Quantum Physics

Question 1

What is meant by a photon?

Answer 1

It is a quantum(definite) amount of energy of electromagnetic radiation.

Question 2

Energy = hf = hc/Λ How is hf = hc/Λ derived?

Answer 2

v = fΛ Since v is the speed of light, represented by c as a photon is a quantum if energy of electromagnetic radiation, f= c/Λ

Question 3

The least amount of energy an em radiation of frequency f can have is _____.

Answer 3

Hf, the energy of one photon

Question 4

Derive rate of emission of photons given : Power = P Speed of light = c Wavelength of light = Λ

Answer 4

Using P = E/t Number of photons emitted per second = 1/t = dn/dt = P/E = PΛ/hc n is the number of photons

Question 5

Electrons are released from the surface if the metal when ______.

Answer 5

The frequency of the incident electromagnetic radiation has a frequency higher than the threshold frequency of the metal.

Question 6

Define work function.

Answer 6

It is the minimum energy necessary to release an electron from the surface of the emitter material.

Question 7

Threshold frequency is the minimum frequency of the electromagnetic radiation, below which, _____.

Answer 7

No photoelectric will be emitted regardless of intensity of the radiation

Question 8

Stopping potential is the minimum_____.

Answer 8

Potential difference between the emitter and the collector which stood the most energetic photo electron

Question 9

Why is KEmax = eVs?

Answer 9

Stopping potential is the minimum potential difference the emitter and the collector which stops the most energetic photoelectron. 1ev is the amount of kinetic energy an electron attains when it accelerates in a potential difference of 1v. The most energetic photoelectron is the one with the highest KE (because of higher f). Hence stopping potential difference required is Vs to stop it. Why it is multiplied by e: Change in energy = q x change in v

Question 10

According to the quantum theory, light has a ______.

Answer 10

Particle nature.

Question 11

Why is there a range of kinetic energies of emitted photoelectrons for each metal even if the incident photons have the same frequency?

Answer 11

When a photon collides with an electron in a one to one interaction, the electron can escape if hf > work function. Max KE the electron can have is equal to hf - work function if it is emitted from the surface. For collisions below the surface, work is done by the electron to reach the surface before it can be emitted. Hence the KE of such an emitted electron is less than the max KE. the electrons will therefore be emitted with a range of KE.

Question 12

Why are photoelectrons emitted instantaneously as Long as incident frequency is above threshold frequency?

Answer 12

Electron emission is due to the direct interaction between one photon and one electron, which absorbs all photon energy at an instant or none at all. If the energy absorbed us greater than work function, the electron may escape from the surface instantaneously.

Question 13

Why is maximum KE of photoelectrons dependent on the frequency of the incident em radiation but independent on the intensity of the radiation?

Answer 13

Vs = KEmax Increasing frequency of light increases the energy of each photon as E = hf. As electrons are emitted from the one to one interaction where the electron absorbs all the photon energy or none at all, the maximum KE of the photoelectrons will increase. Therefore a larger stopping potential is required to reduce the photocurrent to zero. However, increasing the light intensity simply increases the number of photons falling on the metal per unit time, resulting in an increase in emission of photoelectrons and hence photocurrent I due to one to one interaction. The incident photons still impart the same amount of energy hf to every electron as f is constant. Hence the Probability of emission is the same. Hence the maximum kinetic energy and stopping potential do not change.

Question 14

Why does V increase when the photocurrent I is increased?

Answer 14

As hf >/= work function, electrons are emitted in all directions. And at certain p.d., not all electrons emitted reaches C. As V increases, electrons experience larger E force so more electrons can reach C. (Because e force is in the direction of potential difference?) Thus, photocurrent increases.

Question 15

Why is there a saturation photocurrent when V is increased further?

Answer 15

After a certain maximum p.d. , all emitted electrons have reached C so photocurrent is at its max.

Question 16

When VE> VC (negative V?), why does the photo current gradually decrease to zero?

Answer 16

Reverse potential difference results in a decelerating force on electrons Travelling towards c. Those with smaller speeds are unable to reach c. At the Vs, no photoelectrons reaches c as the photoelectrons with maximum KE will stop before reaching C.

