1. Estimate the peak wavelength for radiation from (a) ice at 0 C, (b) a floodlamp at. emission. In what region of the EM spectrum is each?

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1 Physics AP Quantum Worksheet 1. Estimate the peak wavelength for radiation from (a) ice at 0 C, (b) a floodlamp at 3500 K, (c) helium at 4 K, (d) for the universe at T K, assuming blackbody emission. In what region of the EM spectrum is each? 2. What is the energy range (in joules and ev) of photons in the visible spectrum, of wavelength 400 nm to 750 nm? 3. About 0.1 ev is required to break a hydrogen bond in a protein molecule. Calculate the minimum frequency and maximum wavelength of a photon that can accomplish this. 4. What is the momentum of a λ nm X-ray photon? 5. What minimum frequency of light is needed to eject electrons from a metal whose 19 work function is J? 6. What is the maximum kinetic energy of electrons ejected from barium ( W ev) when illuminated by white light, λ 400 to 750 nm? 7. When UV light of wavelength 285 nm falls on a metal surface, the maximum kinetic energy of emitted electrons is 1.40 ev. What is the work function of the metal? 8. What is the longest wavelength photon that could produce a proton antiproton pair? 27 (Each has a mass of kg. ) 5 9. An electron and a positron, each moving at m s, collide head on, disappear, and produce two photons, each with the same energy and momentum moving in opposite directions. What is the energy and momentum of each photon? 10. Calculate the wavelength of a 0.23-kg ball traveling at 0.10 m s. 11. Through how many volts of potential difference must an electron be accelerated to achieve a wavelength of 0.24 nm?

2 An electron has a de Broglie wavelength λ m. (a) What is its momentum? (b) What is its speed? (c) What voltage was needed to accelerate it to this speed? 13. How much energy is needed to ionize a hydrogen atom in the n 2 state? 14. Calculate the ionization energy of doubly ionized lithium, Li 2+, which has Z 3. For atoms with z>1, one could use the following formula to calculate the total atomic energy E n z2 ( 13.6eV ) n 2, where z proton number, and n the energy level. 15. What wavelength photon would be required to ionize a hydrogen atom in the ground state and give the ejected electron a kinetic energy of 10.0 ev? 16. If a 100-W lightbulb emits 3.0% of the input energy as visible light (average wavelength 550 nm) uniformly in all directions, estimate how many photons per second of visible light will strike the pupil (4.0 mm diameter) of the eye of an observer 1.0 km away. 17. By what fraction does the mass of an H atom decrease when it makes an n 3 to n 1 transition? 18. In a particular photoelectric experiment, a stopping potential of 2.10 V is measured when ultraviolet light of wavelength 290 nm is incident on the metal. Using the same setup, what will the new stopping potential be if blue light of wavelength 440 nm is used, instead?

3 Solutions 1. (a) We find the peak wavelength from λ P m K T m K 273 K This wavelength is in the near infrared. (b) We find the peak wavelength from λ P m K T m K 3500 K This wavelength is in the infrared. (c) We find the peak wavelength from λ P m K T m K 4 K This wavelength is in the microwave region. (d) We find the peak wavelength from λ P m K T m K K This wavelength is in the microwave region m 10.6µm m 829 nm m 0.73 mm m 1.06 cm. 2. The energy of the photons with wavelengths at the ends of the visible spectrum are E 1 hf 1 hc λ 1 E 2 hf 2 hc λ 2 ( Jis) m / s ( m) ( Jis) m / s ( m) J; J. Thus the range of energies is J < E < J or 1.7eV < E < 3.1eV.

