Multiple Choice Practice Questions on Atomic Physics and Quantum Effects for AP Physicsw B

“Being ignorant is not so much a shame as being unwilling to learn.”

– Benjamin Franklin

During March and April I have been unexpectedly busy and therefore could not post any thing here. Today I give you a few multiple choice practice questions on atomic physics and quantum effects along with their solution. This section will be interesting to most of you. Questions (with solution) in this section were posted on many occasions on this site. You may access them by clicking on the label ‘atomic physics and quantum effects’ below this post. Here are the additional questions:

(1) The energy of the electron in the hydrogen atom in the ground state is E₁. When the hydrogen atom is in its first excited state, the energy of its electron is E₂. Suppose we have a large number of hydrogen atoms in the ground state and radiation of frequency (E₂ – E₁) is incident on these atoms. What happens?

(a) All atoms will jump to the first excited state.

(b) All atoms will continue to remain in the ground state.

(c) Some of the atoms will jump to the first excited state.

(d) Most of the atoms will jump to the second excited state.

(e) Most of the atoms will get ionised.

Since the energy of each photon in the incident radiation is equal to the energy difference between the ground state and the first excited state of the hydrogen atom, the probability of absorption of energy is high so that some of the hydrogen atoms will jump to the first excited state. The correct option is (c).

(2) The simple Bohr model is applicable to which one of the following?

(a) Atoms with light nuclei

(b) Atoms with heavy nuclei

(c) Helium atoms

(d) Doubly ionised lithium

(e) Uranium atoms

Doubly ionized lithium has just one electron moving around the nucleus and hence it is hydrogen-like. Therefore the simple Bohr model is applicable to it [Option (d)].

(3) A proton is moving through a region of space where a constant electric field acts along its direction of motion. The de Broglie wave length of the proton

(a) decreases with time

(b) increases with time

(c) remains constant

(d) varies in a sinusoidal manner

(e) goes on increasing first and soon becomes constant

The de Broglie wave length λ is given by

             λ = h/p where h is Planck’s constant and and p is the momentum.

Since the proton is accelerated by the electric field, its momentum goes on increasing with time. Therefore, the de Broglie wave length λ of the proton decreases with time [Option (a)].

(4) An electron is projected at an acute angle with respect to a uniform magnetic field so that it travels along a helical path. The de Broglie wave length of the electron

(a) varies periodically

(b) increases with time

(c) remains constant

(d) decreases with time

(e) goes on decreasing first and attains a constant value

A static magnetic field cannot change the kinetic energy of a charged particle. The momentum of the electron therefore remains unchanged. Therefore the de Broglie wave length of the electron remains constant.

AP Physics B – Some Interesting Multiple Choice Practice Questions on Atomic and Nuclear Physics

“Non-violence leads to the highest ethics, which is the goal of all evolution. Until we stop harming all other living beings, we are still savages.”

– Thomas A. Edison

The sections ‘atomic physics and quantum effects’ and ‘nuclear physics’ in the AP Physics B syllabus will appear to be interesting to most of the AP Physics B aspirants. Today we will discuss a few practice questions (multiple choice) in these sections:

(1) The de Broglie wave length of a particle with kinetic energy E is λ. If its kinetic energy is increased  to 4E, its de Broglie wave length will be

(a) λ/4

(b) λ/2

(c) λ

(d) 2λ

(e) 4λ

Since the kinetic energy is directly proportional to the square of momentum, the momentum of the particle is doubled when its kinetic energy is quadrupled.

[Note that kinetic energy E = p²/2m where p is the momentum and m is the mass].

The de Broglie wave length λ is given by

             λ = h/p where h is Planck’s constant and p is the momentum.

Therefore, when p is doubled, λ is halved [Option (b)].

