavelength in Paschen Series n1=2 and n2=
The shortest wavelength in Paschen Series is therefore 818 nm.