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

Equations to be remembered in atomic physics and quantum effects were discussed on this blog in the post dated 23^rd February 2008. Subsequently some typical multiple choice practice questions (with solution) were discussed in the posts dated 25^th February 2008 and 1^st March 2008, followed by a couple of free response practice questions (with model answers) in the post dated 8^th March 2008. Some more multiple choice practice questions (with solution) were discussed in the post dated 12^th March 2009. All these posts can be accessed by clicking on the label ‘atomic physics’ below this post.

Today we will discuss a few more multiple choice practice questions in this section:

(1) A beam of α-particles is directed towards a gold foil. A, B, C, and D are the paths of α-particles incident on the gold foil (fig.) and A₁, B₁, C₁, and D₁ are the paths of the α-particles scattered by the gold foil. Pick out the correct statement:

(a) The number of α-particles in A₁ will be minimum and those in C₁ will be maximum

(b) The number of α-particles in A₁ will be maximum and those in C₁ will be minimum

(c) The number of α-particles in B₁ will be minimum and those in C₁ will be maximum

(d) The number of α-particles in C₁ will be minimum and those in D₁ will be maximum

(e) The number of α-particles in C₁ will be maximum and those in D₁ will be minimum

The volume of the nucleus in a an atom is very much insignificant compared to the volume of the atom. The space within an atom is mostly empty and the nucleus is often compared to a ‘fly in a cathedral’. So most of the α-particles incident on the gold foil proceed undeviated. Some of the α-particles are deviated by the Coulombian repulsion. The probability for a head on collision which makes the α-particle retrace its path is least. So the number of α-particles in C₁ will be maximum and those in D₁ will be minimum [Option (e)].

(2) An α-particle of mass m, charge 2e and speed v proceeds to make a head on collision with the nucleus of an atom of atomic number Z. What is the distance of closest approach of the α-particle?

(a) Ze²/4πε₀mv²

(b) Ze²/πε₀mv²

(c) Z²e²/πε₀mv²

(d) Z²e⁴/πε₀mv²

(e) Z²e²/4πε₀mv²

When the α-particle approaches the nucleus its kinetic energy decreases because of the electrostatic repulsion and its electrostatic potential energy increases. At the distance of closest approach (d) the entire kinetic energy gets converted into electrostatic potential energy and so we have

(½) mv² = (1/4πε₀) (Ze×2e/d)

[Note that the nuclear charge is Ze].

This gives d = 4Ze²/4πε₀mv² = Ze²/πε₀mv²

(3) Consider a system consisting of an electron and a photon. The photon collides with the electron and gets scattered. In this process the kinetic energy of the electron is found to be increased. Compared with the incident photon, the scattered photon has

(a) larger energy

(b) smaller speed

(c) smaller wave length

(d) smaller momentum

(e) larger speed

By the law of conservation of energy, the scattered photon has smaller energy E. The momentum of the photon is E/c where c is the speed of the photon. Since the speed c is constant, the momentum of the scattered photon must be smaller [Option (d)].

(4) Light of frequency 1.8 times the threshold frequency is incident on a photosensitive surface. If the frequency of the incident light is halved and the intensity of the light is doubled, the photoelectric current will be

(a) unchanged

(b) halved

(c) doubled

(d) quadrupled

(e) zero

Photo electrons will not be emitted if the frequency of the incident light is less than the threshold frequency. Therefore, the photoelectric current will be zero for all incident intensities [Option (e)].

(5) A proton, a deuteron (nucleus of deuterium or heavy hydrogen) and an α-particle are accelerated by the same potential difference. Their velocities will be in the ratio

(a) 1:1:1

(b) 1:2:4

(c) √2:1:1

(d) 1:1:√2

(e) 1:2:2

The kinetic energy acquired is given by

(½)mv² = qV where m is the mass, v is the velocity and q is the charge of the particle accelerated by a potential difference V.

Therefore, v = √(2qV/m).

The velocity is thus directly proportioal to √(q/m).

The ratio of velocities of the proton, deuteron and α-particle is:

1 : (1/√2) : √(2/4) = 1 : (1/√2) : (1/√2) = √2:1:1

[Remember that the deuteron has twice the mass of the proton and the same charge as the proton where as the α-particle has four times the mass of the proton and twice the charge of the proton].