Question 17

Explain why photocurrent is directly proportional to intensity of incident radiation using both the wave theory and quantum theory.

Answer 17

Wave theory: intensity increases –> more energy absorbed by electrons –> more would be emitted and hence higher photocurrent. Quantum theory: intensity increases –> number of photons increases proportionally –> more photoelectrons emitted –> higher photocurrent. Probability of emission depends on energy of incident photons.

Question 18

Explain hf = work function + eVs Hf = photon energy eVs = work done by the field to stop the most energetic electron from reaching the collector.

Answer 18

By the principle of conservation of energy, hf is converted to work done to emit the electron from the surface plus work fine by the field to stop the most energetic electron from reaching the conductor.

Question 19

In a graph of Vs against f graph, what is the significance of the gradient and intercept?

Answer 19

hf = work function + eVs Vs = h/e f - h/e f0 (f0 is the threshold frequency) Gradient = h/e Intercept = h/e f0

Question 20

In a Vs against f graph, what will happen when: A) intensity increases B) work function increases

Answer 20

A) position of the graph remains the same since the same amount of energy hf is imparted to every electron because frequency of light is constant. Hence, maximum Ke of photoelectrons and hence Vs do not change. B) graph shifts right since work function = hf0

Question 21

Explain the process that causes the rate of emission of electrons to decrease when the frequency increases but intensity remains the same.

Answer 21

Rate of emission of electrons(and hence photocurrent) = p/hf = E/hft (Intensity should remain the same as it is directly proportional to photocurrent so it will affect photocurrent)

Question 22

What is the difference between intensity and power?

Answer 22

Power is the energy emitted per unit time while intensity is the number of photons transferred per unit time.

Question 23

If the rate of emission of electrons to decrease when the frequency increases, why can the photo current increase as f increases when intensity is the same?

Answer 23

Photon energy(E increases more than f increases) is higher and hence the quantum efficiency is increased.

Question 24

For a particle if momentum p (or mv) that exhibits the wave behaviour, it will have an associated wavelength, the de Broglie wavelength given by _____.

Answer 24

Wavelength = h/p For a photon, the associated momentum p can be determined by : p = h/wavelength Photons have no mass but have momentum.

Question 25

Why can wave behaviour be observed in electrons?

Answer 25

The de Broglie wavelength of an electron is in the order of magnitude 10^-10m which is comparable to the interacting spacing in any Latrice structure. Hence the electron diffracts significantly and forms an interference pattern.

Question 26

Why is wave behaviour not observed in ordinary objects?

Answer 26

Ordinary objects have a large mass and hence large momentum, and hence the de Broglie wavelength is very small and cannot be measured so the wave behaviour cannot be observed.

Question 27

Why are the electron diffraction patterns concentric rings of varying diameters instead of a 1-D series if bright spots?

Answer 27

The graphite film has a 2-D regular arrangement of carbon atoms. Diffraction is significant if wavelength is comparable to slit opening. As de Broglie wavelength of electron is comparable (when appropriate accelerating p.d. Is applied) to the separation between graphene atoms, diffraction is significant and interference can take place.

Question 28

Why is the spacing between the concentric rings less in an electron diffraction pattern when potential difference increases?

Answer 28

De Broglie wavelength = h/p = h/mv V increases as p.d. Increases, de Broglie wavelength of each electron decreases. Since interatomic spacing of carbon atoms in graphene film is constant and as dsin theta = nlambda(in waves), a decrease in de Broglie wavelength will lead to a smaller spacing between concentric rings.

Question 29

How does an electron transition into different energy levels?

Answer 29

By losing energy by emitting a photon of frequency f (higher to lower) or gaining energy by absorbing a photon of frequency f (lower to higher) EH -EL = hf

Question 30

An electron can transition into a lower energy level due to:

Answer 30

1. Spontaneous emission 2: stimulated emission

Question 31

An electron can transit into a higher energy level in an atom by:

Answer 31

1. Collision with another energetic atom or fast moving electron 2. Absorbing energy by heating 3. Spontaneous absorption of a photon (the difference in energy levels must match exactly the energy of the photon)

Question 32

Name the differences in emission line spectra and absorption line spectra.

