IIT-JEE 2009- Multiple Correct Choice Type Questions on Electromagnetic Induction, Resonance Column & Kepler’s Laws

Today we will discuss three Multiple Correct Choice Type questions which appeared in the IIT-JEE 2009 question paper. Even though these questions are named ‘multiple correct choice’ type the number of correct choices can be one or more. These questions are simple, in line with the current trend, as you can see:

(1) Two metallic rings A and B, identical in shape and size but having different resistivities ρ_A and ρ_B, are kept on top of two identical solenoids as shown in the figure. When current I is switched on in both the solenoids in identical manner, the rings A and B jump to heights h_A and h_B respectively, with h_A > h_B. The possible relation(s) between their resistivities and their masses m_A and m_B is(are)

(A) ρ_A > ρ_B and m_A = m_B

(B) ρ_A < ρ_B and m_A = m_B

(C) ρ_A > ρ_B and m_A > m_B

(D) ρ_A < ρ_B and m_A < m_B

The voltages induced in the rings are equal. Since ring A jumps to greater height compared to ring B, the current in ring A must be greater. This means that the resistivity of ring A is less than that of ring B. So ρ_A < ρ_B.

Ring A can jump to a greater height if the mass of ring A is equal to or less than that of B. So the correct options are (B) and (D).

(2) A student performed the experiment to measure the speed of sound in air using resonance air-column method. Two resonances in the air-column were obtained by lowering the water level. The resonance with the shorter air-column is the first resonance and that with the longer air-column is the second resonance. Then

(A) the intensity of the sound heard at the first resonance was more than that at the second resonance

(B) the prongs of the tuning fork were kept in a horizontal plane above the resonance tube

(C) the amplitude of vibration of the ends of the prongs is typically around 1 cm

(D) the length of the air-column at the first resonance was somewhat shorter than 1/4^th of the wavelength of the sound in air.

Option (A) is correct. (The second resonance will give sound of smaller intensity since the air column is longer and hence there is greater chance for the attenuation of the sound before and after the reflection at the water surface).

Option (B) is incorrect since the prongs are to be kept in a vertical plane to transfer energy efficiently into the resonance column.

Option (C) also is incorrect since the amplitude of vibration of the ends of the prongs is typically of the order of a millimetre only.

Option (D) is correct since the anti-node is obtained slightly above the end of the resonance tube and λ/4 = l + e where λ is the wave length of sound, l is the resonating length of the tube and e is the end correction.

(3) Under the influence of the Coulomb field of charge +Q, a charge –q is moving around it in an elliptical orbit. Find out the correct statement(s)

(A) the angular momentum of the charge –q is constant

(B) the linear momentum of the charge –q is constant

(C) the angular velocity of the charge –q is constant

(D) the linear speed of the charge –q is constant.

The situation here is similar to that of a hydrogen atom (or, the motion of a planet around the sun) which is central field motion under an inverse square law force. The only correct option is (A) which high lights the constancy of angular momentum (Kepler’s law).

[Note that the torque is zero and hence the angular momentum is constant].

AIEEE 2008- An Imaginary Question on Bohr Model

The following question was included in the AIEEE 2008 question paper:

Suppose an electron is attracted towards the origin by a force k/r where ‘k’ is a constant and ‘r’ is the distance of the electron from the origin. By applying Bohr’s model to this system, the radius of the n^th orbital of the system is found to be ‘r_n’ and the kinetic energy of the electron to be ‘T_n’. Then which of the following is true?

(1) T_n α 1/n², r_n α n²

(2) T_n independent of n, r_n α n

(3) T_n α 1/n, r_n α n

(4) T_n α 1/n, r_n α n²

The force k/r supplies the centripetal force for the circular motion of the electron so that we have

k/r = mv²/r where ‘m’ is the mass and ‘v’ is the speed of the electron.

Therefore, mv² = k which is independent of the quantum number ‘n’. The kinetic energy T_n of the electron is ½ mv² which is therefore independent of the quantum number ‘n’.

Also, v = √(k/m).

The angular momentum of the electron in the n^th orbit of radius r_n is mvr_n and in the Bohr model mvr_n = nh/2π where ‘h’ is Planck’s constant. Substituting for ‘v’ we have

m×√(k/m) ×r_n = nh/2π

This gives r_n α n. So the correct option is (2).

In the above question the attractive force on the electron was imagined to be inversely proportional to the distance just for the sake of testing your problem solving skill. In a real hydrogen atom the force is certainly inversely proportional to the square of the distance. You will find questions on real Bohr model on this site by clicking on the label ‘Bohr model ‘ below this post.

