PHY 111 - Ch 1 PDF

Chapter 1 Light Quanta Blackbody radiation curves (BBRC) ❑ If temperature (T) increase, the peak of the BBRC o moves to greater intensities o toward shorter wavelengths. ❑ Electromagnetic waves (EMWs) emit from a hot body due to the oscillation of electric charges in the molecules of th...

Chapter 1 Light Quanta Blackbody radiation curves (BBRC) ❑ If temperature (T) increase, the peak of the BBRC o moves to greater intensities o toward shorter wavelengths. ❑ Electromagnetic waves (EMWs) emit from a hot body due to the oscillation of electric charges in the molecules of the material. Wave model for black body (Rayleigh and Jeans Rule ) ❑ Wave theory predicted that, the variation of the spectral radiancy (intensity) with λ at a given temperature is ❑ Intensity of electromagnetic radiation tend to zero when wavelength increase (infrared region, IR), but also approaches infinity when wavelength decreases (Ultraviolet region, UV). 𝟐𝝅𝒄𝒌𝑻 𝑺= 𝝀𝟒 k is Boltzmann constant ❑ The wave theory agree with experiment at very long Wavelength and disagree at shorter wavelengths Blackbody radiation Blackbody ❑ It absorbs all thermal radiation incident on its surface and emits radiation of wavelengths (l) that depends only on its temperature (T) ❑ Its thermal emissivity (e) = 1 ❑ it can be formed by making a cavity within a body  Wien’s law The relationship between the temperature T of a black body, and wavelength λ max at which the intensity is maximum is inversely proportional 𝑻 ⋉ 𝟏Τ𝝀𝐦𝐚𝐱 𝝀𝐦𝐚𝐱 × 𝑻 = 𝟐𝟖𝟗𝟖 × 𝟏𝟎−𝟔 𝒎. 𝑲 Wilhelm Wien German physicist Nobel Prize 1911 Problems λmax = 550 nm At what temperature is cavity radiation most visible to human eye if the eye is most sensitive to yellow – green light, wavelength = 550 nm? T=?? solution 𝝀𝐦𝐚𝐱 × 𝑻 = 𝟐𝟖𝟗𝟖 × 𝟏𝟎−𝟔 𝒎. 𝑲 𝟐𝟖𝟗𝟖 𝑻= −𝟑 = 𝟓𝟐𝟔𝟗. 𝟎𝟗𝑲 𝟓𝟓𝟎 × 𝟏𝟎 The effective surface temperature of the sun is 5800 K, at what wavelength would you expect the sun to radiate most strongly? In what region of the spectrum is this? Why, then, does the sun appear yellow? T = 5800 K λmax=?? solution Visible region (yellow color) 𝝀𝐦𝐚𝐱 × 𝑻 = 𝟐𝟖𝟗𝟖 × 𝟏𝟎−𝟔 𝒎. 