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

AP Physics B– Atomic Physics and Quantum Effects – Equations to be Remembered

The topics included under atomic physics and quantum effects are the following:

(i) Photons, the photoelectric effect, Compton scattering, x-rays

(ii) Atomic energy levels

(iii) Wave-particle duality

Remember the following:

(1) Energy of a photon, E = 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. [Note that the speed and the wave length of light depend on the medium where as the frequency is independent of the medium].

Since the energy of a photon of wave length 1000 Angstrom is very nearly 12.4 electron volt, in the case of any photon, the product of energy in eV and the wave length in Ǻ is 12400. This can be used to calculate the energy or wave length of a photon.

(2) Momentum of photon, p = hν/c = h/λ = E/c

(3) Einstein’s photoelectric equation relates the energy of incident photon to the maximum kinetic energy (K_max) of the photo electron and the work function (Ф) of the photosensitive surface and can be written as

hν = K_max + Ф

Or, hν = K_max + hν₀ where ν₀ is the threshold frequency (minimum frequency of light to initiate photo electron emission from the surface).

The above equation can be written as K_max = h(ν – ν₀) = hc[(1/λ) – (1/λ₀)] where λ₀ is the threshold wave length (maximum wave length of light to initiate photo electron emission from the surface).

The cut-off or stopping potential is the minimum negative potential to be applied on the anode (plate) so that the photoelectric current becomes zero (stops).

In terms of the stopping potential V_s, Einstein’s equation can be written as

eV_s = h(ν – ν₀), since K_max = eV_s.

(4) de Broglie wave length is l = h/p = h/mv where λ is the wave length associated with a particle of momentum ‘p’.

[In the case of an electron accelerated by a small voltageV (so that relativistic mass increase is negligible), the wave length in Angstrom is very nearly √(150/V)]

(5) The minimum wave length (λ) of X-rays produced by an X-ray tube is inversely proportional to the anode voltage V.

Since the entire energy of the impinging electron is converted in to the energy of the X-ray photon of minimum wave length, the energy of the photon must be V electron volt. Therefore, the minimum wave length (λ_min ) in Angstrom of X-rays produced by an X-ray tube operating with anode voltage V volt must be given by

λ_min×V = 12400 [See (1) above].

(6) In Compton effect, the change (dλ) in the wave length of the scattered x-ray photon is given by

dλ = (h/mc) (1– cosφ) where ‘h’ is Planck’s constant, ‘m’ is the mass of the electron, ‘c’ is the speed of light in free space and ‘φ’ is the angle of scattering.

The above equation shows that the maximum change in wave length of the scattered X-ray photon is 2h/mc which happens when the photon is turned back (φ = 180º).

(7) In a hydrogen like atom of atomic number Z, the energy (E_n) of the electron in the n^th orbit is given by

E_n = – 13.6 Z²/n² electron volt. Here n = 1,2,3,4 etc.

Note that the energy is inversely proportional to the square of the quantum number n.

The above energy is the total energy of the electron and is made of negative potential energy and positive kinetic energy. The potential energy value is twice the kinetic energy value (as in the case of planetary motion) and hence the total energy is negative.

(8) Momentum of the electron in the n^th orbit is nh/2π.

(9) The Bohr radius (r₀) is the radius of the innermost orbit in a hydrogen atom and is given by

r₀ = h²ε₀/πme² = 0.53 Ǻ (nearly)

(10) The radius (r) of the n^th orbit in a a hydrogen like atom is given by

r = n²r₀/Z

Note that the orbital radius is directly proportional to the square of the quantum number n.

(11) The orbital period T is related to the orbital radius r as

T² α r³, as in planetary motion.

(12) When an electron undergoes a transition from an orbit of energy E₂ to an orbit of energy E₁, the frequency of the radiation emitted is given by Bohr’s frequency condition,

ν = (E₂ – E₁)/h

Since the emitted photon has linear momentum p = hν/c = h/λ = E/c, the atom receives an equal and opposite recoil momentum.

[When an atom is excited by absorbing a photon of energy hν, the electron in the atom undergoes transition from lower energy E₁ to higher energy E₂ and the same frequency condition (Bohr’s) holds].

(13) Rydberg’s relation for the wave number`ν (number of waver per metre) of the spectral line emitted by a hydrogen atom is

`ν = 1/ λ = R[(1/n₁²) – (1/n₂²)] where n₁ and n₂ are the quantum numbers of the inner and outer orbits.

(14) The important spectral series in the case of hydrogen atom are Lyman, Balmer, Paschen, Brackett and Pfund series. Of these, the Balmer series contains spectral lines in the visible region.

Wave numbers of the spectral lines in the Lyman series are obtained by putting n₁ =1 and n₂ = 2,3,4,5…etc. in the Rydberg relation (since the spectral lines in this series arise due to transitions from outer orbits to the innermost orbit).

For Balmer series, n₁ = 2 and n₂ = 3,4,5,6.…etc. (since the spectral lines in this series arise due to transitions from outer orbits to the second orbit).

For Paschen series, n₁ = 3 and n₂ = 4,5,6.…etc. (since the spectral lines in this series arise due to transitions from outer orbits to the third orbit).

For Brackett series, n₁ = 4 and n₂ = 5,6,7..…etc. (since the spectral lines in this series arise due to transitions from outer orbits to the fourth orbit).