IIT-JEE 2008: Linked Comprehension Type Multiple Choice Questions on Bohr Model of Hydrogen-like Atoms

The IIT-JEE Physics question paper is no more a nightmare to students aspiring for a seat in one of the IIT’s, consequent on the changes effected from 2007. If you have good grasp of fundamentals, you should definitely take up the challenge for brightening your future. Students appearing for other entrance examinations such as Physics GRE, AP Physics, AIEEE, KEAM and the like also will find the IIT-JEE question paper very useful for their preparation.

Here are three multiple choice questions on Bohr atom model which appeared under Linked Comprehension Type in IIT-JEE 2008 question paper:

Paragraph for Question Nos. 1 to 3

In a mixture of H – He⁺ gas (He⁺ is singly ionized He atom) H atoms and He⁺ ions are excited to their respective first excited states. Subsequently, H atoms transfer their total excitation energy to He⁺ ions (by collisions). Assume that the Bohr model of atom is exactly valid.

(1) The quantum number n of the state finally populated in He⁺ ions is

(A) 2

(B) 3

(C) 4

(D) 5

The singly ionised helium atom is hydrogen like since it contains a single electron. In a hydrogen like atom the energy (E) of the electron in the n^th orbit is given by

E = – 13.6 Z²/n² electron volt where Z is the atomic number.

For hydrogen atom the energies are – 13.6 eV, – 3.4 eV, – 1.51 eV and – 0.85 eV for values of n equal to 1, 2, 3 and 4 respectively

For the helium ion the energies are four fold, equal to – 54.4 eV, – 13.6 eV, – 6.04 eV and – 3.4 eV for values of n equal to 1, 2, 3 and 4 respectively

The energy difference between the electrons in first excited states of hydrogen atom and helium ion is (– 3.4 eV) – (– 13.6 eV) = 10.2 eV.

On collision, the helium ion absorbs this energy and occupies the 3^rd excited state (n = 4)

of energy – 13.6 eV + 10.2 eV = –3.4 eV.

Therefore the correct option is (C).

(2) The wave length of light emitted in the visible region by He⁺ ions after collision with H atoms is

(A) 6.5×10^–7 m

(B) 5.6×10^–7 m

(C) 4.8×10^–7 m

(D) 4.0×10^–7 m

The visible photons have wave lengths in the range 4000 Ǻ to 7000 Ǻ and their energies lie in the range 3.1 eV to 1.771 eV.

[It will be useful to remember the range of energies in eV. You will definitely remember the wave length range of visible photons which you can convert into energy range using the equation, λ E = 12400  where λ is the wave length in Angstrom and E is the energy in electron volt].

The energy difference between the He⁺ ion levels for n = 3 and n = 4 correspond to visible photon energy: – 3.4 eV– (– 6.04 eV) =  2.64 eV

The wave length of the light emitted in the visible region is therefore 124000/2.64 = 4800 Ǻ, nearly. The answer is thus 4.8×10^–7 m [Option (C)].

(3) The ratio of the kinetic energy of the n = 2 electron for the H atom to that of the He⁺ ion is

(A) ¼

(B) ½

(C) 1

(D) 2

The kinetic energy of the electron in the n^th orbit is 13.6 Z²/n².

[Note that the kinetic energy and the total energy of the electron are equal in amount. But the kinetic energy is positive where as the total energy is negative].

The required ratio is Z_H²/Z_He² = ¼.

All posts on Bohr model of hydrogen atom on this site can be accessed by clicking on the label ‘Bohr model’ below this post.

You can find a few posts on hydrogen atom at apphysicsresources.blogspot.com also. One such post is here.

Two Questions (MCQ) on Bohr Atom Model

You can find the earlier questions (with solution) on Bohr atom model posted on this site by clicking on the label ‘Bohr model’ or ‘hydrogen atom’ below this post. You can get them also by using the search option at the top of this page. Today I give you two questions which are meant for gauging the depth of your understanding of Bohr’s theory.

(1) When the electron in a hydrogen atom of mass M undergoes transition from an orbit of higher quantum number n₂ to an orbit of lower quantum number n₁, the recoil velocity acquired by the atom is (Rydberg’s constant = R, Planck’s constant = h)

(a) (R/hM) (1/n₁² – 1/n₂²)

(b) (hR/M) (n₂ – n₁)

(c) 1/hRM (1/n₁² – 1/n₂²)

(d) h/RM) (1/n₁² – 1/n₂²)

(e) (hR/M) (1/n₁² – 1/n₂²)

The wave number of the photon emitted because of the electron transition is

ν' = 1/λ = R(1/n₁² – 1/n₂²) where λ is the wave length of the photon and R is Rydberg’s constant.