𝑲 Because it radiates at the 𝟐𝟖𝟗𝟖 × 𝟏𝟎−𝟔 wavelength of the yellow color ∴ 𝝀𝒎𝒂𝒙 = = 𝟎. 𝟒𝟗𝟗𝟕 × 𝟏𝟎−𝟔 𝒎 𝟓𝟖𝟎𝟎 ∴ 𝝀𝒎𝒂𝒙 = 𝟒𝟗𝟗. 𝟕 𝒏𝒎  Example Find the peak wavelength of the radiation emitted by each of the following: a) Human body when the skin temperature is 35 ℃ (b) The tungsten filament, of a light bulb which operates at 𝟐𝟎𝟎𝟎 °𝑲 Solution a) Form the wine's displacement low we have. 𝛌𝒎𝒂𝒙. 𝑻 = 𝟐. 𝟖𝟗𝟖 𝝁𝒎. 𝑲 Body temperature = 273 + 35=308 oK 𝛌𝒎𝒂𝒙 × 𝟑𝟎𝟖 = 𝟐. 𝟖𝟗𝟖 × 𝟏𝟎−𝟔 𝒎. 𝑲 ∴ 𝛌𝒎𝒂𝒙 = 𝟗. 𝟒 × 𝟏𝟎−𝟔 𝒎 = 𝟗. 𝟒 𝝁𝒎 This radiation is in the infrared region (IR) and this is λ the basics for the instrument (night version instrument) which can detect persons at dark. b) For tungsten filament. 𝛌𝒎𝒂𝒙. 𝑻 = 𝟐. 𝟖𝟗𝟖 𝝁𝒎. 𝑲 𝛌𝒎𝒂𝒙 × 𝟐𝟎𝟎𝟎 = 𝟐. 𝟖𝟗𝟖 × 𝟏𝟎−𝟔 𝒎. 𝑲 ∴ 𝛌𝒎𝒂𝒙 = 𝟏. 𝟒 × 𝟏𝟎−𝟔 𝒎 = 𝟏. 𝟒 𝝁𝒎 Stefan–Boltzmann law This law states the relation between the thermal power of a black body and its temperature, 𝑷 = 𝝈𝑨𝒆 𝑻𝟒𝟐 − 𝑻𝟒𝟏 𝑷 is the Power radiated (W) 𝝈 is the Stefan's Constant 5.67 x 10-8 (W m-2 K-4) 𝑨 is the Surface area of the body (m²) Stefan–Boltzmann 𝑻𝟐 is the Temperature of the body (K) Swedish scientist 𝑻𝟏 is the room Temperature (K) Nobel Prize 1903 𝒆 is the thermal emissivity of the body has a value changes from 0 to 1, depending on the nature of blackbody surface Example An atomic oscillator emits radiation of 700 nm wavelength how much energy is associated with a photon of light of this wavelength? Solution: The energy of the photon is given by: 𝑬 = 𝒉 𝒇 = 𝒉𝒄Τ𝝀 𝟔. 𝟔𝟐𝟔 × 𝟏𝟎−𝟑𝟒 𝑱. 𝒔 × 𝟑 × 𝟏𝟎𝟖 𝒎𝒔−𝟏 𝑬= −𝟗 = 𝟐. 𝟖𝟒 × 𝟏𝟎−𝟏𝟗 𝑱 𝟕𝟎𝟎 × 𝟏𝟎 𝒎 Thus, the photon of light is indeed a small bundle of energy. Problems An oven with an inside temperature 227 oC is in a room having a temperature 27 oC. There is a small opening of area 5 cm2 in one side of the of the oven. How much net power is transferred from the oven to the room? Consider both oven and room as cavities. T2 = 500 ͦK, T1 = 300 ͦK, A = 5 cm2 , e = 1, σ = 5.67 x 10-8 (W m-2 K-4) solution P=??? 𝑻𝟏 = 𝟐𝟕 + 𝟐𝟕𝟑 = 𝟑𝟎𝟎 𝑲 𝑻𝟐 = 𝟐𝟐𝟕 + 𝟐𝟕𝟑 = 𝟓𝟎𝟎 𝑲 𝑷 = 𝝈𝑨𝒆 𝑻𝟒𝟐 − 𝑻𝟒𝟏 𝑷 = (𝟓. 𝟔𝟕 × 𝟏𝟎−𝟖 )(𝟓 × 𝟏𝟎−𝟒 )(𝟏) (𝟓𝟎𝟎)𝟒 − (𝟑𝟎𝟎)𝟒 = 𝟏. 