Answer 32

Emission line spectra have coloured lines corresponding to a particular frequency or wavelength of photons emitted while absorption line spectra have dark lines which do so. Formation of emission: gas atoms are excited, photons are emitted Formation of absorption: gas atoms are in Ground state, photons are absorbed

Question 33

Explain the existence of a continuous spectrum and cut-off wavelength in X-ray Spectra.

Answer 33

1. A fast moving electron of energy Ko collides with one of the atoms in the heavy metal e.g. tungsten or copper.
2. The electron loses a fraction of its energy to produce an X-ray of photon energy hf and is scattered at at an angle.
3. Since different electrons experience different losses in energy, the photons generated due to the collisions of different electrons have different energies and thus different frequencies and wavelengths.
4. The difference wavelengths form the continuous spectrum of X-ray radiation.

Question 34

How is the loss in KE of bombarding electrons derived?

Answer 34

• • (1/2 mv² - 1/2 mu²)*
• = 1/2 mu² - 1/2 mv²*

where u is the initial velocity and v is the final velocity of the electron and m is the mass of the electron

Note: this is loss in energy and not change in energy so it is -(1/2 mv_final² - 1/2mv_initial²).

Question 35

How is cut-off wavelength of X-ray spectra derived from the energy loss in KE of the bombarding electron?

Answer 35

• energy of photon = loss in KE of bombarding electron*
• = -(1/2 mv² - 1/2 mu²)*
• = 1/2 mu² - 1/2 mv²*

where u is the initial velocity and v is the final velocity of the electron and m is the mass of the electron

The cut-off wavelength is the minimum wavelength λ_min. for λ to be minimum, f is the maximum since f = v/**λ. Thus we consider the maximum energy hf.

If the electron loses all its KE, v = 0

• maximum energy of photon = hf_max = KE of electron = 1/2 mu² = qV since all electrons carry qV of energy.*
• hf_max = qV*
• hc/λ_min = qV*
• λ_min = hc/qV*

Question 36

How are X-rays produced?

Answer 36

A beam of high energy (KE) electrons strikes on a heavy metal target, causing X-rays to be emitted.

The electrons are obtained by heating a fillament,causing it to emit electrons and a potential difference between the filament and the anode (copper mounted on the target of very high potential)

This causes the electrons to accelerate and attain a high velocity and thus KE.

Question 37

When the energy of the beam of electrons striking the metal target increases (via increase in potential difference), how is the X-ray spectrum affected?

Answer 37

Relative intensity increases at all points but the K_α​ and K_ß has the same wavelength.

Question 38

Explain the characteristic spectrum of heavy metals.

Answer 38

When the bombarding electron has sufficient energy, it knocks a deep-lying electron near the nucleus out of the inner shell of that atom, creating a hole in the K-shell, defined by n=1.

An electron of higher energy falls to the K-shell to fill the gap, emitting an X-ray photon in the process.

Since the emitted X-ray photons have precise energies which are varying, they form a characteristic spectrum of x-ray radiation for that metal.

Question 39

Explain how how the K_α​ and K_ß​ come about.

Answer 39

When the bombarding electron has sufficient energy, it knocks a deep-lying electron near the nucleus out of the inner shell of that atom, creating a hole in the K-shell, defined by n=1.

An electron of higher energy falls to the K-shell to fill the gap, emitting an X-ray photon in the process.

If the electron falls from the L-shell (defined as n=2), the emitted radiation gives a K_α​ line.

If the electron falls from the M-shell (defined as n=3), the emitted radiation gives a K_ß​ ​line.

Question 40

State Heisenberg’s uncertainty principle.

Answer 40

In general, neither the position nor the momentum of a particle can be determined simultaneously with great precision.

ΔxΔp ≥ h/(4π)

where Δx is the uncertainty in the position of the particle and Δp is the uncertainty in the momentum of the particle.

Note: These uncertainties are inherent in a particle and not due to the uncertainties of measuring instruments.

Question 41

Explain the significance of Heisenberg’s uncertainty principle in determing the location and energy of an electron.

Answer 41

Δx is the region the electron is confined within, which is roughly the width of a typical atom so the electron can be anywhere in that region.