For Pfund series, n₁ = 5 and n₂ = 6,7,8..…etc. (since the spectral lines in this series arise due to transitions from outer orbits to the fifth orbit).

In the next post, I’ll discuss questions from this section. Meanwhile, find some useful multiple choice questions here.

AP Physics B & C- Answer to Free-Response Question on Magnetic Fields

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

- Albert Einstein

 

A free response question (for AP Physics B & C aspirants) involving magnetic fields was posted on 11^th February 2008 for your practice. This was the question:

(a) An electron enters an environment of a non-uniform static magnetic field varying in magnitude and direction from point to point, and comes out of it following an irregular path. Would its final speed be equal to the initial speed if it suffered no collisions with anything during its irregular motion in the magnetic field? Justify your answer.

(b) In a cloud chamber photograph of the electron-positron pair production by a high energy gamma ray photon, two circular tracks emerging from a common point can be seen. The tracks curve in opposite directions in a plane normal to the magnetic field applied in the chamber. But if this event is photographed in a liquid hydrogen bubble chamber, two spiral tracks (instead of circular tracks) are seen. Explain why.

(c) Explain why a charged particle projected into a uniform magnetic field at an arbitrary angle θ with respect to the field follows in general, a helical path. Derive an expression for the ‘pitch’ of the helix.

(d) A positively charged particle of mass ‘m’ and charge ‘q’ traveling with constant velocity ‘v’ in the positive X-direction enters a uniform magnetic field B directed along the negative Z-direction extending from x = x₁ to x = x₂ (fig.). Derive an expression (in terms of the given parameters) for the minimum velocity required for the particle to cross the magnetic field.

As promised, I give the answer below:

(a) The magnetic force on a moving charge is always perpendicular to the magnetic field and hence no work is done by the field. The kinetic energy and the speed of the charged particle therefore remain unchanged.

[You might have noted the adjective ‘static’ with the magnetic field. If the magnetic field has a time variation, there will be induced electromotive force, which will change the speed of the charged particle (as in a betatron)].

(b) In a liquid, the charged particles lose energy by collision to a much greater extent than in a gas. (They cause much greater ionization in a liquid medium). The velocity of the charged particle goes on decreasing and therefore, the radius (r) of the circular path goes on decreasing in accordance with the expression for the radius,

r = mv/qB where ‘m’ is the mass, ‘q’ is the charge and ‘v’ is the speed of the particle and B is the magnetic flux density.Instead of a circular path, a spiral path is therefore followed by the particle.

(c) When a charged particle is projected with velocity ‘v’ in to a magnetic field at some arbitrary angle θ with respect to the field, the velocity component v cosθ, which is along the direction of the magnetic field makes the particle move forward where as the velocity component v sinθ, which is perpendicular to the direction of the magnetic field makes the particle move along a circle of radius ‘r’ given by the equation,

(mv² sin² θ)/r = q(v sinθ)×B

[We have equated the centripetal force to the magnetic force].

Therefore, r = (mv sinθ)/qB

The charged particle therefore moves along a helical path within the magnetic field.

The ‘pitch’ of the helix is obtained by multiplying the forward velocity v cosθ with the period (T) of the circular component of motion. The period of the circular component of motion is given by

T = (2πr/v sinθ) = [2π(mv sinθ/qB)] /v sinθ) = 2πm/qB

Therefore, pitch of the helix = T×v cosθ = (2πm/qB)×v cosθ = (2πmv cosθ)/qB

(d) The charged particle will move along a semicircular path just before crossing the magnetic field (fig.). The direction of the curve is obtained by applying Fleming’s left hand rule for the magnetic force. If the velocity of the particle is slightly greater than the velocity for traversing the semicircle, it will cross the magnetic field. Therefore, the required limiting velocity is the velocity for the motion along the semicircle within the magnetic field region.

The radius of the semicircle is x₂ – x₁. Therefore, the limiting velocity ‘v’ is given by

mv²/r = qvB where r = x₂ – x₁

Therefore, mv²/(x₂ – x₁) = qvB, from which v = qB(x₂ – x₁)/m

AP Physics B & C- Free-Response Practice Question on Magnetic Fields

Here is a free-response question meant for those appearing for AP Physics B as well as AP Physics C examination:

(a) An electron enters an environment of a non-uniform static magnetic field varying in magnitude and direction from point to point, and comes out of it following an irregular path. Would its final speed be equal to the initial speed if it suffered no collisions with anything during its irregular motion in the magnetic field? Justify your answer.

(b) In a cloud chamber photograph of the electron-positron pair production by a high energy gamma ray photon, two circular tracks emerging from a common point can be seen. The tracks curve in opposite directions in a plane normal to the magnetic field applied in the chamber. But if this event is photographed in a liquid hydrogen bubble chamber, two spiral tracks (instead of circular tracks) are seen. Explain why.

(c) Explain why a charged particle projected into a uniform magnetic field at an arbitrary angle θ with respect to the field follows in general, a helical path. Derive an expression for the ‘pitch’ of the helix.

(d) A positively charged particle of mass ‘m’ and charge ‘q’ traveling with constant velocity ‘v’ in the positive X-direction enters a uniform magnetic field B directed along the negative Z-direction extending from x = x₁ to x = x₂ (fig.). Derive an expression (in terms of the given parameters) for the minimum velocity required for the particle to cross the magnetic field.

The above question carries 15 points (3+2+5+5). Try to answer it within 15 minutes. I’ll be back soon with a model answer for you.