𝟓𝟒𝑾 Example If the stars behave like black body, for the sun 𝛌𝒎𝒂𝒙 = 𝟓𝟏𝟎 𝒏𝒎 whereas for the north-star 𝛌𝒎𝒂𝒙 = 𝟑𝟓𝟎𝒏𝒎. Find: (a) The surface temperature of these stares. (b) Using Stefan's Boltzmann law and the result obtained above, determine the power radiated from 1 cm2 of the star surface. Given σ = 5.67×10-8 W/m2.k4 Answer a) Form the wine's displacement law for the sun. 𝛌𝒎𝒂𝒙 ∙ 𝑻 = 𝟐. 𝟖𝟗𝟖 𝝁𝒎. 𝑲 𝟓𝟏𝟎 × 𝟏𝟎−𝟗 ∙ 𝑻 = 𝟐. 𝟖𝟗𝟖 × 𝟏𝟎−𝟔 𝒎. 𝑲 ∴ 𝑻 = 𝟓𝟕𝟎𝟎 °𝑲 Form the wine's displacement law for the north-star. 𝟑𝟓𝟎 × 𝟏𝟎−𝟗 ∙ 𝑻 = 𝟐. 𝟖𝟗𝟖 × 𝟏𝟎−𝟔 𝒎. 𝑲 ∴ 𝑻 = 𝟖𝟑𝟎𝟎 °𝑲 b) For the sun 𝑰 = 𝝈 𝑻𝟒 𝑷𝒐𝒘𝒆𝒓 = 𝑰 𝑨 = 𝑨 ∙ 𝝈 ∙ 𝑻𝟒 = 𝟓. 𝟔𝟕 × 𝟏𝟎−𝟗 × 𝟓𝟕𝟎𝟎𝟒 × 𝟏 × 𝟏𝟎−𝟒 𝒎𝟐 = 𝟓. 𝟗 × 𝟏𝟎𝟑 𝑾 For the north-star: 𝑷𝒐𝒘𝒆𝒓 = 𝑨 ∙ 𝝈 ∙ 𝑻𝟒 = 𝟓. 𝟔𝟕 × 𝟏𝟎−𝟗 × 𝟖𝟑𝟎𝟎𝟒 × 𝟏 × 𝟏𝟎−𝟒 𝒎𝟐 = 𝟐. 𝟕𝟏 × 𝟏𝟎𝟒 𝑾 photon model for black body Planck derived a formula for the spectral radiancy that fitted the experimental data perfectly at all wavelengths and for all temperatures 𝟐𝝅𝒉𝒄𝟐 𝟏 Max Planck 𝑺(𝝀) = German physicist 𝝀𝟓 𝒆(𝒉𝒄Τ𝝀𝒌𝑻) − 𝟏 Nobel Prize 1918 Where h is plank constant Plank's Quantum Hypothesis 1- blackbody radiation is due to oscillating molecules at its surface 2- Energy of molecules En have only discrete values Given by: 𝑬𝒏 = 𝒏𝒉𝒇 f is frequency, and n is quantum number. 3- The molecules emit or absorb energy in discrete packets (photons) by jumping between quantum states. 𝜟𝑬 = 𝒉𝑪Τ𝝀 = 𝑬𝒖𝒑𝒑𝒆𝒓 − 𝑬𝒍𝒐𝒘𝒆𝒓 Example: Show that Planck's law tends to Rayleigh-Jeans law for long wavelength 200 angstrom (°𝑨) Solution: 𝟐𝝅𝒉 𝒄𝟐 𝟏 Planks law 𝑰 𝛌, 𝐓 = 𝟓 × 𝒉𝒄Τ𝛌𝒌𝑩 𝑻 𝛌 𝒆 −𝟏 For long 𝛌 , the exponential term in the denominator can be approximated by the first tow terms in the 𝒙𝟐 series: 𝒆𝒙 = 𝟏 + 𝒙 + + ⋯ … ….. 𝟐! if x small 𝒙 𝒉𝒄Τ𝛌𝒌𝑻 𝒉𝒄 𝒆 =𝟏+𝒙 ∴𝒆 =𝟏+ 𝛌𝒌𝑩 𝑻 𝟐𝝅𝒉 𝒄 𝟐 𝟏 ∴ 𝑰 𝛌, 𝐓 = × 𝛌 𝟓 𝒉𝒄 𝟏+ −𝟏 𝛌𝒌𝑩 𝑻 𝟐𝝅𝒉 𝒄𝟐 𝛌𝒌𝑩 𝑻 𝒌𝑩 𝑻 𝑰 𝛌, 𝐓 = 𝟓 × = 𝟐𝝅𝒄 𝟒 𝛌 𝒉𝒄 𝛌 Is Rayleigh -Jeans law  Light quanta (Photon) light consists of small particles are called light quanta (Photons), each with a specified energy and momentum. ❑ The energy of a single photon (E) is 𝒉𝑪 𝑬 = 𝒉𝒇 = 𝝀 𝒉𝑪 ❑ The photon has momentum (P) is 𝑬= 𝝀 𝒉 𝑬 𝑬 = 𝒎𝒄𝟐 𝑷= = 𝝀 𝒄 𝒑=𝒎𝒄 𝑬 = 𝒑.𝒄 Where h is the Plank constant 𝑬 𝒑= h = 6.63 x 10-34 J. s 𝟏. 𝟎 𝒆𝑽 = 𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 𝑱 𝒄 𝒉 𝒑= 𝝀 𝒉 = 𝟒. 𝟏𝟒 × 𝟏𝟎−𝟏𝟓 𝒆𝑽. 𝒔 Example Yellow light from a sodium vapor lamp has an effective Wavelength of 589 nm. What is the energy of the corresponding photons? 