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## Karnataka 2nd PUC Physics Question Bank Chapter 12 Atoms

### 2nd PUC Physics Atoms NCERT Text Book Questions and Answers

Question 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 ………….. 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) No different from

(b) Thomson’s model, Rutherford’s model

(c) Rutherford’s model

(d) Thomson’s model, Rutherford’s model

(e) both the models.

Question 2

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:

The nucleus of hydrogen contains only one proton. Its mass is 1.67 x 10^{-27} kg. The mass of an alpha-particle is 6.64 x 10^{-27} kg. When the alpha particles are incident on a thin sheet of solid hydrogen [in place of the gold foil], even in a head-on collision, the alpha-particle will not be scattered back. It is because the scattering particle is more massive than the target particle.

Question 3.

What is the shortest wavelength present in the Paschen series of spectral lines?

Answer:

The wavelengths of the spectral lines in Paschen series are given by

Question 4.

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:

Question 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?

Answer:

The kinetic energy of the electron in an orbit is numerically equal to its total energy. Therefore, kinetic energy of the electron = 13.6 eV The potential energy of the electron in a orbit is equal to twice, its total energy

= -13.6 ×2

= -27.2 eV.

Question 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.

Answer:

The energy of the electron in ground state

E_{e }= -13.6 eV

Also, energy of the electron in n=4 level,

Question 7.

(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) The speed of electron in the n^{th} orbit of a H-atom is given by

From the equation (i), it follows that the velocity of electron in an orbit of a hydrogen atom is inversely proportional to its quantum number (n).

(b) The orbital period of an electron in the n^{th }orbit of a H-atom is given by

It follows that orbital period of an electron in H-atom is directly proportional to the cube of its quantum number.

Question 8.

The radius of the innermost electron orbit of a hydrogen atom is 5.3 x 10^{-11} What are the radii of the n = 2 and n =3 orbits?

Answer:

The radius of an orbit in hydrogen atom is directly proportional to the square of its quantum number i.e.

Question 9.

A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Answer:

The first, second and third excited energies for hydrogen atom are

E’ _{1} = E_{2} – E_{1} = -3.4 – [-13.6] = 10.2 eV,

E’_{2} = E_{3} – E_{1}= -1.51 – [-13.6] = 12.09 eV,

and E’_{3} = E_{4} – E_{1} = -0.85 – [-13.6] = 12.75 eV

It follows that when a 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature, the H – atoms at the most will be raised to n = 3 level. The H – atoms may come to ground state directly or via n = 2 level. Therefore, the wavelengths emitted will lie in Balmer series [for de-excitation from n = 3 level to n = 2 and 1 level] and Lyman series [for deexcitation from n = 2 level to n = 1 level].

Question 10.

In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 x 10^{11} m with orbital speed 3 x 10^{4} m/s. (Mass of earth = 6.0 x 10^{24})

Answer:

Here, m = 6.0 x 10^{24} kg; r = 1.5×10^{n}m

and v = 3 x 10^{4} ms^{-1
}The angular momentum of the earth,

mvr = 6.0×10^{24 }x 3 x 10^{4 }x 1.5 x 10^{11
}= 2.7 x 10^{40} kg ms^{-1 }

In accordance with the Bohr’s model

### 2nd PUC Physics Atoms Additional Exercise

Question 11.

Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better.

(a) Is the average angle of deflection of α-particles by a thin gold foil predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?

(b) Is the probability of backward scattering (i.e., scattering of α-particles at angles greater than 90°) predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?

(c) Keeping other factors fixed, it is found experimentally that for small thickness t, the number of α-particles scattered at moderate angles is proportional to t. What clue does this linear dependence on t provide?

(d) In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of a -particles by a thin foil?

Answer:

(a) Thomson’s atom model cannot explain the scattering of α -particles through large angles and particularly through 180°. It is because, the light particles like electrons cannot deflect the massive a -particles through 180°.

(b) Thomson’s model predicts the backward scattering of a -particles much less than that predicted by Rutherford’s model. The reason is that in Thomson’s model, positive charge is uniformly distributed over whole of the nucleus instead of being concentrated in a small part.

(c) The chance of single collision increases linearly with the number of targeted atoms and hence linearly with thickness. That the number of α -particles scattered at moderate angles is proportional to t, suggests that the scattering is predominantly due to single collision.

(d) On the basis of Thomson’s atom model, the scattering angle of a -particles due to single collision will be very small. It is because, mass of atom is distributed all over it. The observed average scattering angle can be explained only by considering multiple scattering. Therefore, it is wrong to ignore multiple scattering in Thomson’ atom model.

Since the mass of the atom is concentrated in the form of nucleus in Ruther ford’s atom model, most of the scattering comes through a single collision and multiple scattering can be ignored.

Question 12.

The gravitational attraction between electron and proton in a hydrogen atom is weaker than the Coulomb attraction by a factor of about 10^{-40}. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.

Answer:

Let m and m be the masses of electron and proton respectively. If the electron and proton were bound only by gravitational attraction, then

According to Bihr’s quantistion conditions,angular momentum of the electron.

Question 13.

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:

When a hydrogen atom de-excites from level n_{1} to n_{p} the frequency of radiation emitted is given by

Hence, when n is large, the frequency of radiation emitted is equal to the classical frequency of revolution of the electron in the orbit.