The momentum of the photon is p = h/λ = hR(1/n₁² – 1/n₂²) where h is Planck’s constant.

When the photon is emitted with this momentum, the atom recoils (like a gun firing a bullet) with an equal and opposite momentum. Therefore, the recoil velocity of the atom is given by

v = p/M = (hR/M)(1/n₁² – 1/n₂²).

(2) If the radius of the innermost electron orbit in a hydrogen atom is R₁, the de Broglie wave length of the electron in the second excited state is

(a) πR₁

(b) 3πR₁

(c) 4πR₁

(d) 6πR₁

(e) 9πR₁

The wave length of the electron in the n^th orbit is given by

λ = 2πR_n/n where R_n is the radius of the n^th orbit.4

[This follows because the angular momentum of the electron in the n^th orbit is

mvR_n = nh/2π.

Therefore, de Broglie wave length, λ = h/mv = 2πR_n/n ]

The second excited state has quantum number n = 3 (Third orbit). The radius of the 3^rd orbit in terms of the radius R₁ of the first orbit is given by (remembering R_n = n² R₁)

R₃ = 9R₁

Therefore, λ = 2πR_n/n = 2π×9R₁/3 = 6πR₁

[It will be convenient to remember that the de Broglie wave length of the electron in the n^th orbit is n times the the wave length in the innermost orbit].

You will find some useful posts on Atomic Physics and Quantum effects at apphysicsresources

Two Questions (MCQ) on Angular Momentum

The following two questions are similar in that both require the calculation of angular momentum in central field motion under inverse square law forces.

(1) An artificial satellite of mass ‘m’ is orbiting the earth of mass ‘M’ in a circular orbit of radius ‘r’. If ‘G’ is the gravitational constant, the orbital angular momentum of the satellite is

(a) [GMm²r]^1/2 (b) [GMmr]^1/2 (c) [GMm/r]^1/2 (d) [GMm²/r²]^1/2 (e) [GMm²r³]^1/2

The orbital angular momentum of a satellite is mvr where ‘v’ is the orbital speed. [Angular momentum = Iω = mr²ω = mr²v/r = mvr where ‘I’ is the moment of inertia and ‘ω’ is the angular velocity of the satellite].

The centripetal force required for the circular motion of the satellite is supplied by the gravitational pull so that we have

mv²/r = GMm/r²

From this, m²v²r² = GMm²r so that angular momentum mvr = [GMm²r]^1/2

(2) In a hydrogen atom in its ground state, the electron of mass ‘m’ is moving round the proton in a circular orbit of radius ‘r’. The orbital angular momentum of the electron is (with usual meaning for symbols)

(a) [m²e²r/4πε₀]^1/2 (b) [m²er/4πε₀]^1/2 (c) [me²r²/4πε₀]^1/2

(d) [me²r/4πε₀]^1/2 (e) [me²r³/4πε₀]^1/2

The steps for finding the orbital angular momentum of the electron are similar to those in question No.1, with the difference that the centripetal force is supplied in this case by the electrostatic attractive force between the proton and the electron.

We have mv²/r = (1/4πε₀)e²/r² from which m²v²r² = (1/4πε₀) ×me²r, so that orbital angular momentum, mvr = [me²r/4πε₀]^1/2

Bohr Model of Hydrogen Atom – An IIT-JEE 2007 question

The following MCQ appeared in IIT-JEE 2007 question paper:

The largest wave length in the ultraviolet region of the hydrogen spectrum is 122 nm. The smallest wave length in the infra red region of the hydrogen spectrum (to the nearest integer) is

(a) 802 nm (b) 823 nm (c) 1882 nm (d) 1648 nm

The wave number 1/λ, which is the number of waves per meter length in the case of hydrogen spectrum is given by Rydberg’s relation,

1/λ = R(1/n₁² – 1/n₂²) where R is Rydberg’s constant and n₁ and n₂ are integers.

Ultraviolet radiations are obtained in the Lyman series of hydrogen spectrum when electron transitions take place from higher orbits (of quantum number n>1)to the innermost orbit (of quantum number n=1). So, for the Lyman series, n₁=1 and n₂ = 2,3,4,…etc. The largest wave length in the Lyman series is obtained when the transition is from 2^nd orbit (n₂=2) to the first orbit (n₁=1).