(2) Particles A and B have masses m and 4m respectively but they carry the same charge. When they are accelerated by the same voltage, their de Broglie wave lengths are in the ratio

(a) 1 : 1

(b) 2 : 1

(c) 4 : 1

(d) 1 : 4

(e) 1 : 8

Let V represent the common accelerating voltage and q represent the common charge of the particles. If p₁ and p₂ are the momenta of the particles A and B, on equating their kinetic energies, we have

             p₁²/2m = p₂²/(2×4m)

[Remember that the kinetic energy of a particle of charge q accelerated by a voltage V is qV. The kinetic energies of A and B are equal since they have the same charge and they are accelerated by the same voltage]

The above equation gives

             p₁/ p₂ = ½

Since the de Broglie wave length λ is given by

             λ = h/p where h is Planck’s constant and p is the momentum, the ratio of the de Broglie wave lengths of A and B is given by

              λ₁/ λ₂ = p₂/ p₁ = 2, as given in option (b).

(3) A metallic surface is found to emit photo-electrons when monochromatic light rays of frequencies n₁ and n₂ (n₂ > n₁) are incident on it. If the maximum values of kinetic energy of the photo-electrons emitted in the two cases are in the ratio 1 : 3, the threshold frequency of the metallic surface is

(a) (3n₁ – n₂)/ 2

(b) (2n₁ – n₂)/ 2

(c) (3n₁ – n₂)/ 3

(d) (n₂ – n₁)/ 3

(e) (n₂ – n₁)/ 2

If the threshold frequency is n₀ and the maximum values of kinetic energy in the two cases are E₁ and E₂ respectively, we have

             hn₁ = hn₀ + E₁ and

             hn₂ = hn₀ + E₂

Therefore, E₁/E₂ = (n₁ – n₀)/(n₂ – n₀)

Since the ratio is 1 : 3 we have

             (n₁ – n₀)/(n₂ – n₀) = 1/3

Or, 3n₁ – 3n₀ = n₂ – n₀

This gives n₀ = (3n₁ – n₂)/ 2, as given in option (a).

(4) The energy that must be added to an electron to reduce its de Broglie wave length from 2 nm to 1 nm is

(a) half the initial energy

(b) equal to the initial energy

(c) twice the initial energy

(d) thrice the initial energy

(e) four times the initial energy

The de Broglie wave length λ is given by

             λ = h/p where h is Planck’s constant and p is the momentum.

Since the de Broglie wave length of the electron is to be reduced to half the initial value (from 2 nm to 1 nm), the momentum of the electron is to be doubled. But when the momentum is doubled, its kinetic energy becomes four times the initial value. Therefore the energy that must be added is three times the initial energy [Option (d)].

(5) An alpha particle of mass m and speed v  proceeds directly towards a heavy nucleus of charge Ze. The distance of closest approach of the alpha particle is directly proportional to

(a) 1/Ze²

(b) 1/Ze

(c) v

(d) m

(e) 1/v²

The alpha particle has to move towards the nucleus with difficulty, doing work against the electrostatic repulsive force. When the alpha particle reaches the distance of closest approach, the entire kinetic energy gets converted into electrostatic potential energy. Therefore we have

             ½ mv² = (1/4πε₀)(2Ze²/r) where r is the distance of closest approach.

[Remember that the charge on the alpha particle is 2e].

This gives r = Ze²/πε₀mv²

This shows that r is directly proportional to 1/v².

AP Physics B - Multiple Choice Practice Questions on Atomic Physics and Quantum Effects

"Men often become what they believe themselves to be. If I believe I cannot do something, it makes me incapable of doing it. But when I believe I can, then I acquire the ability to do it even if I didn't have it in the beginning

– Mahatma Gandhi

 

AP Physics 2012 exams are just a few days away. Your final preparations for the exam must be in full swing and there is no time to waste. Today I give you a few multiple choice practice questions on atomic physics and quantum effects. Questions in this section posted earlier on this site (with solution) can be accessed by clicking on the label ‘atomic physics and quantum effects’ given below this post. Or, you may try a search for ‘atomic physics and quantum effects’ using the search box provided on this page.

