NCERT solutions for class 12 physics chapter 12 Atoms: Do you know that an atom is electrically neutral, which tells that atoms contain an equal number of positive and negative charges. The solutions of NCERT class 12 physics chapter 12 atoms explain questions related to different models of the atom, their drawbacks, hydrogen spectra, and Debroglie hypothesis. CBSE NCERT solutions for class 12 physics chapter 12 atoms will help you in board exam preparation. The solutions of NCERT plays a major role in making the concepts easy and hence will help in competitive exams also. Some of the important formulas of the chapter atoms which will help in NCERT solutions for class 12 physics chapter 12 atoms are given below.
The various series in line spectrum of atomic hydrogen are:
Where n is the quantum number
The total energy of the electron in the stationary states of the hydrogen atom in electronvolt is given by
Another important formula which is used in NCERT solutions for class 12 physics chapter 12 atoms is the
De Broglie wavelength
Where ‘h’ is Planck's constant and mv is the momentum. This relation can be modified in terms of the kinetic energy of the particle. Try to derive it yourself.
Q 12.1 Choose the correct alternative from the clues given at the end of the each statement:
(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.
On repeating the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil we would have different observations than Rutherford, as the alpha particles won't be scattered much because of being heavier than the nucleus of the Hydrogen atom. Therefore we would not be able to confirm the presence of almost the entire mass of the atom at its centre.
Q 12.3 What is the shortest wavelength present in the Paschen series of spectral lines?
The Rydberg's formula for the hydrogen atom is
Where R is Rydberg constant for the Hydrogen atom and equals to 1.110^{7} m^{-1}
For shortest wavelength in Paschen Series n_{1}=2 and n_{2}=
The shortest wavelength in Paschen Series is therefore 818 nm.
Frequency of radiation consisting of photons of energy E is given by
E=2.3 eV
Plank's constant(h)=6.6210^{-34 }Js
The ground state energy E=-13.6 eV.
The kinetic energy= -E=13.6 eV
Also ground state energy = Kinetic energy+Potential energy
E=K+U
U=E-K
U=-13.6-13.6
U=-27.2 eV
The kinetic and potential energies are 13.6 eV and -27.2 eV respectively.
The initial energy of the electron is E_{1}
E_{1 }=-13.6 eV
The energy of the electron when it is excited to level n=4 is E_{2}
E_{2}=-0.85 eV
The difference between these two energy levels is equal to the energy of the photon absorbed by the electron.
The energy of the photon E = E_{2 }- E_{1}
E = -0.85 -(-13.6)
E = 12.75 eV
The wavelength of the photon can be calculated using relation
hc=1240 eV
The wavelength and frequency of the photon absorbed by the hydrogen atom are 98.6 nm and 3.0410^{15} Hz respectively.
As per Bohr's model the angular momentum of electrons in each orbit is constant and a multiple of
(i)
The electrostatic force of attraction between the electron and the nucleus provides the required centripetal force for the circular motion of the electron.
(ii)
Using equation (i) and (ii) we get
v_{1}=2.1810^{6} ms^{-1}
v_{2}=1.0910^{6} ms^{-1}
v_{3}=7.2810^{5} ms^{-1}
(b) calculate the orbital period in each of these levels.
Orbital period (T_{n }) is defined as time taken by the electron to complete one revolution around the nucleus and is given by
T_{1}=1.5310^{-16} s
T_{2}=1.2210^{-15} s
T3=4.1210^{-15} s
The radius of the orbit is proportional to the square of n.
For n=2 the radius of the orbit is
For n=3 the radius of the orbit is
Since the energy of the electron beam is 12.5 eV the Hydrogen atoms will get excited to all requiring energy equal to or less than 12.5 eV
E_{1 }=-13.6 eV
E_{3 }= -1.5 eV
E_{3} -E_{1} = 12.1 eV
E_{4}= -0.85 eV
E_{4}-E_{1}=12.75 eV
Therefore the electron can reach maximum upto the level n=3.