𝑪 = 𝟑𝒙𝟏𝟎𝟖 𝒎/𝒔 𝝀 = 𝟓𝟖𝟗𝒏𝒎, 𝒉 = 𝟒. 𝟏𝟒𝒙𝟏𝟎−𝟏𝟓 𝒆𝑽. 𝒔 𝑬 =? ? ? Solution 𝒉𝑪 𝟒. 𝟏𝟒 × 𝟏𝟎−𝟏𝟓 𝒆𝑽. 𝒔 𝟑 × 𝟏𝟎𝟖 𝒎/𝒔 𝑬= = 𝝀 𝟓𝟖𝟗 × 𝟏𝟎−𝟗 𝒎 𝑬 = 𝟐. 𝟏𝟏 𝒆𝑽 Example During radioactive decay, a certain nucleus emits a gamma ray whose photon energy is 1.35 MeV. (a) to what wavelength does this photon correspond ? (b) what is the momentum of this photon? 𝑪 = 𝟑𝒙𝟏𝟎𝟖 𝒎/𝒔, 𝑬 = 𝟏. 𝟑𝟓𝑴𝒆𝑽, 𝒉 = 𝟒. 𝟏𝟒𝒙𝟏𝟎−𝟏𝟓 𝒆𝑽. 𝒔 𝝀 =? ? ? , 𝑷 =? ? ? Solution 𝒉𝑪 (𝟒. 𝟏𝟒𝒙𝟏𝟎−𝟏𝟓 )(𝟑𝒙𝟏𝟎𝟖 ) −𝟏𝟑 𝒎 𝝀= = = 𝟗. 𝟐𝒙𝟏𝟎 𝑬 𝟏. 𝟑𝟓𝒙𝟏𝟎𝟔 𝑬 𝟏. 𝟑𝟓𝒙𝟏𝟎𝟔 −𝟑 𝒆𝑽. 𝒔/𝒎 𝑷= = = 𝟒. 𝟓𝒙𝟏𝟎 𝑪 𝟑𝒙𝟏𝟎𝟖 Problems Show that the energy E of a photon (in eV) is related to its wavelength λ (in nm) 𝟏𝟐𝟒𝟎 by 𝑬 (𝒆𝑽) = 𝝀(𝒏𝒎) solution 𝒉𝒄 (𝟒. 𝟏𝟒 × 𝟏𝟎−𝟏𝟓 𝒆𝑽. 𝒔)(𝟑 × 𝟏𝟎𝟖 𝒙𝟏𝟎𝟗 𝒏𝒎/𝒔) 𝑬 = 𝒉𝒇 = = 𝝀 𝟏𝟐𝟒𝟎 𝝀 𝑬(𝒆𝑽) = 𝝀(𝒏𝒎) The orange-colored light from a highway sodium lamp has a wavelength of 589 nm. How much energy is possessed by an individual photon from such a lamp? λ = 589 nm solution E=?? 𝟏𝟐𝟒𝟎 𝟏𝟐𝟒𝟎 𝑬= = = 𝟐. 𝟏𝒆𝑽 𝝀(𝒏𝒎) 𝟓𝟖𝟗 Problems A particular x – ray photon has a wavelength of 35 pm. calculate the photon’s (a) energy, (b) frequency, and (c) momentum. λ = 35 pm = 0.035 nm, 𝐡 = 𝟒. 𝟏𝟒𝐱𝟏𝟎−𝟏𝟓 𝐞𝐕. 𝐬 𝐄 =? ? , 𝐟 =? ? , 𝐏 =? ? solution 𝟏𝟐𝟒𝟎 𝟏𝟐𝟒𝟎 𝑬= = = 𝟑𝟓𝟒𝟐𝟖. 𝟔𝒆𝑽 𝝀(𝒏𝒎) 𝟎. 𝟎𝟑𝟓 𝒄 𝟑 × 𝟏𝟎𝟖 𝟏𝟖 𝑯𝒛 𝒇= = = 𝟖. 𝟓𝟕 × 𝟏𝟎 𝝀 𝟑𝟓 × 𝟏𝟎−𝟏𝟐 𝒉 𝟒. 𝟏𝟒 × 𝟏𝟎−𝟏𝟓 −𝟒 𝒆𝑽. 𝒔/𝒎 𝒑= = = 𝟏. 𝟏𝟖𝟑 × 𝟏𝟎 𝝀 𝟑𝟓 × 𝟏𝟎−𝟏𝟐  Photoelectric effect What is the Photoelectric Effect? ❑ When light incident on a metallic surface, the surface can emit electrons (Photoelectrons) ❑ Light energy convert to electric energy. 𝑬𝒑𝒉𝒐𝒕𝒐𝒏 = 𝒉𝒇 > 𝚽 𝑲. 𝑬. 𝜱 Work function 𝑬 𝒑𝒉𝒐𝒕𝒐𝒏 = 𝜱 + 𝑲. 𝑬 ❑ When a beam of light falls on a metal plate in evacuated tube ❑ Photoelectron are ejecting from the metal surface if 𝑬 𝒊𝒏𝒄𝒊𝒅𝒆𝒏𝒕 ≥ 𝚽 work function of the metal (𝛟) ❑ Photoelectrons are collected by the collector (C). ❑ A suitable potential difference (V) between emitter (E) and (C) sweeps up these photoelectrons. ❑ If reverse the potential of circuit and increase no photoelectrons move to the collector (Stopping Potential) Stopping potential (Vs) It is negative voltage at which no photocurrent or photoelectrons not move ❑ At stopping potential (Vs) 𝐌𝐢𝐱𝐦𝐢𝐮𝐦 𝐤𝒊𝒏𝒆𝒕𝒊𝒄 𝒆𝒏𝒆𝒓𝒈𝒚 𝑲𝒎𝒂𝒙 𝒆𝒍𝒄𝒕𝒓𝒊𝒄 𝑷𝒐𝒕𝒆𝒏𝒕𝒊𝒂𝒍 𝒆𝒏𝒆𝒓𝒈𝒚(𝑼 = 𝒒𝑽) 𝑲𝒎𝒂𝒙 = 𝒆𝑽𝒔 𝐧𝐞𝐠𝐚𝐭𝐢𝐯𝐞 (𝐕) Positive (𝑽) 𝟏 ← 𝑲𝒎𝒂𝒙 = 𝒎𝒆 𝝊𝟐 = 𝒆𝑽𝒔 𝟐 𝒎𝒆 = 𝟗. 