(1) The adjoining figure shows the graphical relation between the frequency of incident radiation and the magnitude of stopping potential in the case of metals A and B. Note that the straight line graphs have the same slope. Which one among the following statements is correct in this case?

(a) A and B have the same work function

(b) A and B have the same threshold wave length

(c) Maximum kinetic energy of photoelectron in the case of metals A and B is directly proportional to the increment in frequency of incident radiation over the threshold frequency.

(d) Metal B is a better photosensitive material than metal A.

(e) For a given change in the frequency of incident radiation, the changes in stopping potentials are different for metals A and B

Since we have straight line graphs for both metals, maximum kinetic energy of photoelectron is directly proportional to the increment in frequency of incident radiation over the threshold frequency. Therefore option (c) is correct.

(2) The de Broglie wave length of a particle can be reduced to half its initial value by changing its kinetic energy to

(a) half the initial value

(b) twice the initial value

(c) three times the initial value

(d) four times the initial value

(e) a quarter of the initial value

Kinetic energy E of a body of mass m is given by

             E = p²/2m where p is the momentum

[This follows from E = ½ mv² = m²v²/2m. Here v is the velocity and mv = p]

The above relation shows that the kinetic energy becomes four times when the momentum is doubled.

The de Broglie wave length λ is given by

             λ = h/p where h is Planck’s constant.

Therefore, the de Broglie wave length of a particle can be reduced to half its initial value by changing its momentum to twice the initial value. Evidently the kinetic energy of the particle the becomes four times the initial value [Option (d)].

[Suppose the above question is modifie as follows:

A particle has de Broglie wave length λ when its kinetic energy is E. What additional kinetic energy is to be aded to it in order to reduce the de Broglie wave length to λ/2?

(a) E

(b) 2E

(c) 3E

(d) 4E

(e) E/4

The answer is 3E since you are asked to find the additional kinetic energy].

(3) Uranium (atomic number 92) has an isotope of mass number 235. It can undergo successive disintegrations to get transformed into lead (₈₂Pb²⁰⁷). How many α-particles and β-particles are emitted during this transformation?

(a) α = 7, β = 4

(b) α = 4, β = 3

(c) α = 7, β = 0

(d) α = 7, β = 7

(e) α = 4, β = 7

Beta particle emission does not affect the mass number. In order to reduce the mass number by 28 (from 235 to 207), the number of α-particles to be emitte must be 7. Since each α-particles carries two fundamental units of positive charge, the atomic number of the end product gets reduced by 14. But the final product (₈₂Pb²⁰⁷) has its atomic number reduced by 10 only. The extra 4 units must be obtained by the emission of four β-particles. The correct option therefore ia (a).   

[Note that when a β-particles (electron) is emitted from the nucleus, the nuclear charge increases by one unit. This happens as a result of the transformation of a neutron in the nucleus into a proton].

(4) Fundamental forces in nature are gravitational force, electromagnetic force, nuclear force and weak force. If these forces act over very short distances of the order of nuclear dimensions, how do you arrange them in decreasing order (starting with the strongest?

(a) Gravitational force, electromagnetic force, nuclear force, weak force

(b) Electromagnetic force, gravitational force, nuclear force, weak force

(c) Electromagnetic force, nuclear force, weak force, gravitational force

(d) Gravitational force, nuclear force, electromagnetic force, weak force

(e) Nuclear force, electromagnetic force, weak force, gravitational force

The correct option is (d).

[Don’t get carried away by the term ‘weak force’. The weakest force is gravitational force where as the strongest is nuclear force].

(5) Two protons are separated by a distance of 50 Ǻ. If the electromagnetic force between them is F₁ and the nuclear force between them is F₂, which one among the following is the most reliable statement?

(a) F₁ >> F₂

(b) F₂ >> F₁

(c) F₁ > F₂

(d) F₂ > F₁

(e) F₂ = F₁

This question is similar to question No. (4) in the sense that it is meant for checking your knowledge of nuclear physics. The correct option is (a). Nuclear  force is a very short range force. At a separation of 50 Ǻ which is very large compared to the size of a nucleus, nuclear force (strong interaction) between two protons is negligible compared to the electrostatic force. Therefore the correct option is (a).