During de-excitations, the electron can jump directly from n=3 to n=1 or it can first jump from n=3 to n=2 and then from n=2 to n=1
Therefore two wavelengths from the Lyman series and one from the Balmer series will be emitted
To find the wavelengths emitted we will use the Rydberg's Formula
where R is the Rydberg's constant and equals 1.09710^{7} m^{-1}
For n_{1}=1 and n_{2}=3
Emitted wavelength is 102.5 nm
For n_{1}=1 and n_{2}=2
Emitted wavelength is 121.54 nm
For n_{1}=2 and n_{2}=3
Emitted wavelength is 656.3 nm
n = 2.5610^{74}
Therefore the quantum number that characterises the earth’s revolution around the sun in an orbit of radius m with an orbital speed
is 2.5610^{74}
The average angle of deflection of -particles by a thin gold foil predicted by both the models is about the same.
The probability of backward scattering predicted by Thomson’s model is much less than that predicted by Rutherford’s model.
Scattering at moderate angles requires head-on collision the probability of which increases with the number of target atoms in the path of -particles which increases linearly with the thickness of the gold foil and therefore the linear dependence between the number of -particles scattered at a moderate angle and the thickness t of the gold foil.
It is completely wrong to ignore multiple scattering for the calculation of the average angle of scattering of -particles by a thin foil in Thomson's model as the deflection caused by a single collision in this model is very small.
As per the bohrs model
(i)
If the proton and the electron were bound only by the gravity the gravitational force between them will provide the centripetal force required for circular motion
(ii)
From equation (i) and (ii) we can calculate that the radius of the ground state (for n=1) will be
The above value is larger in order than the diameter of the observable universe. This shows how much weak the gravitational forces of attraction as compared to electrostatic forces.
Using Bohr's model we have.
Since n is very large 2n-1 can be taken as 2n and n-1 as n
The frequency of the emission caused by de-excitation from n to n-1 would be
The classical frequency of revolution of the electron in the nth orbit is given by
The above is the same as the frequency of the emission during de-excitation from n to n-1.
Using dimensional analysis we can see that the quantity to be constructed and consisting of m_{e}, e and c will also have and will be equal to
and has numerical value 3.510^{-14 }which is much smaller than the order of atomic radii.
Using dimensional analysis we can see that the quantity to be constructed and consisting of m_{e}, e and h will also have and will be equal to
and has a numerical value of approximately 6.65710^{-10 }which is about the order of atomic radii.
Q 12.15 (a) The total energy of an electron in the first excited state of the hydrogen atom is about –3.4 eV.
What is the kinetic energy of the electron in this state?
Since we know that kinetic energy is equal to the negative of the total energy
K=-E
K=-(-3.4)
K=3.4 eV
Q 12.15 (b) The total energy of an electron in the first excited state of the hydrogen atom is about –3.4 eV.
What is the potential energy of the electron in this state?
Total Energy= Potential energy + Kinetic Energy
E=U+K
U=E-K
U=-3.4-3.4
U=-6.8 eV
Q 12.15 (c) The total energy of an electron in the first excited state of the hydrogen atom is about - 3.4eV.
Which of the answers above would change if the choice of the zero of potential energy is changed?
The total energy would change if the choice of the zero of potential energy is changed.
We never speak of Bohr's quantization postulate while studying planetary motion or even motion of other macroscopic objects because they have angular momentum very large relative to the value of h. In fact, their angular momentum is so large as compared to the value of h that the angular momentum of the earth has a quantum number of order 10^{70} . Therefore the angular momentum of such large objects is taken to be continuous rather than quantized.
As per Bohr's quantization postulate
Similarly, like the case in a simple hydrogen atom, the electrostatic force acts centripetally
From the above relations, we can see that in Bohr's model the Radius is inversely proportional to the mass of the orbiting body and Energy is directly proportional to the mass of the orbiting body.
In case of hydrogen, atom r_{1} is 5.310^{-11 }m
Therefore in case of a muonic hydrogen atom
r_{1} = 2.5610^{-13 }m
In case of the hydrogen atom, E_{1} is -13.6 eV
Therefore in case of a muonic hydrogen atom
E_{1}=207(-13.6)
E_{1}=2.81 keV
For CBSE board exam on an average 3 to 4 marks questions are asked from the chapter. The solutions of NCERT class 12 physics chapter 12 atoms will help in securing full marks in board exam for this chapter. As far as the NEET exam is considered up to 2 questions are expected from the chapter. The NCERT solutions for class 12 physics chapter 12 atoms will also help to perform better in exams like JEE Main and other competitive exams.
Q 12.8 The radius of the innermost electron orbit of a hydrogen atom is . What are the radii of the n = 2 and n =3 orbits?
Q 12.6 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.
Q 12.5 The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?