4 3. We find the minimum frequency from E min hf min ; ( 0.1 ev )( J ev ) J s The maximum wavelength is λ max c f min ( m s) ( Hz) m. 4. The momentum of the photon is kg m s. p h λ J s m f min, which gives f min Hz. 5. At the threshold frequency, the kinetic energy of the photoelectrons is zero, so we have KE hf W 0; hf min W 0 ; ( J s) f min J, which gives f min Hz. 6. The photon of visible light with the maximum energy has the minimum wavelength: hf max hc λ min ( Jis) m / s ( J/eV )( m) The maximum kinetic energy of the photoelectrons is 3.11 ev. KE max hf W ev 2.48 ev 0.63 ev. 7. The energy of the photon is hf hc λ ( m / s) ( m) Jis J/eV We find the work function from KE max hf W 0 ; 4.36 ev ev 4.36 ev W 0, which gives W ev.

5 8. The photon with the longest wavelength has the minimum energy to create the masses: hf min hc λ max 2m 0 c 2 ; ( J s) λ max 2( kg) ( m s), which gives λ max m. 9. Since an electron and a positron have identical masses, the energy of each photon will consist of the total energy from one electron or positron, as given by E m 0 c2 1 v2 c 2. For v< 0.001c, this is essentially identical to! 0.51 MeV $ E m 0 c 2 # " c 2 & % c MeV. The momentum of each photon is p E 0.51 MeV. c c 10. We find the wavelength from ( λ h p h mv J s) ( 0.23 kg) 0.10 m s 11. We find the speed from m. λ h p h mv ; m J s ( kg)v, which gives v m s. Because this is much less than c, we can use the classical expression for the kinetic energy. The kinetic energy is equal to the potential energy change: ev KE 1 2 mv ( kg) ( m s) J 26.2 ev. Thus the required potential difference is 26 V.

6 12. (a) We find the momentum from p h λ J s m kg m s. (b) We find the speed from λ h p h mv ; m J s ( kg)v, which gives v m s. (c) With v< 0.005c, we can calculate KE classically: KE 1 2 mv2 equals ( m s) J, which converted to electron-volts kg ( ) ev. This is the energy gained by an electron as it is ( J ev ) accelerated through a potential difference of 6.0 V. 13 To ionize the atom means removing the electron, or raising it to zero energy: ( 13.6 ev ) ( 13.6 ev ) E ion 0 E n 3.4 ev. n Doubly ionized lithium is like hydrogen, except that there are three positive charges ( Z 3) in the nucleus. The square of the product of the positive and negative charges appears in the energy term for the energy levels. We can use the results for hydrogen, if we replace e 2 by Ze 2 : E n Z ( ev ) 3 n ev n 2 ( 122 ev ). n 2 We find the energy needed to remove the remaining electron from " % $ 122 ev ' E 0 E 1 0 $ ( 1) 2 ' 122 ev. # $ & '

7 15. The energy of the photon is hf E ion + KE 13.6 ev ev 23.6eV. We find the wavelength from λ hc hf ( m / s) ( 23.6 ev) Jis J/eV 16. The energy of the photon is hf hc λ ( m/s) ( m) Jis m 52.7nm J. Because the light radiates uniformly, the intensity at a distance L is I P, so the rate at which energy enters the pupil is 2 4πL E t I πr 2 Pr 2 4L. 2 Thus the rate at which photons enter the pupil is n t! " # E t hf $ % & Pr 2 4L 2 hf 2 2 ( J) ( 100 W) m m photons s. 17. The decrease in mass occurs because a photon has been emitted: Δm m 0 " ΔE % $ # c 2 ' & m 0 ΔE m 0 c 2 )" 1% ( 13.6 ev ) $ # 1 2 & ' " 1 %, + $ # 3 2 '. * + & ev

8 18. The stopping potential is the voltage that gives a potential energy change equal to the maximum kinetic energy: KE max ev 0 hc λ W 0 ; J s ( 1 e) ( 2.10 V) J ev ( m s) m W 0, which gives W J. With the 440-nm light, then, ( m s) ev J s m J J J 0.64 ev J ev The potential difference needed to cancel an electron kinetic energy of 0.64 ev is 0.64 V.