𝟏𝟏 × 𝟏𝟎−𝟑𝟏 𝒌𝒈 is mass of the electron 𝝊 is the speed of the photoelectron 𝒆 = 𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 𝑪 is charge of the electron Work function (f) ∴ 𝚽 = 𝒉𝒇𝒐 = 𝒉𝒄Τ𝝀𝒐 It is minimum energy needed to remove the electron from surface of metal. 𝒇 ≺ 𝒇𝑶 𝒇 = 𝒇𝑶 𝒇 ≻ 𝒇𝑶 𝒉𝒇 ≤ 𝚽 Cut off frequency (fo) ❑ Emission of electrons from the surface can only occur if 𝒉𝒇 ≥ 𝝋 ❑ Cut-off frequency is the minimum frequency of the incident light needed to emit photoelectrons ∴ 𝚽 = 𝒉𝒇𝒐 = 𝒉𝒄Τ𝝀𝒐 Photoelectric Current & Voltage Graph ❑ The photocurrent increases with intensity of incident light ❑ No current (photocurrent =0 ) flows at −𝐕 ≤ −𝑽𝒔 ❑ Vs is independent of the intensity of incident light –Vs ❑ kinetic energy of photoelectrons increases with frequency of light 𝑲𝒎𝒂𝒙 ∝ 𝑬 = 𝒉𝒇 ❑ kinetic energy 𝑲𝒎𝒂𝒙 = 𝟎 of photoelectrons at cutoff frequency (𝒇𝒐 ) is zero 𝒉𝒇𝒐 = 𝝓 + 𝑲𝒎𝒂𝒙 = 𝟎 Einstein’s Explanation 𝒉𝒄 𝟏𝟐𝟒𝟎 ❑The photon’s energy is 𝑬 = 𝒉𝒇 = = 𝒆𝑽 𝝀 𝝀 (𝒏𝒎) ❑ Each photon gives this energy to an electron in the metal ❑This energy equal to: 𝒉𝒇 = 𝚽 + 𝑲𝒎𝒂𝒙 ❑Where 𝑲𝒎𝒂𝒙 is the maximum kinetic energy of the photoelectron German Physicist (1879–1955) ❑ 𝚽 is called the work function of the metal Nobel Prize in Physics 1921 Einstein Photoelectric Equation 𝐄 = 𝛟 + 𝐊 𝐦𝐚𝐱 Energy of Work function Maximum Kinetic energy incident photon of the metal of ejected electron 𝐜 𝟏𝟐𝟒𝟎 𝐜 𝟏𝟐𝟒𝟎 𝟏 𝐄 = 𝐡𝐟 = 𝐡 = 𝒆𝑽 𝛟 = 𝐡𝐟𝑶 = 𝐡 = 𝒆𝑽 𝐊 𝐦𝐚𝐱 = 𝐦𝐯 𝟐 = 𝐞𝐕𝐬 𝛌 𝝀 (𝒏𝒎) 𝛌𝑶 𝝀𝒄 (𝒏𝒎) 𝟐  𝑲 𝒎𝒂𝒙 = 𝒆𝑽𝒔 𝐞 = 𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 𝑪 is charge of electron when 𝑲 𝒎𝒂𝒙 = 𝒆𝑽𝒔 measure by Joel 𝐞 = 𝟏 is charge of electron when 𝑲 𝒎𝒂𝒙 = 𝒆𝑽𝒔 measure by electron volt (eV) Notice It must be using the unit of 𝑲 𝒎𝒂𝒙 in Joel to calculate the speed of photoelectron 𝑲 𝒎𝒂𝒙 = 𝒆𝑽𝒔 = 𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 𝑪 𝑽𝒔 𝒗𝒐𝒍𝒕 = … … …. 𝑱𝒐𝒆𝒍 = or 𝟏 (𝑬 − 𝜱) 𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 𝑪 = 𝒎𝒆 𝒗𝟐 𝟐 Problem 8 The energy needed to remove an electron from metallic sodium is 2.28 eV. Does sodium show a photoelectric effect for red light, with λ = 680 nm? (ii) what is the cutoff wavelength for photoelectric emission from sodium? c = 3𝑥108 𝑚𝑠 −1 𝜆 = 680 𝑛𝑚, ℎ = 4.14𝑥10−15 𝑒𝑉. 