Question 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^{-10}m).

(a) Construct a quantity with the dimensions of length from the fundamental constants e, m_{e}, 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 non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr to 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, m_{e}, and e and confirm that its numerical value has indeed the correct order of magnitude.

Answer:

(a) The quantity with the dimensions of length constructed from the fundamental constants e, m and c is

(b) The quantity with the dimensions of length constructed from the fundamental constants h, e, and m is the radius of the innermost orbit (n = 1) of the H – atom. It is given by,

It is also called Bohr’s radius

Substituting the values of h, m_{e} and 4π∈_{o} in equation (1),

Question 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:

We know that K.E and P.E of the electron in the n^{th} energy state of hydrogen atom are respectively given by

The first excited state corresponds to n = 2 level. Now, total energy of electron in

n = 2 state is given to be – 3.4 eV.

(a) Therefore, K.E of the electron in the first excited state (n = 2) of hydrogen atom

(b) E of the electron in the first excited state (n = 2) of the hydrogen atom.

(c) If the zero of potential energy is chosen differently, K.E does not change. The P.E and total energy will change

Question 16.

If Bohr’s quantisation postulate (angular momentum = nh/2 π) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantisation of orbits of planets around the sun?

Answer:

The mass of a planet is very large as compared to that of electron and accordingly, its angular momentum is very large. For \(\frac { h }{ 2\pi } \) example, in terms of the angular, from the

Bohr’s quantisation relation,\(\frac { nh }{ 2\pi } \) it follows that for the orbital motion of the earth, n = 10^{70}. For such a large value of n, the difference in the energies of the successive energy levels are so small as compared to the energies associated with the energy levels that these levels form an energy continuum, i.e the levels appear to be continous.

Question 17.

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 207m_{e }orbits around a proton].

Answer:

Let m be the mass of the electron. the firest Bohr radius of H- atom is given by,

in a muonic hydrogen atom, in place of an electron, a negatively charged muon having mass 207 m. revolves around the nucleus of hydrogen i.e. proton.

If r_{1}‘ is first Bohr’s radius of muonic hydrogen atom,

### 2nd PUC Physics Atoms Additional Questions and Answers

Question 1.

What is the ratio of radii of the orbits corresponding to first excited state and ground state of hydrogen atom?

Answer:

Radius of n^{th} orbit, r_{n} α n^{2
}The ground state corresponds to n = 1 and the

first excited state corresponds to n = 2

Question 2.

The ground state energy of the hydrogen atom is -13.6eV. What are the kinetic and potential energies of the electron in this state?

Answer:

The potential energy (E_{p}) of the electron in an orbit is equal to twice its total

energy (E)

= -13.6 x 2 = -27.2eV

The kinetic energy (E_{k}) of the electron in an orbit is equal to negative of its total energy (E)

∴ E_{k} – E

= – (-13.6) = 13.6 eV.

Question 3.

The radius of 1st electron orbit of hydrogen atom is 5.3 x 10^{-11} m. What is the radius of 2^{nd} orbit?

Answer:

Question 4.

What is energy possessed by an electron in n = ∞ ?

Answer:

Energy of electron in n^{th} orbit of hydrogen atom is given by

Question 5.

If 13.6eV energy is required to ionise the hydrogen atom, then energy required to remove an electron from n = 2 is

(A) 2 eV

(B) 0 eV

(C) 3.4 eV

(D) 6.8 eV

Answer:

Required energy

Question 6.

The transition from state n = 3 to n = 1 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in transition from

(A) n = 2 to n = 1

(B) n = 3 to n = 2

(C) n = 4 to n = 2

(D) n = 5 to n = 4

Answer:

Infrared region lies for the Bracket series, i.e. the transition from n = 5 to n = 4.

Question 7.

Which of the following transition in a hydrogen atom emits the photon of highest frequency?

(A) n = 2 to n = 6

(B) n = 2 to n = 1

(C) n = 6 to n = 2

(D) n = 1 to n = 2

Answer:

\(v={ R }_{ c }\left( \frac { 1 }{ { n }_{ i }^{ 2 } } -\frac { 1 }{ { n }_{ f }^{ 2 } } \right) \)

Substituting various values of n and n_{p} it is observed, maximum frequency is obtained for transition from n = 2 to n = 1.

Question 8.

When a hydrogen atom is raised from ground to excited state,

(A) The P.E increases and K.E decreases

(B) The P.E decreases and K.E increases

(c) Both P.E and K.E increase

(d) Both P.E and K.E decreases

Answer:

(B) The P.E decreases and K.E increases

Question 9.

In which of these, will radius of first orbit be minimum?

(A) H atom

(B) ^{2}_{1}H atom

(C) He^{+} ion

(D) Li^{2+} ion

Answer:

Question 10.

The minimum wavelength of the x-rays produced by electrons accelerated through a potential difference of V(volt) is directly proportional to

(A) \(\sqrt { v } \)

(B) \({ v }^{ 2 }\)

(C) \(\frac { 1 }{ \sqrt { v } } \)

(D) \(\frac { 1 }{ v } \)

Answer:

(D) \(\frac { 1 }{ v } \)