RI 19 Quantum Physics tutorial solution

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 1 Tutorial 19 Quantum Physics Suggested Solutions D1 (a) Since p I λ= = × dN hcPA dt , the rate of incident photons 69 p 15 1 34 8 (210)(12 10 )(254 10 ) 3.22 10 s(6.63 10 )(3.00 10 ) IAλ −− − − ××= = = ××× dN dt hc (b) Since e 0i = ×dN edt , the rate of emission of photoelectrons 10 91e0 19 4.8 10 3.00 10 s1.60 10 i − − − ×= = = ×× dN dt e (c) (i) Quantum yield 9 7e 15 p 3.00 10 9.3 103.22 10 −×= = = ×× dN dtQ dN dt (ii) Any two of the following: • P hotons are mostly reflected off the metal surface and so are not involved in photoemission. • Besides energy, a photon imparts momentum to the electron. These electrons are generally “back-scattered” (away from surface). To escape from the metal, these electrons must acquire sufficient energy by colliding with other electrons. This process has a very low probability. • After acquiring energy from a photon, an electron near the surface of the metal will lose energy through multiple collisions with other electrons or ions as it migrates towards the surface of the metal. (d) The wavelength of 313 nm is greater than the threshold wavelength. Hence, photons of such radiation have insufficient energy to overcome the work function of the metal. D2 (a) Energy of photon 34 8 19 7 (6.63 10 )(3.00 10 ) 7.8 10 J2.55 10λ − − − ××= = = ×× hcE (b) (i) The photoelectric current depends on the rate of emission of photoelectrons, which in turn depends on the intensity of the incident radiation. Increasing V accelerates the electrons but does not affect the rate of emission of photoelectrons. Hence, current reaches a maximum value no matter how large V is made. (ii) Electrons are emitted with a range of kinetic energy (up to K.E.max). Some energetic electrons still have sufficient K.E. to overcome the retarding p.d. to reach the gauze.

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 2 (c) (i) The rate of incidence photon increases with increasing intensity of light. Increasing rate of incident photon increases the rate of emission of photoelectrons (assuming constant quantum yield) and hence maximum photoelectric current increases. (ii) The stopping potential sV is the minimum value of the potential difference between the gauze (collector) and plate (emitter) needed to prevent the most energetic photoelectrons from reaching the gauze. s maxK.E.eV = K.E. of most energetic photoelectrons is dependent only on the frequency of the light and the work function of the surface but is independent of the intensity of light. So the stopping potential remains unchanged. (d) Since nickel has a greater work function, K.E. of most energetic photoelectrons is lower. Hence a lower stopping potential is needed to prevent the most energetic photoelectrons from reaching the gauze. Since the intensity and frequency of the light did not change, the saturated photocurrent remains constant. (Not in syllabus: Typically, quantum yield decreases with increasing work function as a smaller fraction of the incident photons is capable of ejecting electrons from the nickel surface. As a result, there is a lower rate of emission of photoelectron and hence a lower maximum photoelectric current.) D3 (a) 1.0 VV =− (b) Maximum K.E. 19 maxK.E. 1.0 eV 1.60 10 JSeV −= = = × (c) Work function max 34 8 19 19 9 K.E. (6.63 10 )(3.00 10 ) 1.60 10 3.85 10 J365 10 λ − −− − Φ= − ××= ×× =×× hc (d) Rate of emission of photoelectrons 6 13 1e0 19 4.0 10 2.5 10 s1.60 10 i − − − ×= = = ×× dN dt e Hence rate of incidence of photons p 13 1e3 7.5 10 s −= ×= ×dN dN dt dt i V A (d) Not in syllabus