𝑠, Φ = 2.28𝑒𝑉 Does sodium show a photoelectric effect, 𝝀𝑪 = ??? Solution 𝒉𝒄 𝟏𝟐𝟒𝟎 𝟏𝟐𝟒𝟎 𝑬= = = = 𝟏. 𝟖𝟐 𝒆𝑽 < 𝜱 𝝀 𝝀 (𝒏𝒎) 𝟔𝟖𝟎 so, sodium doesn’t show the photoelectric for red light 𝒉𝒄 𝜱 = 𝒉𝒇𝒐 = 𝝀𝒐 𝒉𝒄 𝟏𝟐𝟒𝟎 𝝀𝒐 = = = 𝟓𝟒𝟑. 𝟖𝟔𝒏𝒎 𝜱 𝟐. 𝟐𝟖 Problem 9 Find the maximum kinetic energy of photoelectrons if the work function of the material is 2.3 eV and the frequency of the radiation is 3 × 1015 𝐻𝑧. 𝒇 = 𝟑 × 𝟏𝟎𝟏𝟓 𝑯𝒛, 𝒉 = 𝟒. 𝟏𝟒𝒙𝟏𝟎−𝟏𝟓 𝒆𝑽. 𝒔, 𝚽 = 𝟐. 𝟑𝒆𝑽 𝑲𝒎𝒂𝒙 =? ? ? Solution 𝑬 𝒑𝒉𝒐𝒕𝒐𝒏 = 𝜱 + 𝑲𝒎𝒂𝒙 𝑲𝒎𝒂𝒙 = 𝒉𝒇 − 𝜱 𝑲𝒎𝒂𝒙 = = 𝟒. 𝟏𝟒 × 𝟏𝟎−𝟏𝟓 𝟑 × 𝟏𝟎𝟏𝟓 − 𝟐. 𝟑 𝑲𝒎𝒂𝒙 = 𝟏𝟎. 𝟏𝟎𝟖 𝒆𝑽 Problem 11 The work function of tungsten is 4.5 eV. Calculate the speed of the fastest of the photoelectrons emitted when photons of energy 5.8 eV are incident on a sheet of tungsten. 𝑬 = 𝟓. 𝟖𝒆𝑽, 𝜱 = 𝟒. 𝟓𝒆𝑽 𝒎 = 𝟗. 𝟏𝟏 × 𝟏𝟎−𝟑𝟏 𝝊 =? ? ? Solution 𝟏 𝟏 𝑬 − 𝜱 = 𝒎𝒆 𝒗𝟐 𝑬 = 𝜱 + 𝑲𝒎𝒂𝒙 𝑬 = 𝜱 + 𝒎𝒆 𝒗𝟐 𝟐 𝟐 𝟐(𝑬 − 𝜱) 𝟐 𝟓. 𝟖 − 𝟒. 𝟓 × 𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 𝑱 𝒗= ∴𝒗= 𝒎𝒆 𝟗. 𝟏𝟏 × 𝟏𝟎−𝟑𝟏 𝒌𝒈 𝒗 = 𝟔. 𝟕𝟔 × 𝟏𝟎𝟓 𝒎/𝒔 Problem 12 Light of wavelength 200 nm falls on an aluminum surface. In Aluminum 4.2 eV are required to remove an electron. What is the kinetic energy of (a) the fastest and (b) the slowest emitted photoelectrons? (c) what is the stopping potential? (d) what is the cutoff wavelength for aluminum? 𝝀 = 𝟐𝟎𝟎𝒏𝒎, 𝒉 = 𝟒. 𝟏𝟒𝒙𝟏𝟎−𝟏𝟓 𝒆𝑽. 𝒔, 𝜱 = 𝟒. 𝟐𝒆𝑽 𝑲𝒎𝒂𝒙 , 𝑲𝒔 , 𝑽𝒔 , 𝝀𝒐 = ? ? ? Solution 𝒉𝒄 𝒉𝒄 ∴ 𝑬 = 𝜱 + 𝑲𝒎𝒂𝒙 ∴ = 𝜱 + 𝑲𝒎𝒂𝒙 ∴ 𝑲𝒎𝒂𝒙 = −𝜱 𝝀 𝝀 (a) 𝟏𝟐𝟒𝟎 ∴ 𝑲𝒎𝒂𝒙 = − 𝟒. 𝟐 = 𝟐. 𝟎 𝒆𝑽 𝟐𝟎𝟎 (b) ∴ 𝑲. 𝑬 𝒔𝒍𝒐𝒘𝒆𝒔𝒕 = 𝒁𝒆𝒓𝒐 (c) ∴ 𝑲𝒎𝒂𝒙 = 𝒆𝑽𝒔 ∴ 𝟐 (𝒆𝑽) = 𝒆 × 𝑽𝒔 ∴ 𝑽𝒔 = 𝟐 𝒗𝒐𝒍𝒕 (d) 𝒉𝒄 𝒉𝒄 𝟏𝟐𝟒𝟎 𝜱= 𝝀𝒐 = = = 𝟐𝟗𝟓. 𝟐𝟒𝒏𝒎 𝝀𝒐 𝜱 𝟒. 𝟐 Problem 13 (a)If the work function for a metal is 1.8 eV, what would be the stopping potential for light having a wavelength of 400 nm? (b) what would be the maximum speed of the emitted photoelectrons at the metal’s surface? 𝝀 = 𝟒𝟎𝟎𝒏𝒎, 𝒉 = 𝟒. 𝟏𝟒𝒙𝟏𝟎−𝟏𝟓 𝒆𝑽. 𝒔, 𝜱 = 𝟏. 𝟖 𝒆𝑽 𝒎 = 𝟗. 𝟏𝟏𝒙𝟏𝟎−𝟑𝟏 𝒌𝒈 𝑽𝒔 , 𝝊𝒎 = ? ? Solution 𝟏𝟐𝟒𝟎 ∴ 𝑬 = 𝜱 + 𝑲𝒎𝒂𝒙 ∴ = 𝟏. 𝟖 + 𝑲𝒎𝒂𝒙 ∴ 𝑲𝒎𝒂𝒙 = 𝒆 𝑽𝒔 = 𝟏. 𝟑 𝒆𝑽 𝟒𝟎𝟎 𝑽𝒔 = 𝟏. 𝟑 𝒗𝒐𝒍𝒕 𝒉𝒄 𝟏 𝑲𝒎𝒂𝒙 = − 𝜱 = 𝒎𝒆 𝒗𝟐 𝝀 𝟐 𝟐 𝑲𝒎𝒂𝒙 𝟐 × 𝟏. 