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 – Practice Questions (MCQ) on Wave Motion (including sound)

"Only two things are infinite, the universe and human stupidity, and I'm not sure about the former."

– Albert Einstein

Today we will discuss a few multiple choice practice questions involving wave motion. It will be useful to access earlier posts in this section by clicking on the label ‘waves’ or ‘wave motion (including sound)’. Here are the questions:

(1) When the atmospheric pressure is P the minimum resonating length of air column in a resonance column apparatus is L with a tuning fork of frequency f. When the atmospheric pressure increases by 0.5% the consequent change in the minimum resonating length will be

(a) 0.25%

(b) 0.5%

(c) 1%

(d) 2%

(e) zero

The speed v of sound in a gas is given by Newton-Laplace equation:

             v = √(γP/ρ) where γ is the ratio of specific heats of the gas, P is the pressure of the gas and ρ is the density of the gas.

When pressure P increases, the density ρ also increases by the same proportion so that the quantity P/ρ is a constant. Therefore, the speed of sound is unchanged with changes in pressure. This means that the wave length λ of sound and the resonating length are unchanged [Option (e)]

[Note that the minimum resonating length is λ/4]

(2) A tuning fork of frequency 512 Hz resonates with the air column in a pipe of length 15 cm closed at one end. This tuning fork does not resonate with the air column in any shorter pipes. For which one of the following closed pipe lengths will this tuning fork exhibit resonance with the air column?

(a) 20 cm

(b) 30 cm

(c) 45 cm

(d) 55 cm

(e) 60 cm

Since the tuning fork does not resonate with the air column in any shorter pipes, the resonating length is equal to λ/4 where λ is the wave length of sound in air. In this case there is a node at the closed end of the pipe and the next antinode is at the open end (Fig.). If the length of the pipe used is an odd multiple of λ/4, resonance will occur since in such cases the open end can be an antinode. Thus resonance is possible only if the closed pipe lengths are λ/4, 3λ/4, 5λ/4, 7λ/4 etc.

[Note that a standing wave can be obtained in a closed pipe with consequent resonance only if the the closed end is a noe an the open end is an antinode].

Since λ/4 = 15 cm we have 3λ/4 = 45 cm, 5λ/4 = 75 cm, 7λ/4 = 105 cm and so on. Option (c) gives the length of the pipe as 45 cm and it is the correct one.

(3) The frequencies of the 2^nd and 3^rd overtones of a vibrating string are 3f/4 and 2f respectively. The fundamental frequency of vibration of the string is

(a) f/8

(b) f/6

(c) f/4

(d) f/2

(e) f

The natural frequencies of vibration of a string are integral multiples of the fundamental (lowest) frequency. If the fundamental frequency is n, the frequency of the first overtone (or, the 2^nd harmonic) is 2n. The frequency of the 2^nd overtone (or, the 3^rd harmonic) is 3n and the frequency of 3^rd overtone (or, the 4^th harmonic) is 4n.

Therefore, considering the 2^nd overtone, we have

             3n = 3f/4 from which n = f/4.

(4) Sound waves producing interference have their amplitudes in the ratio 3 : 2. The intensity ratio of maximum and minimum of interference fringes is

(a) 27 :  8

(b) 25 : 1

(c) 3 : 2

(d) 9 : 4

(e) 6 : 4

The  resultant amplitudes at the interference maximum and the interference minimum are in the ratio (3+2) : (3 – 2) since the waves are in phase at the interference maximum and 180º out of phase at the interference minimum. Since the intensity is directly proportional to the amplitude, the intensity ratio of maximum and minimum of interference fringes is (3+2)² : (3 – 2)² = 25 : 1

Multiple Choice Practice Questions on AP Physics B Thermodynamics

“This time, like all times, is a very good one if we but know what to do with it”

– Ralph Waldo Emerson

You will find many practice questions (with solution) on thermodynamics on this site. You can access them by clicking on the label ‘thermodynamics’ below this post. Today we shall discuss a few more multiple choice practice questions in this section for the benefit of AP Physics B aspirants.

(1) Keeping the pressure constant, the temperature of m kg of a gas is raised through ΔT º C. If the specific heats of the gas at constant volume and constant pressure are C_v and C_p respectively, the increase in internal energy of the gas is

(a) m C_p ΔT

(b) m (C_p – C_v) ΔT

(c) m C_v ΔT

(d) m (C_p + C_v) ΔT

(e) m (C_p + C_v) ΔT/2

If the temperature increase occurs at constant volume, the entire energy supplied to the gas is used up in increasing the internal energy of the gas. The energy required for increasing the temperature of one kilogram of the gas through 1º C at constant volume is the specific heat of the gas at constant volume (C_v). When the temperature increase occurs at constant pressure, the gas has to expand and therefore has to do external work. The specific heat of the gas at constant pressure is equal to the energy supplied for increasing the temperature of one kilogram of the gas through 1º C at constant pressure. But the increase in the internal energy of 1 kg of the gas on heating through 1º C in this case also is C_v.

Therefore, the increase in internal energy (ΔU) of m kg of the gas when its temperature is raised through ΔT º C is given by

             ΔU = m C_v ΔT

(2) In the above question the work done by the gas is

(a) mC_pΔT

(b) m(C_p – C_v)ΔT

(c) mC_vΔT

(d) m(C_p + C_v)ΔT

(e) Zero

Keeping the pressure constant, when the temperature of m kg of the gas is raised through ΔT º C, the total energy required is mC_pΔT. Out of this the energy used in increasing the internal energy of the gas is mC_vΔT. The energy spent for doing work is therefore given by the difference between the two values and is equal to m(C_p – C_v)ΔT.

(3) A sample of gas contained in a cylinder undergoes a cyclic process shown by the adjoining PV diagram. Among the following statements which one is correct?