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 3 (e) D4 (a) (i) Threshold frequency 14 0 6.4 10 Hzf = × (ii) From the graph, 19 maxK.E. 4.8 eV 7.7 10 J −= = × (b) Einstein’s photoelectric equation max 0K.E. hf hf= − Rearranging, ( ) 19 34max 14 0 K.E. 7.68 10 6.62 10 J s18.0 6.4 10h ff − −×= = = ×− − × D5 (a) (i) Energy of photon 34 8 19 7 (6.63 10 )(3.00 10 ) 4.42 10 J4.5 10λ − − − ××= = = ×× hcE (ii) Since p I= = × dNPA E dt , the rate of incident photons 4 p 17 1 19 (700)(1 10 ) 1.58 10 s4.42 10 I − − − ×= = = ×× dN A dt E Recall: Intensity Nhf tA= If frequency is decreased, and rate of photon incidence (N/t) stays constant, then intensity will decrease. But because N/t is constant, the rate of electron emission is also constant, hence the photocurrent stays constant. If frequency is decreased, and intensity stays constant, the rate of photon incidence (N/t) must increase. This leads to an increase in rate of electron emission, and hence higher photocurrent. This question here (a very old A-level question) is vague with what quantity is being kept constant. Future questions should be more specific. Note: if you face such vague questions in exams, write down your assumption(s) and proceed accordingly. V/ V I/ µA 10−1 2 2 4 1. 2. I / µA I / µA Rate of photon incidence (N/t) constant Intensity constant

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 4 (b) (i) de Broglie relation h pλ = (ii) Momentum of photon 34 27 1 7 6.63 10 1.47 10 kg m s4.5 10 hp λ − −− − ×= = = ×× (c) (i) Change in momentum of each photon on reflection is 2p− . Rate of change in momentum p 17 27 10 ( 2 ) (1.58 10 )(2)(1.47 10 ) 4.66 10 N − − = ×− = − × × = −× dNdp pdt dt Therefore the change in momentum in 1 second 104.66 10 N sp −∆= − × (ii) Photons experienced a force when they are reflected off the surface. By Newton’s 3rd law, the photons exert an equal but opposite force on the surface. The force per unit area exerted on the surface is the radiation pressure. D6 (a) (i) Wavelength of γ-ray photon is 34 8 12 6 19 (6.63 10 )(3.00 10 ) 1.06 10 m(1.17 10 )(1.60 10 )λ − − − ××= = = ××× hc E (ii) Using de Broglie equation, the momentum of the photon is 34 22 1 12 6.63 10 6.24 10 kg m s1.06 10 hpγ λ − −− − ×= = = ×× (b) (i) (ii) By the principle of conservation of momentum, mv p γ= or 22 31 26 6.24 10 6.27 10 m s9.95 10 pv m γ − − − ×= = = ×× γ-ray photonnickel nucleus v

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 5 (c) The angle between the final direction of motion of the nucleus and that of the emitted photon will only be either zero or 180° if the second nickel nucleus emits photon parallel to its momentum. If photon is emitted in other directions, then the initial momentum of the nucleus must be added to the final momentum of the nucleus determined in (b). Hence the angle between zero and 180°. D7 (a) (i) The lowest and most stable energy state. 1n = (ii) The energy needed to cause an electron in an atom in a lower energy state to transit to a higher energy state. (iii) The minimum energy needed to completely remove the valence electron of an atom in the ground state. 1n = to n =∞ (b) (i) No transition (ii) 1n = to 2n = (iii) No transition (iv) 1n = to 3n = (c) (i) No transition (ii) 1n = to 2n = (iii) 1n = to 2n = (iv) 1n = to 2n = or 1n = to 3n = D8 (a) (i) Level 5 to Level 1 (ii) 10 (b) 4 spectral absorption lines. (c) (i) No transition (ii) Level 2 to Level 1 (iii) Level 3 to Level 1, Level 3 to Level 2, Level 2 to Level 1 (d) (i) Ionisation energy of sodium (ii) To produce spectral line emission but no free electrons, the bombarding electrons must possess energy from 193.38 10 J−× to 198.21 10 J−× . The corresponding accelerating p.d. is 2.11 V p.d. 5.13 V≤< . γ-ray photonnickel nucleus p pi momentum acquired due to emission of photon

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 6 (e) (i) Let 1 589.0 nmDλ = and 2 589.6 nmDλ = . The corresponding energy of the photons are 19 1 1 3.3769 10 JD D hcE λ −= = × ; 19 2 2 3.37