𝟑 × 𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 𝑱 𝒗= = 𝒎𝒆 𝟗. 𝟏𝟏 × 𝟏𝟎−𝟑𝟏 𝒌𝒈 𝒗 = 𝟔. 𝟕𝟔 × 𝟏𝟎𝟓 𝒎/𝒔 Problem 15 The stopping potential for photoelectrons emitted from a metal surface illuminated by light of wavelength 491 nm is 0.71 V. When the incident wavelength is changed to a new value, the stopping potential is found to be 1.43 V. (a) what is his new wavelength? (b) what is the work function for the surface? 𝝀𝟏 = 𝟒𝟗𝟏𝒏𝒎, 𝒉 = 𝟒. 𝟏𝟒𝒙𝟏𝟎−𝟏𝟓 𝒆𝑽. 𝒔, 𝑽𝒔𝟏 = 𝟎. 𝟕𝟏𝑽, 𝑽𝒔𝟐 = 𝟏. 𝟒𝟑𝑽 𝝀𝟐 , 𝜱 = ? ? Solution 𝒉𝒄 𝒉𝒄 𝟏𝟐𝟒𝟎 = 𝜱 + 𝒆𝑽𝒔𝟏 → 𝜱= − 𝒆𝑽𝒔𝟏 = − 𝟏 × 𝟎. 𝟕𝟏 = 𝟏. 𝟖𝟕 𝒆𝑽 𝝀𝟏 𝝀𝟏 𝟒𝟖𝟏 𝒉𝒄 𝒉𝒄 𝟏𝟐𝟒𝟎 ∴ = 𝜱 + 𝒆𝑽𝒔𝟐 → 𝝀𝟐 = = = 𝟑𝟕𝟔. 𝟎𝟔 𝒏𝒎 𝝀𝟐 𝜱 + 𝒆𝑽𝒔𝟐 𝟏. 𝟖𝟕 + 𝟏 × 𝟏. 𝟒𝟑  Compton Effect What is Compton Effect? ❑ When X-ray photon (l) collides with a target of graphite, the electrons ejected from the atm or molecule with scattered photon of wavelength (𝝀𝒔𝒄 ). ❑ The scattered photon has longer a wavelength (𝝀𝒔𝒄 ) than the incident wavelength, ❑ The shift in wavelengths (Dl = lsc-l) was measured experimentally. Experimental Results ✓ Wavelength of scattered photon found to be different from that of incident photon and it depends on detection angle f: ✓ The shift in wavelengths (Dl = lsc-l) increase with angle of detecting. Explanation of wave model; The incident photon causes the electron to oscillate with the same frequency. The electron is charged particle, so its oscillation radiates an electromagnetic wave with the same frequency. So, there is no shift in frequency or wavelength Dl = lsc-l = 0 These result is opposite of experimental observation Compton explanation (Photon model) o X-ray incident photon has energy (𝑬 = 𝒉 𝒄/𝝀) and momentum (𝑷 = 𝒉/𝝀) colliding with free electron o X-ray photon gives part of its energy to electron, in the form of kinetic energy 𝑬 = 𝒎𝒄𝟐 o The rest part of X-ray photon energy is converted to scattered photon of energy 𝑬 = 𝒉 𝒄/𝝀𝒔𝒄 𝝀𝒔𝒄 > 𝝀 o So, the shift in wavelength is (Dl) 𝚫𝝀 = 𝝀𝒔𝒄 − 𝝀 Compton equation 𝒉 𝜟𝝀 = 𝝀𝒔 − 𝝀 = 𝟏 − 𝐜𝐨𝐬 𝝋 = 𝝀𝒄 𝟏 − 𝐜𝐨𝐬 𝝋 𝒎𝑪 ∆λ is called Compton shift & m is the mass of the electron λc is called Compton wavelength = 2.43 pm Grazing collision when (𝝋 = 𝟎) the incident photon pass without any scattering ( ∆𝝀 = 𝟎) Head on collision when (𝝋 = 𝟏𝟖𝟎°) the incident photon reverse its direction 𝜟𝝀 = 𝝀𝒄 𝟏 − 𝑪𝒐𝒔 𝟏𝟖𝟎 = 𝟐𝝀𝒄 Derivation of Compton equation Apply energy and linear momentum conservation rule between the photon and electron. i) Energy conservation equation. 