(a) Work is done on the gas during the process shown by AB

(b) Work done by the gas during the cycle ABCA is negative

(c) BC represents an isochoric  process

(d) CA represents an isobaric process

(e) Work done on the gas during the process CA is zero

During the process shown by AB the gas expands. Therefore work is done by the gas (and not on the gas). Statement (a) is therefore incorrect.

Since the cycle ABCA is clockwise, work done by the gas is positive. Statement (b) is therefore incorrect.

[Work done by the gas during its expansion represented by the area under the curve AB is greater than the work done on the gas during its contraction represented by the area under the curve BC]

During the process BC the volume of the gas changes and hence the process is not isochoric.

[Isochoric process is also known as isovolumetric process]

During the process CA the pressure of the gas changes and hence the process is not isobaric.

Work done on the gas uring the process CA is zero since CA represents an isochoric process. Therefore option (e) is correct.

[Note that work is done only if volume changes].

(4) A Carnot engine operates using a high temperature source at 400 K and a low temperature sink at 300 K. How much more efficient will this engine be if the temperature of the sink is reduced to 200 K?

(a) Twice as efficient

(b) Three times as efficient

(c) Four times as efficient

(d) Five times as efficient

(e) Six times as efficient

Efficiency (η) of Carnot engine is given by

            η = (Q₁ – Q₂)/Q₁ = (T₁ – T₂)/T₁ where Q₁ is the quantity of heat absorbed from the source at the source temperature T₁ and Q₂ is the quantity of heat liberated to the sink at the sink temperature T₂.

The efficiency η₁ when the source and sink are at 400 K and 300 K respectively is given by

            η₁ = (400 – 300)/400 = ¼

The efficiency η₂ when the source and sink are at 400 K and 200 K respectively is given by

             η₂ = (400 – 200)/400 = ½

The engine is therefore twice as efficient [Option (a)].

(5) A fixed mass of gas does 25 J of work on its surroundings and transfers 15 J of heat to the surroundings. The internal energy of the gas

(a) decreases by 15 J

(b) increases by 15 J

(c) remains unchanged

(d) decreases by 40 J

(e) increases by 40 J

The internal energy of the gas decreases when it does work on the surroundings and also when it transfers heat to the surroundings. The total decrease in the internal energy of the gas is therefore equal to 25 J + 15 J = 40 J [Option (d)].

You will find similar questions with solution here.

AP Physics B - Multiple Choice Practice Questions on Doppler effect

In the post dated 9^th April 2009 the essential points to be noted about Doppler effect were given. You may click here to access that post. All posts related to Doppler effect can be accessed by clicking on the label ‘Doppler effect’ below this post.

Today we shall discuss a few multiple choice practice questions involving Doppler effect:

(1) The figure above shows a car sounding its horn and moving with uniform velocity along a straight road. At time t = 0 the car is at position A and at time t = t₁ the car moves past a stationary listener L. At time t = t₂ the car is at position B. Which one among the following graphs best represents the variation of the frequency (pitch) of the horn with time as heard by the listener?

 

When the car moves towards the listener the number of complete sound waves reaching the ear of the listener in one second is greater than the number reaching his ears when the car is at rest. The frequency of the horn as heard by the listener in this case is higher than the real requency of the horn. When the car moves away from the listener the number of complete sound waves reaching the ears of the listener in one second is less than the number of waves produced by the horn in one second.

[All the sound waves produced by the horn in one second.cannot reach the listener’s ears within one second since the car is moving away].

Therefore, the frequency of the horn as heard by the listener is less than the real frequency after the car moves past the listener.

The graph which best represents the variation of the frequency with time is (a).

(2) In the above question suppose the car is moving with uniform acceleration. Which one among the graphs will best represent the variation of the frequency (pitch) of the horn with time as heard by the listener?

In this case the pitch of the sound as heard by the listener will go on increasing when the car moves towards the listener since the relative velocity of the car with respect to the listener goes on increasing. After the car moves past the listener the pitch of the sound as heard by the listener will go on decreasing since the source of sound (horn) is moving away from the listener with continuously increasing velocity.

Therefore, the graph which best represents the variation of the frequency with time in this case is (e).

(3) A stationary police car at position A (Fig.) sounds its horn when the air is still and a listener L on the road hears the frequency of the note as n. A steady wind now blows continupusly from B to A with speed equal to 2 % of the speed of sound in air. What will be the frequency of the sound of  the siren as heard by the listener when the wind is blowing?

(a) 0.98 n

(b) 1.02 n

(c) n

(d) 0.99 n

(e) 1.01 n

The number of complete sound waves emitted per second by the siren will be n whether the wind is present or absent. Since the steady wind is blowing continuously a stationary listener will receive just n complete waves per second. The correct option is (c).

[If the sound source and listener are moving relative to each other, the listener will hear one shifted frequency (Doppler shift) in still air and a different shifted frequency if wind is blowing. However, if the source and the listener are at rest relative to each other, there is no Doppler shift when the wind is blowing. If you are confused, you may use the expression for the apparent frequency as heard by the listener:

Let the source S of sound (fig.) move with velocity v_S, the listener move with velocity v_L and the wind blow with velocity v_w, all in the same direction as shown in the figure.

            Wind→ w         S●→v_S        L●→v_L

The apparent frequency (n’) of sound is then given by

            n’ = n(v + w – v_L)/ (v + w – v_S)

where n is the real frequency of the sound and v is the velocity of sound.

If v_L and v_S are zero we obtain n’ = n

When there is relative motion between the source and the listener, you will have to take the sign of the wind velocity w as positive, as given in the above expression, when the wind blows from source to the listener. If the wind blows from listener to source, the sign of w is to be taken as negative]

You will find more questions (with solution) in this section at www.physicsplus.in .