(6) Light beams with photon energies 1.5 eV and 3 eV are allowed to fall in succession on a photosensitive surface of work function 1 eV. The ratio of the maximum speeds of the emitted electrons will be

(a) 1 : 2

(b) 1 : 1

(c) √2 : 1

(d) 1 : √2

(e) 1 : 4

The maximum kinetic energy of the emitted electron is equal to the difference between the energy of the incident photon and the work function of the photosensitive surface. Therefore we have

(½)mv² = E – Φ where m is the mass and v is the maximum velocity of the photo electron, E is the energy of the incident photon and Φ is the work function of the photosensitive surface.

Considering the emission by the two light beams, we have

(½)mv₁² /(½)mv₂² = (1.5 – 1)/(3 – 1) = 1/4

[We could substitute the photon energies and the work functions in electron volts since we require the ratio of velocities].

Therefore, the ratio of the maximum velocities of the emitted electrons is given by

v₁/v₂ = √(1/4) = 1/2 [Option (a)].

AP Physics B– Atomic Physics and Quantum Effects– Multiple Choice Practice Questions

-->

The essential points you need to remember in atomic physics and quantum effects (especially to make you strong in working out multiple choice questions in the stipulated time) were given in the post dated 23^rd February 2008. Some typical multiple choice questions (with solution) were discussed in the posts dated 25^th February 2008 and 1^st March 2008. Two free response questions (with model answers) in this section were discussed in the post dated 8^th March 2008. You can access all these posts by clicking on the label ‘atomic physics’ below this post.

Today we will discuss some more multiple choice questions in this section:

(1) The photoelectric work function of a metal surface is Φ. The thermionic work function of the same metal surface will be

(a) less than Φ

(b) equal to Φ

(c) greater than Φ

(d) dependent on the method of heating used

(e) independent of the photoelectric work function

The difference between photoelectric emission and thermionic emission is in the method of supplying energy to the electrons in the metal to make them come out of the metal. The minimum energy required for the electrons to escape from the surface of a given metal is a constant. This means that the photoelectric work function of a given metal surface is equal to its thermionic work function [Option (b)].

-->

(2) The adjoining figure shows four energy levels of an atom. The wave lengths emitted due to transitions 1, 2, 3 and 4 (fig.) are λ₁, λ₂, λ₃ and λ₄ and the corresponding frequencies are ν₁, ν₂, ν₃ and ν₄ respectively. Then

(a) λ₁ = λ₂ + λ₃ + λ₄

(b) λ₄ = λ₁ + λ₂ + λ₃

(c) ν₁ = ν₂ + ν₃ + ν₄

(d) ν₄ = ν₁ + ν₂ + ν₃

(e) λ₄ = λ₁ λ₂ λ₃ /(λ₁ + λ₂ + λ₃)

From Bohr’s frequency condition we have

hν₄ = hν₁ + hν₂ + hν₃ where h is Planck’s constant.

Therefore ν₄ = ν₁ + ν₂ + ν₃

(3) In the above question the emitted wave lengths are related as

(a) λ₁ = λ₄ – (λ₂ + λ₃)

(b) λ₄ = λ₁ – (λ₂ + λ₃)

(c) 1/λ₁ = 1/(λ₂ + λ₃ + λ₄)

(d) 1/λ₄ = 1/(λ₁ + λ₂ + λ₃)

(e) λ₄ = λ₁λ₂λ₃ /(λ₁λ₂ + λ₂λ₃ + λ₃λ₁)

As shown above, ν₄ = ν₁ + ν₂ + ν₃

Or, c/λ₄ = c/λ₁ + c/λ₂ + c/λ₃ where c is the speed of light.

Therefore, 1/λ₄ = 1/λ₁ + 1/λ₂ + 1/λ₃ = (λ₂λ₃ + λ₃λ₁ + λ₁λ₂) /λ₁λ₂λ₃ from which

λ₄ = λ₁λ₂λ₃ /(λ₁λ₂ + λ₂λ₃ + λ₃λ₁)

(4) What is the change in the angular momentum of the electron in the hydrogen atom when it undergoes a transition to emit the H_α line in the Balmer series? (Planck’s constant = h).

(a) h/2π

(b) h/π

(c) 2h/π

(d) 3h/2π

(e) zero

When the electron is in the orbit of quantum number n its angular momentum is nh/2π. The H_α line results due to the transition from the 3^rd orbit to the 2^nd orbit. The angular momentum therefore changes from 3h/2π to 2h/2π so that the change in the angular momentum is h/2π.

(5) In an X-ray tube the accelerating voltage is V volt. If the electronic charge is e coulomb, Planck’s constant is h joule second and the speed of light in free space is c ms^–1, the range of wave lengths possible in the continuous X-ray spectrum is

(a) zero to infinity

(b) zero to hc/eV

(c) zero to eV/hc

(d) hc/eV to infinity

(e) eV/hc to infinity

The minimum wave length λ_min of the X-rays produced is given by

hc/λ_min = eV

We have equated the entire energy eV of the electron to the energy hν (= hc/λ) of the photon. This gives λ_min = hc/eV.