The smallest wave length in the infra red region is obtained when electron transition occurs from the outermost orbit (n₂ = ∞) to the third orbit (n₁ = 3) and this spectral line is the shortest wave length line in the Paschen series ( for which n₁ = 3 and n₂ = 4,5,6….etc.).

For the largest wave length ultraviolet line we have (expressing the wave length in nanometer),

1/122 = R(1/1² – 1/2²) = 3R/4

For the smallest wave length(λ') infrared line we have

1/λ' = R(1/3² – 1/∞) = R/9

Dividing the first equation by the second, λ'/122 = 3×9/4, from which λ' = 823 nm.

Let us consider another similar question which appeared in Kerala Medical Entrance 2001 question paper:

Given that the longest wave length in Lyman series is 1240 Ǻ, the highest frequency emitted in Balmer series is

(a) 8×10¹⁴ Hz (b) 8×10¹² Hz (c) 8×10¹⁰ Hz (d) 8×10³ Hz (e) 8×10² Hz

In the Rydberg’s relation, 1/λ = R(1/n₁² – 1/n₂²), n₁=1 and n₂ = 2,3,4,…etc., for the Lyman series. For the Balmer series (which is in the visible region), n₁=2 and n₂ = 3,4,5….etc.

The longest wave length in Lyman series is obtained for n₂ = 2 and the highest frequency (shortest wavelength) in Balmer series is obtained for n₂ = ∞.

Rydberg’s relation for the above two cases are (expressing wave lengths in Angstrom),

1/1240 = R(1/1² – 1/2²) = 3R/4 and

1/ λ' = R(1/2² – 1/∞) = R/4

Dividing the first equation by the second,

λ'/1240 = 3, from which λ' = 3720 Ǻ.

The frequency (ν) of this line is given by

ν = c/λ' = (3×10⁸) /(3720×10^–10) = 8×10¹⁴ Hz.

