1. Position: # (your seat #)……………………………………….. Your Name: Last, First………………………………………… Name of Workshop Instructor: Last, First…………………….. Exam: # (#3)…………………………………………………… Date: April xx, 200x………………………………………… Q1 (3 points). An ordinary incandescent lightbulb emits: A. Polarized lightB. Partially polarized light C. Unpolarized light D. Polarized and unpolarized light Please choose the correct answer and explain it briefly. Answer: C is correct. 123 4 5679 101112 Use the following constants, if necessary: 1 eV=1.6  10 -19 J; m e =9.1  10 -31 Kg; E 1 =-13.6 eV c=3  10 8 m/s; h=6.62  10 -34 Js; r 1 =0.529  10 -10 m 1 nm=10 -9 m R=1.097  10 7 m -1 131415

2. Q3 (4 points). The photon energy is proportional to its A. FrequencyB. Wavelength C. Period Please choose the correct answer and explain it briefly. Answer: A. is correct. Q2 (3 points). What is the photoelectric effect? A.The phenomenon that when light shines on a metal surface, photons are emitted B.The phenomenon that when light shines on a metal surface, nuclei are emitted C. The phenomenon that when light shines on a metal surface, electrons are emitted Please choose the correct answer and explain it briefly. Answer: C is correct.

3. Q4 (10 points). A light ray is incident on a potassium surface. The work function W 0 for potassium is 2 eV. A.If the light is red, with a wavelength of 700 nm, are any electrons emitted? If so, what is their maximum kinetic energy? What is their maximum speed? B. The same as part A, but replace the red light by green light with a wavelength of 550 nm. C. The same as part A, but replace the red light by blue light with a wavelength of 400 nm. Answer: eV is smaller than W 0 for potassium. A.No electrons will be emitted if the red light is shinned on a potassium surface B. E=2.25 eV is greater than W 0 for potassium The maximum kinetic energy is 2.25-2= 0.25 eV E kinetic =m e v 2 /2=0.25 eV Maximum speed is v max =2.96×10 5 m/s C. E= 3.1 eV E kinetic =1.1 eV v max =6.22×10 5 m/s

4. Q6 (10 points). An X-ray flux consists of photons with a wavelength of 0.3 nm. How many of the X-ray photons must strike a surface in one second to deliver a power of 1×10 -12 W? A. 1B. 1510 C. 20000D. 4×10 8 Reminder: Power is Energy divided by Time. Answer: B is correct. Power is defined as P=E/t, therefore energy E=P×t=10 -12 Ws= 10 -12 J The number of photons is Q5 (7 points). What is the energy in eV of an UV photon of wavelength λ = 300 nm? A. 4.1×10 -2 eVB. 4.1 eV C. 41 eVD. 410 eV Answer: =6.62×10 -19 J Since 1 eV=1.6×10 -19 eV E = 4.1 eV B is correct.

5. Q8 (12 points). What is the maximum energy for an electron bound to the nucleus in the Bohr model? A. – infinityB. +infinity C. -13.6 eVD. 0 Answer: D is correct. Q7 (3 points). What is the wavelength, λ, of an electron traveling at 2 x 10 6 m/s? A. 2  10 -5 m B. 3.6  10 -8 m C. 3.6  10 -10 m D. 2  10 -25 m Answer: C is correct.

6. Q10 (12 points). What state (n = ?) would the Bohr atom be in to have a diameter of 1 mm? A. 124B. 2700 C. 3074D. 4348 Answer: r n =n 2 r 1 Where r 1 =0.529×10 -10 m r n =0.5×10 -3 m n 2 =r n /r 1 =(0.5×10 -3 m)/(0.529×10 -10 m)=0.945×10 7 n=3.074×10 3 =3074 C is correct Q9 (9 points). How much energy in Joules does it take to excite a hydrogen atom into its first excited state? The energy corresponding to the ground state is -13.6 eV. A. 10.2 JB. 1.6  10 -19 J C. 1.63  10 -18 J D. 3.4 eV Answer: E 1 =-13.6 eV E 2 =-E 1 /2 2 =-E 1 /4=-3.4 eV E=E 2 -E 1 =-3.4 eV - (-13.6 eV)=10.2 eV Since 1 eV=1.6×10 -19 J, E=16.32×10 -19 J=1.6×10 -18 J Therefore, C is correct.

7. Q11 (10 points). What is the shortest wavelength in the Balmer series? For the Balmer series, the lines are transitions from the level n=2 to the next excited states, namely n=3, 4, 5, 6. A. 0.15  10 -6 mB. 0.36  10 -6 m C. 0.41  10 -6 mD. 0.66  10 -6 m Q12 (7 points). Which line in the figure represents the binding energy? Here, r is the separation between nuclei and the vertical axis is the atom-atom potential energy A) DB) E C) F Answer: Line D is the binding energy, as this corresponds to a minimum of potential energy. Therefore, A is correct. Answer: Short wavelength is large energy. Therefore 1/ =R(1/2 2 -1/6 2 )=1.097×10 7 ×(0.250-0.027)=0.244×10 7 m -1 Therefore, =4.1×10 -7 m=0.41×10 -6 m C is correct.

8. Q 13 (7 points). Which line in the below figure represents a region of attraction? A) AB) B C) C Q 14 (5 points). What is the difference between a covalent bond and an ionic bond? Answer: 1. A covalent bond is formed when the electrons are shared between atoms. 2. An ionic bond is formed when the electrons are transferred from one atom to the other. Q 15 (5 points). Which of the following is a very weak interaction? A. CovalentB. Ionic C. HydrogenD. van der Waals Please explain your answer. Answer: D is correct. The van der Waals forces are very weak (-0.5 to -1 kcal/mol). Answer: Line B represents a region of attraction, as the potential energy increases with the distance. Therefore, B is correct.