By finding Δp, the KE can be determined using KE = 1/2 mv² and p = mv. Using the uncertainty principke, the energy of the electron can be determined to a reasonable order of magnitude.

Question 42

Discuss using the Heisenberg uncertainty principle to explain why there must be some molecular motion even at T = 0K or absolute zero.

Answer 42

A state of no molecular motion would mean that every molecule was at rest (Δp=0) at a fixed position (Δx=0).

But the uncertainty principle says that the product of the two uncertainties has to be a minimum of h/(4π) and hence can never be zero. Thus there must be some molecular motion.

Question 43

State the time-energy uncertainty principle.

Answer 43

ΔEΔt ≥ h/(4π)

The uncertainty ΔE depends on the time interval Δt the system remains in the given state.

Question 44

The time-energy undertainty principle implies that the greater the time interval, the ____ the energy uncertainty.

Answer 44

ΔEΔt ≥ h/(4π)

hence the anser is smaller.

Question 45

Given ΔE, E and λ, determine Δλ. Where Δλ is the wavelength spread.

Answer 45

1. Differentiate E= hc/λ. Result: ΔE/Δλ = -hc*(1/λ²)
2. By multiplying Δλ and 1/E on both sides, we get: ΔE/E = Δλ/λ
3. Substitute the values of ΔE, E and λ.

Question 46

What is the probability density distribution of a wave function?

Answer 46

It is the square of the wave function (ψ²)

Question 47

What is a potential barrier?

Answer 47

It is the potential energy barrier U_o where U_o > E of the electron, where it will not be able to pass through and will bounch back.

Question 48

The wave function ψ describes the _____ of the electron.

Answer 48

location

Question 49

How does an electron appear on the other side of a potential barrier?

Answer 49

An electron is a matter wave(?) and has a wave function. Thus there is a possibility of the electron appearing on the other side of a barrier with a smaller amplitude. This is beacuse there is a small probability of |ψ²|(non- zero) for an electron to be on the other side

Question 50

What happens to the amplitude of an electron in the barrier?

Answer 50

It decreases exponentially.

Question 51

After tunnelling through the barrier, the electron has the same _____, _____ and _____.

Answer 51

Energy, wavelength, momentum

Question 52

What is the transmission coefficient and how is it related to reflection coefficient?

Answer 52

Transmission coefficent T is the probability that the particle penetrates the potential barrier and reflection coefficient R is the probability of the particle being reflected by the potential barrier. T + R = 1

Question 53

When does the relationship T ∝ e^-2kd apply approximately?

d is the barrier thickness and k= √([8π²m(U-E)]/h²) where m is the mass of the particle and (U - E) is energy difference, which is the work function of the metal.

Answer 53

For a high or wide barrier, where kd << 1.

Question 54

What is the significance of T in the relationship T ∝ e^-2kd being exponential?

Answer 54

T is very sensitive to change in mass m and energy difference/work function (U-E) (since k= √([8π²m(U-E)]/h²), and barrier thickness d.

Question 55

How does the scanning tunnelling microscope (STM) work in terms of quantum tunnelling?

Answer 55

The gap between the needle-like probe and the metallic specimen acts as a potential barrier. When a potential difference is applied at the tip, electrons tunnel between the tip and surface produccing an electric tunnelling current across the gap.

Magnitude of the tunneling current is extremely sensitive to the distance between between the probe and metal so it is monitored as the tip moves across the metal surface.

I ∝ e^-2kd

Question 56

In T ∝ e-2kd, what is k and d?

Answer 56

k is the reciprocoal of the decay length and d is the barrier thickness.

k= √([8π²m(U-E)]/h²)

Question 57

Name and describe an example of quantum tunneling in nature.

Answer 57

The alpha-decay of Uranium isotope.

In the radioactive decay of U-238, an alpha particle (emitted particle) is ejected from the nucleus with large kinetic energy. The potential energy barriers are the strong nuclear force if the particle is in the nucleus or the repulsive coulomb force after it is outside the nucleus.

The energy of the alpha particle within the nucleus is smaller than the potential energy barrier, but but it is able to tunnel out of the nucleus, which has a barrier of 30 MeV, as a free particle with only 4 MeV.