𝒉𝑪 𝒉𝑪 𝟐 𝟏 = + 𝒎𝒄 𝜸 − 𝟏 ∵ 𝜸 = 𝝀 𝝀𝒔 𝒗𝟐 𝟏− 𝟐 𝑪 𝒉 𝒉 ∴ = +𝒎𝒄 𝜸 −𝟏 (𝟏) 𝝀 𝝀𝒔 𝜸𝒎𝒗 𝐬𝐢𝐧 𝜽 ii) The momentum equation 𝒉 𝒉 X – components: = 𝐜𝐨𝐬 𝝋 + 𝜸𝒎𝒗 𝐜𝐨𝐬 𝜽 (𝟐) 𝝀 𝝀𝒔 𝜸𝒎𝒗 𝐜𝐨𝐬 𝜽 𝒉 𝒉 Y – components: 𝐬𝐢𝐧 𝝋 = 𝜸𝒎𝒗 𝐬𝐢𝐧 𝜽 (𝟑) 𝒉 𝝀𝒔 𝐜𝐨𝐬 𝝋 𝝀𝒔 𝝀 From equations 1, 2, & 3 𝒉 𝒉 𝝀𝒔 𝜟𝝀 = 𝝀𝒔 − 𝝀 = 𝟏 − 𝐜𝐨𝐬 𝝋 𝝀𝒔 𝐬𝐢𝐧 𝝋 𝒎𝒄 Example 6 X ray of wavelength 22 pm (photon energy = 56 keV) are scattered from a carbon target, radiation being viewed at 85o to the incident beam. (a) what is the Compton shift? (b) what percentage of this initial energy does an incident x ray photon loose? 𝝀 = 𝟐𝟐𝒑𝒎, 𝑬 = 𝟓𝟔𝒌𝒆𝑽, 𝝀𝒄 = 𝟐. 𝟒𝟑𝒑𝒎, 𝜱 = 𝟖𝟓° 𝑬 − 𝑬𝑺 𝜟𝝀, =? ? 𝑬 Solution (a) 𝜟𝝀 = 𝝀𝒄 𝟏 − 𝐜𝐨𝐬 𝝋 = 𝟐. 𝟒𝟑 𝟏 − 𝐜𝐨𝐬 𝟖 𝟓° = 𝟐. 𝟐𝟏 𝒑𝒎 𝑪 𝑪 𝑬−𝑬𝑺 𝒉𝒇−𝒉𝒇𝒔 − 𝝀 𝝀𝒔 𝝀𝒔 −𝝀 𝝀𝒔 −𝝀 𝒃 = = =𝝀 × = 𝑬 𝒉𝒇 𝑪 𝝀𝝀𝒔 𝝀𝒔 𝝀 ∴ ∆𝝀 = 𝝀𝒔 − 𝝀 ∴ 𝝀𝒔 = ∆𝝀 + 𝝀 𝑬 − 𝑬𝑺 𝝀𝒔 − 𝝀 𝜟𝝀 𝟐. 𝟐𝟏 ∴ = = = = 𝟎. 𝟎𝟗𝟏 = 𝟗. 𝟏% 𝑬 𝝀𝒔 𝝀 + 𝜟𝝀 𝟐𝟐 + 𝟐. 𝟐𝟏 Problem 17 Photons of wavelength 2.4 pm are incident on a target containing free electrons. (a) find the wavelength of a photon that is scattered at 30 ͦ from the incident direction (b) do the same for a scattering angle of 120ͦ. λ = 2.4 pm, λc = 2.43 pm, 𝝋 = 𝟑𝟎° ͦ, ͦ 𝝋 = 𝟏𝟐𝟎° λs1, λs2 solution 𝜟𝝀 = 𝝀𝒔 − 𝝀 = 𝝀𝒄 𝟏 − 𝐜𝐨𝐬 𝝋 𝝀𝒔𝟏 − 𝟐. 𝟒 = 𝟐. 𝟒𝟑 𝟏 − 𝑪𝒐𝒔𝟑𝟎𝝄 ⇒ 𝝀𝒔𝟏 = 𝟐. 𝟕𝟑𝒑𝒎 𝝀𝒔𝟐 − 𝟐. 𝟒 = 𝟐. 𝟒𝟑 𝟏 − 𝑪𝒐𝒔𝟏𝟐𝟎𝝄 ⇒ 𝝀𝒔𝟐 = 𝟔. 𝟎𝟒𝒑𝒎 Problem 17 A 0.511 MeV gamma ray photon is Compton scattered from a free electron in an aluminum block (a) what is the wavelength of the incident photon? (b) what is the wavelength of the scattered photon? (c) what is the energy of the scattered photon? Assume a scattering angle of 90o E = 0.511 MeV, λc = 2.43 pm, 𝝋 = 𝟗𝟎° Req: λ, λs , Es Solution 𝟏𝟐𝟒𝟎 𝟏𝟐𝟒𝟎 𝑬 𝒆𝑽 = = 𝟎. 𝟓𝟏𝟏 × 𝟏𝟎𝟔 ∴ 𝝀 𝒏𝒎 = = 𝟐. 𝟒𝒙𝟏𝟎 −𝟑 𝒏𝒎 𝝀 𝒏𝒎 𝟎. 𝟓𝟏𝟏 × 𝟏𝟎𝟔 𝝀 = 𝟐. 𝟒𝒙𝟏𝟎−𝟑 𝒏𝒎 = 𝟐. 𝟒𝒑𝒎 𝜟𝝀 = 𝝀𝒔 − 𝝀 = 𝝀𝒄 𝟏 − 𝐜𝐨𝐬 𝝋 𝜟𝝀 = 𝝀𝒔 − 𝟐. 𝟒 = 𝟐. 𝟒𝟑 𝟏 − 𝑪𝒐𝒔𝟗𝟎𝝄 ⇒ ∴ 𝝀𝒔 = 𝟒. 𝟖𝟑𝒑𝒎 𝟏𝟐𝟒𝟎 𝟏𝟐𝟒𝟎 𝟔 𝑬𝒔 = = −𝟑 = 𝟎. 𝟐𝟔 × 𝟏𝟎 𝒆𝑽 = 𝟎. 𝟐𝟔𝑴𝒆𝑽 𝝀𝒔 (𝒏𝒎) 𝟒. 𝟖𝟑 × 𝟏𝟎  𝒉𝒄 𝝀 𝒎 𝒄𝟐 (𝜸 − 𝟏 ) 𝟏 𝜸= 𝒉𝒄 𝒗𝟐 𝟏− 𝝀𝒔 𝑪𝟐