Additional MCQ on Bohr Model of Hydrogen Atom

The following questions are in continuation of the post dated August 26, 2006 (Questions on Bohr Atom Model):
(1) Suppose the energy required to remove all the three electrons from a lithium atom in the ground state is ‘E’ electron volt. What will be the energy required (in electron volt) to remove two electrons from the lithium atom in the ground state?
(a) 2E/3 (b) E – 13.6 (c) E – 27.2 (d) E – 40.8 (e) E – 122.4
The energy of the electron in a hydrogen like atom in the ground state is – 13.6Z² electron volt. Therefore, after removing two electrons from the lithium (Z=3) atom, the third electron has energy equal to – 13.6×3² eV = 122.4 eV. The energy needed to remove two electrons from the lithium atom in the ground state is therefore equal to (E –122.4) eV.
(2) How many revolutions does the electron in the hydrogen atom in the ground state make per second? (h = 6.63×10^-34 Js, mass of electron = 9.11×10^-31 kg, Bohr radius = 0.53 A.U.)
(a) 6.55×10^-15 (b) 3.28×10^-15 (c) 3.28×10^-16 (d) 1.64×10^-15 (e) 9.11×10^-15
The angular momentum (Iω) of the electron is an integral multiple of h/2π. Therefore, Iω = nh/2π, from which, for the first orbit (n=1), ω = h/2πI = h/2πm_e r² . The orbital frequency of the electron is given by f = ω/2π = h/4π²m r² = 6.55×10^-15 per second, on substituting for h, m and r.
(3) The ionisation energy of the hydrogen atom is 13.6 eV. If hydrogen atoms in the ground state absorb quanta of energy 12.75 eV, how many discrete spectral lines will be emitted as per Bohr’s theory?
(a) 1 (b) 2 (c) 4 (d) 6 (e) zero
On absorbing 12.75 eV, the energy of the electron in the hydrogen atom will become (–13.6 + 12.75) eV which is – 0.85 eV. This is an allowed state (with n=4) for the electron, since the energy in the 4th orbit is – 13.6/n² = – 13.6/4² = 0.85 eV. From the 4th orbit, the electron can undrego three transitions to the lower orbits (4→3, 4→2, 4→1). From the third orbit, the electron can undergo two transitions (3→2 and 2→1). The electron in the second orbit can undergo one transition (2→1). So, altogether 6 transitions are possible, giving rise to 6 discrete spectral lines [Option (d)].
You can work it out as n(n–1)/2 = 4(4–1)/2 =6.
(4) The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following siatements is true?
(a) Its kinetic energy increases and its potential and total energies decrease.
(b)Its kinetic energy decreases, potential energy increases and its total energy remains the same.
(c) Its kinetic and total energies decrease and its potential energy increases.
(d) Its kinetic, potential and total energies decrease.
This question appeared in IIT 2000 entrance test paper. The correct option is (a). You should note that the kinetic energy is positive while the potential energy and total energy are negative. Further, the kinetic energy and total energy are numerically equal and the numerical value is equal to half the potential energy.
The total energy is –13.6 Z²/n². In lower orbits (with smaller n), the potential energy is smaller since it has a larger negative value. The total energy also is therefore smaller. But, the kinetic energy is greater since it has a larger positive value.
(5) If the binding energy of the electron in a hgydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of Li⁺⁺ is
(a) 122.4 eV (b) 30.6 eV (c) 13.6 eV (d) 3.4 eV
Questions of this type often appear in entrance test papers.
Since Li⁺⁺ is a hydrogen like system with a single electron revolving round a nucleus of proton number Z = 3, the energy of the electron in orbit of quantum number n is
E = –13.6 Z²/n² eV.
The energy in the first excited state (second orbit) is – 13.6×9/4 eV = – 30.6 eV. The energy to be supplied to the electron to remove it from the first excited state is therefore + 30.6 eV [Option (b)].
(6) The wave lengths involved in the spectrum of deuterium (₁D²) are slightly different from that of hydrogen spectrum, because
(a) the attraction between the electron and the nucleus is different in the two cases
(b) the size of the two nuclei are different
(c) the nuclear forces are different in the two cases
(d) the masses of the two nuclei are different.
This MCQ appeared in AIEEE 2003 questionn paper. The answer to this will not be easy if you stick on to the elementary theory of the Bohr model in which the energy (E_n) of the electron of quantum number n (n^th orbit) is given by
E_n = – me⁴/8ε₀n²h² where m is the mass of the electron, e is its charge, ε₀ is the permittivity of free space and h is Planck’s constant.
In the elementary theory we take ‘m’ as the mass of the electron on the assumption that the nucleus has a very large mass compared to the mass of the electron and hence the electron is moving round with the nucleus at the centre. The real situation is that both the electron and the nucleus are moving along circular paths with the centre of mass as the common centre. Instead of the actual mss of the electron, the reduced mass of the electron and the nucleus is to be substituted in the expression for energy. The modified form of the expression is
E_n = – μ e⁴/8ε₀ n²h² where μ is the reduced mass of electron and the nucleus, given by
μ = Mm_e/(M+m_e), M and m_e being the masses of the nucleus and the electron respectively. [Generally, for a hydrogen like system with proton number Z, the expression for energy is E_n = – μ Z²e⁴/8ε₀n²h² ].
The nucleus of deuterium contains a proton and a neutron and has very nearly twice the mass of the hydrogen nucleus (proton). So, the reduced mass and the energy levels of deuterium are slightly greater than those of hydrogen and this is the reason for the difference in wave length. [The wave lengths are slightly shorter].The correct option is (d).
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An important point you should remember in the light of the above discussion is the drastic change in the energy levels and the spectrum of positronium compared to hydrogen. Positronium is a highly unstable neutral atom with an electron revolving round a positron. [ You can as well say, a positron revolving round an electron!]. The concept of reduced mass is absolutely necessary in this case since the positron has the same mass as that of the electron so that the reduced mass of positronium is mm/(m+m) = m/2 where ‘m’ is the mass of positron as well as the electron.
Now consider the following MCQ:
A positronium atom undergoes a transition from the state n = 4 to n = 2. The energy of the photon emitted in this process is
(a) 1.275 eV (b) 2.55 eV (c) 3.4 eV (d) 5.1 eV (e) 13.6 eV
The expression for energy of positronium is E_n = – μ e⁴/8ε₀n²h² where μ is the reduced mass of positron and electron, given by μ = mm/(m+m) = m/2. Therefore, the mass of the electron (m) used in the expression for the energy of a hydrogen atom (Bohr’s theory) is to be replaced by m/2. All energy levels are therefore reduced to half of the hydrogen levels. Since the energies for states n=4 and n=2 for hydrogen are –13.6/16 eV(= –0.85 eV) and –13.6/4 eV (= –3.4 eV) respectively, the energy of the photon emitted in the case of hydrogen is [(–0.85) – (–3.4)] eV = 2.55 eV. In the case of positronium, the energy will be half of this. So, the answer is 2.55/2 eV = 1.275 eV.