AP Physics B – Nuclear Physics- Multiple Choice Practice Questions

Let us discuss some multiple choice questions on nuclear physics which will benefit AP Physics B aspirants:

(1) The ratio of the nuclear radius of ₅₂Te¹²⁵ to that of ₁₃Al²⁷ is

(a) 4

(b) 125/17

(c) ¼

(d) 5/3

(e) 3/5

The nuclear radius of an atom of mass number A is given by

R =R₀A^1/3 where R₀ = 1.2×10^–15 m.

The required ratio is therefore (125/27)^1/3 which is equal to 5/3.

(2) A nucleus of ₉₂U²³⁸ gets converted into a ₉₁Pa²³⁴ nucleus. The particles emitted during this decay are

(a) one α-particle and one positron

(b) one α-particle and one electron

(c) one α-particle and one antineutrino

(d) one α-particle and one neutrino

(e) one α-particle and two β-particles

The mass number decreases by 4 and the atomic number decreases by 1 in the above decay. When an α-particle is emitted, the mass number decreases by 4 and the atomic number decreases by 2. The atomic number can be increased by one from this condition only if an electron also is emitted. The correct option therefore is (b).

(3) The de Broglie wave length of an α-particle of mass m emitted by a nucleus of mass M initially at rest is λ. The de Broglie wave length of the nucleus immediately after the α-emission is

(a) λ

(b) λ (M– m)/m

(c) λm/(M–m)

(d) λ (m/M)²

(e) λ (M/m)²

The nucleus has a recoil momentum on emitting the α-particle. Since the parent nucleus is initially at rest, its recoil momentum has the same magnitude as that of the α-particle but the direction is opposite in accordance with the law of conservation of momentum. The de Broglie wave length λ of the α-particle is given by

λ = h/p where h is Planck’s constant.

Since the recoil momentum of the nucleus has the same magnitude p the de Broglie wave length of the nucleus immediately after the α-emission is λ itself [Option (a)].

(4) Complete the following relation representing one possible fission process:

₀n¹ + ₉₂U²³⁵→ ₃₈Sr⁹⁰ + ----

(a) ₅₄Xe¹⁴⁵

(b) ₅₄Xe¹⁴⁵ + 3 ₀n¹

(c) ₅₄Xe¹⁴³ + 3 ₀n¹

(d) ₅₄Xe¹⁴² + ₀n¹

(e) ₅₄Xe¹⁴² + 3 ₀n¹

The total mass number and the total atomic number on the two sides will match only if the relation is completed with the terms given in option (c).

(5) The mass m of any particle of rest mass m₀ at speed v is given by Einstein’s relativistic relation,

m = m₀/[1– (v²/c²)]^1/2 where c is the speed of light in free space.

The rest energy of an electron is 0.511 MeV. The increase in the energy of the electron when it is accelerated from rest to 80% of the speed of light in free space is (very nearly)

(a) 0.341 MeV

(b) 0.405 MeV

(c) 0.511 Mev

(d) 0.852 Mev

(e) 0.916 Mev

We have m₀c² = 0.511 Mev for the electron.

The total energy at a speed of 0.8 c is m₀c²/[1– (v²/c²)]^1/2 = m₀c²/(0.36)^1/2 since v = 0.8 c.

But m₀c² = 0.511 Mev so that the total energy is 0.511/0.6 = 0.852 MeV nearly.

The increase in the energy is 0.852 – 0.511 = 0.341 MeV.

AP Physics B – Additional Questions (MCQ) on Atomic Physics and Quantum Effects

In continuation of the post dated 25 February 2008, I give below a few more questions on atomic physics and quantum effects

(1) An electron and a proton have the same wave length. Which one of the following statements is correct about them?