All wave lengths above this value are possible so that the correct option is (d).

You will find some useful questions in this section here

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.

AP Physics B– Multiple Choice Questions (for practice) on Atomic Physics and Quantum Effects

As promised in the post dated 23-2-08, I give below some typical multiple choice questions (MCQ) for practice:

(1) Five photons have the following energy values. Which one represents the visible light photon?

(a) 24.8 eV

(b)12.4 eV

(c) 6.2 eV

(d) 2.48 eV

(e) 1.24 eV

Photons in the visible region have wave length range from 4000 Ǻ to 7000 Ǻ (approximately). The product of the energy in eV and the wave length in Angstrom in the case of any photon is 12400 (very nearly).

The energy of the 4000 Ǻ photon in electron volt is 12400/4000 = 3.1 eV.

The energy of the 7000 Ǻ photon in electron volt is 12400/7000 = 1.77 eV.

Therefore, the photon in the visible region is given in option (d).

If you were asked to calculate the wave length of the photon of energy 2.48 eV, you will have

λ = 12400/2.48 = 5000 Ǻ

[You can calculate the wave length using the relation E = hc/λ where E is the energy in joule, h is Planck’s constant, c is the speed of light and λ is the wave length (in metre). In the above problem E = 2.48×1.6×10^–19 joule (on converting the energy in eV into joule), c = 3×10⁸ ms^–1and h = 6.6×10^–34 Js. But this is time consuming].

(2) A laser source gives light output of power P. If the wave length of the laser is λ, the number of photons emitted in a time t in terms of the given parameters and fundamental constants is

(a) Pλt/hc

(b) Phct/λ

(c) Pht/λc

(d) Pλ/hc

(e) Pct/hλ

The energy of a photon is hν = hc/λ where h is Planck’s constant, ν is the frequency of light, c is the speed of light and λ is the wave length.

Number of photons emitted per second = P/(hc/λ) = Pλ/hc.

Therefore, total number of photons emitted in time t = Pλt/hc.

(3) When electromagnetic radiations of wave length λ is incident on a photosensitive surface, the kinetic energy of the photoelectrons emitted from the surface is 2 eV. When the wave length of the incident radiations is 2λ, the kinetic energy of the photoelectrons emitted from the surface is 0.5 eV. The threshold wave length (maximum wave length) for photoelectric emission from the surface is

(a) λ/2

(b) λ

(c) 3λ/2

(d) 2λ

(e) 3λ

From Einstein’s equation, we have for the two cases

hc/λ = hc/λ₀ + 2 eV and

hc/2λ = hc/λ₀ + 0.5 eV

where h is Planck’s constant, c is the speed of light and λ₀ is the threshold wave length. We have expressed the kinetic energy in electron volt itself for convenience, with the understanding that all terms are in electron volt.

Multiplying the second equation by 4 and subtracting the first equation from it, we obtain

hc/λ = 3hc/λ₀ from which λ₀ = 3λ.

(4) According to the Bohr model, electrons of quantum number n = 4 in excited hydrogen atoms can undergo transitions to lower energy states in different ways and give rise to photons of discrete frequencies. How many discrete frequencies are possible in this case?

(a) 3

(b) 4

(c) 6

(d) 8

(e) 12

The possible transitions are shown in the adjoining figure. You will see 6 different transitions. Therefore, there will be six discrete frequencies.

You need not draw an energy level diagram to get the answer for similar questions. {If n is large, the method will be difficult). Remember that the possible number of transitions is n(n – 1)/2.

(5) Singly ionized helium (He⁺) atom is hydrogen like in the sense that a solitary electron revolves around a positively charged nucleus. If the energy of this electron in its first orbit (n = 1) is –54.4 eV, what will be its energy in the first excited state?

(a) –108.8 eV

(b) –27.2 eV

(c) –13.6 eV

(d) –6.8 eV

(e) –3.4 eV

First excited state means the electron is in the second orbit (n = 2).

For a hydrogen like atom, the energy of the electron in the orbit of quantum number ‘n’ is inversely proportional to n². Since the energy in the first orbit is –54.4 eV, the energy in the second orbit will be –54.4/2² eV = –13.6 eV

Few more questions will be posted from this section in the next post.

You will find a useful post on hydrogen like atoms at physicsplus