(a) The momentum of the proton is less than that of the electron

(b) The momentum of the proton is greater than that of the electron

(c) They have the same speed

(d) The kinetic energy of the proton is greater than that of the electron

(e) The kinetic energy of the proton is less than that of the electron

We have de Broglie wave length λ = h/p.

Since ‘h’ is Planck’s constant, the momentum ‘p’ must be the same. The kinetic energy (K) is given by

K = p²/2m where ‘m’ is the mass.

Since the mass of the proton is greater than that of the electron, the kinetic energy of the proton must be less than that of the electron.

(2) A stationary nucleus of mass M emits an electron of mass ‘m’ with a velocity ‘v’. If the recoil velocity of the nucleus is V, the ratio of the de Broglie wave lengths of the nucleus and the electron is

(a) M/m

(b) (M­ – m)/m

(c) V/v

(d) v/V

(e) 1

This is a very simple question. By the law of conservation of momentum, the momentum (p) of the electron is equal in magnitude (and opposite in direction) to the recoil momentum of the nucleus.

Since the wave length, λ = h/p, they have the same wave length and the ratio of wave lengths is 1 [Option (e)].

(3) In an experiment on photo electric effect, a photoelectric target is irradiated with laser beams of various frequencies and in each case the stopping potential is measured. On using the stopping potentials as the Y- coordinates and the frequencies as the X- coordinates, a straight line graph is obtained. If Ф is the work function, ‘e’ is the electronic charge and ‘h’ is Planck’s constant, the slope of this straight line is equal to

(a) h/e

(b) h

(c) Ф

(d) Ф/e

(e) Ф/h

The graph will be as shown in the adjoining figure. Einstein’s photoelectric equation is

hν = Ф + K_max where ν is the frequency of the incident light and K_max is the maximum kinetic energy of the photo electrons, which we can write in terms of the stopping potential V_s as

K_max = eV_s.

Therefore, we have

hν = Ф + eV_s so that V_s = (h/e) ν – Ф/e

This equation is of the form y = mx +c which is the equation of a straight line of slope ‘m’. Therefore, on plotting the stopping potential V_s against the frequency ν, a straight line of slope h/e is obtained.

(4) If an antimatter world exists, hydrogen atoms (to be precise, anti hydrogen atoms) there would be made of positrons revolving round anti protons. If the kinetic energy of the positron in the first orbit (n = 1) of such an anti hydrogen atom is K, what will be the total energy of the positron in the third orbit?

(a) K

(b) – K

(c) – K/3

(d) K/9

(e) – K/9

The motion of the positron will be under the inverse square law force of attraction between the anti proton and the positron and will be similar to the motion of the electron in an ordinary hydrogen atom. The expressions for energies will be the same as in the case of an ordinary hydrogen atom. The kinetic energy in any orbit will be positive where as the total energy will be negative but they will be of the same amount.

Since the energy is inversely proportional to the square of the quantum number ‘n’, the total energy in the third orbit will be –K/3² = –K/9.

[Note that all these results are true for an ordinary hydrogen atom].

(5) The H_α line is the first member of the Balmer series of the hydrogen spectrum and it occurs due to the transition of the electron from the 3^rd orbit to the 2^nd orbit. If its wave length is λ, the wave length of the last member of the Balmer series will be

(a) (4/9) λ

(b) (5/9) λ

(c) (7/9) λ

(d) (1/2) λ

(e) (1/3) λ

On applying Rydberg’s relation to the H_α line, we have

1/λ = R[(1/2²) – (1/3²)] = (5/36)R where R is Rydberg’s constant.

The last member of the Balmer series is due to the transition from the last orbit (n = ∞) to the 2^nd orbit. If its wave length is λ_l, we have

1/λ_l = R[(1/2²) – (1/∞²)] = (1/4)R

Dividing the first equation by the second, we have

λ_l/λ = 20/36

Therefore, λ_l = (5/9) λ.

You can expect some free response questions in this section shortly.

Meanwhile see some multiple choice questions (with solution) at physicsplus: Questions on Bohr Atom Model.