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Name Apr. 27, 2006
Phys 2020, NSCC Exam #3 — Spring 2006
- (12)
MC (16)
Total (102) (!)
Multiple Choice
Choose the best answer from among the four!
- The purpose of the corrective lenses of a farsighted person (trouble seeing near objects) is to
a) Take an object at about 25 cm and make an image at the near point. b) Take an object at the near point and make an image at infinity. c) Take an object at infinity and make an image at the near point. d) Take an object at the near point and make an image at about 25 cm.
- In a single–slit diffraction pattern, the fringes on the screen move farther apart if
a) The wavelength of the light is decreased or the width of the slit is decreased. b) The wavelength of the light is decreased or the width of the slit is increased. c) The wavelength of the light is increased or the width of the slit is decreased. d) The wavelength of the light is increased or the width of the slit is increased.
- Einstein successfully explained the photoelectric effect using the idea of
a) The wave nature of “matter”. b) The quantization of angular momentum. c) The particle nature of light. d) The uncertainty principle.
- When we discuss particle waves, we understand that we are talking about
a) The oscillatory motion of atomic particles. b) Waves of probability. c) The fact that the particles are not points but are “smeared out” in space. d) Particles represent disturbances in some background quantum medium.
- Which of the following sets of qunatum numbers for an electron in the quantum- mechanical H atom is impossible?
a) n = 3, l = 2, ml = 3 b) n = 5, l = 0, ml = 0 c) n = 2, l = 1, ml = − 1 d) n = 4, l = 3, ml = − 3
- Roughly speaking, the size of the nucleus of an atom is about.
a) 10 times larger than the “orbits” of the electrons. b) About 10 times smaller than the orbits of the electrons. c) About 1000 times smaller than the orbits of the electrons. d) About 100,000 times smaller than the orbits of the electrons.
- How do the number of protons and neutrons compare in stable nuclei?
a) For small nuclei there are more protons, while for large nuclei they are about equal. b) For small nuclei there are more neutrons, while for large nuclei they are about equal. c) For small nuclei they are about equal, while for large nuclei there are more protons. d) For small nuclei they are about equal, while for large nuclei there are more neutrons.
- An object sits in front of a converging lens, as shown here; it is between the focal point and the lens. Complete a ray diagram showing the formation of the image. (8)
F F
image object
(Rays added to diagram and extended backwards to meet at a point; this gives the location of the image.)
Is the image on the left or right side of the lens? Is it a real image or a virtual image? (2)
The image is on the left side of the lens. It is a virtual image.
- Suppose an object with a height of 3.0 cm is 30.0 cm in front of a lens with a focal length of 18.0 cm.
Find the location of the image (say if the image is on the left or right side of the lens.) Find the height of the image. Say if the image is upright or inverted and if the image is real or virtual. (9)
Use the lens equation to get di:
1 di
f
do
18 .0 cm
30 .0 cm
= 0.022 cm−^1 =⇒ di = 45.0 cm
This says that the image is on the right (opposite) side of the lens and so the image is real. The magnification is
m = −
di do
hi ho
then the height of the image is
hi = (− 1 .5)(3.0 cm) = − 4 .5 cm
The image is inverted.
- A certain nearsighted person cannot properly focus objects which are farther than 3.0 m from the eye. Give a brief description of how a lens can correct this problem (say something about objects with infinite distance...) and find the focal length of the lens which will correct the problem. (8)
Nearsighted person would like the see objects at infinity so the lens must take an object at infinity and make an image at the far point (so that di = − 3 .0 m) where the person can deal with an object. Putting these values into the lens equation,
1 ∞
(− 3 .0 m)
f
3 .0 m
f
=⇒ f = − 3 .0 m
and so the lens must have a focal length of − 3 .0 m (it is thus a diverging lens).
2.00 m
1.00 mm
- In an (idealized) Young two–slit experiment, coherent light of wavelength 690 nm is incident on the (thin) slits and form an interference pattern on a screen 2.0 m from the slits and parallel to their plane. The distance between any two adja- cent bright fringes is 1.00 mm. Find the separation of the two slits. (6)
The angle to the first-order bright fringe is
tan θ =
1. 00 × 10 −^3
=⇒ θ = 2. 86 × 10 −^2 deg
For the first-order bright fringe,
sin θ = (1)
λ d then we get
d =
λ sin θ
(690 × 10 −^9 m sin θ
= 1. 38 × 10 −^3 m = 1.38 mm
- A photoelectric effect experiment is performed with a certain metal which has a work function of 1.80 eV. a) What is the maximum wavelength of light which can eject electrons from this metal? (5)
The photon must have at least an energy equal to the work function, so its energy would be
E = hf = 1.80 eV = 2. 88 × 10 −^19 J
This gives
f =
E
h
= 4. 35 × 1014 Hz
for which the wavelength is
λ =
c f
- 00 × 10 8 m s
- 35 × 1014 Hz
= 6. 88 × 10 −^7 m = 688 nm
b) Suppose the metal is illuminated with light of wavelength of 440 nm. What is the maxi- mum kinetic energy of the ejected electrons? (5)
The energy of these photons would be
E = hf = h
c λ
= (6. 626 × 10 −^34 J · s)
- 00 × 10 8 m s 440 × 10 −^9 m
= 4. 5 × 10 −^19 J = 2.82 eV
then the maximum kinetic energy of the electrons would be
KEmax = hf − W 0 = (2.82 eV) − (1.80 eV) = 1.02 eV
9.a) When the electron in a hydrogen atom makes a transition from the n = 4 orbit to the n = 3 orbit, what is the wavelength of the photon which is emitted? (6)
λ
= R
( 1 9
) = 5. 33 × 105 m−^1 =⇒ λ = 1. 88 × 10 −^6 m
b) What is the energy of that photon in eV? (3)
E = hf = h
c λ
= (6. 626 × 10 −^34 J · s)
(3. 00 × 10 8 m s
- 88 × 10 −^6 m
= 1. 06 × 10 −^19 J = 0.661 eV
- Calculate the total binding energy and the binding energy per nucleon for the nucleus 20 10 Ne. The^ 20 10 Ne^ atom^ has a mass of 19.992434 u (10)
∆m = 10(mH )+10(mn)− 19 .992434 u = 10(1.0078 u)+10(1.0087 u)− 19 .992434 u = 0.172 u
Then the total binding energy is the energy-equivalent of this mass,
Ebind = ∆mc^2 = (0.172 u)c^2 = (0.172)(931.5 MeV) = 160 MeV
and the binding energy per nucleon is
BE A
160 MeV 20
= 8.03 MeV
You must show all your work and include the right units with your answers!
V = IR Rser = R 1 + R 2 + · · ·
Rpar
R 1
R 2
+ · · · P = V I = I^2 R
λf = v v =
c n
c = 3. 00 × 10 8 m s n 1 sin θ 1 = n 2 sin θ 2
do
di
f
m = −
di do
hi ho
Interf: sin θbr = m
λ d
sin θdk = (m + 12 )
λ d
Diff: sin θdk = m
λ a h = 6. 626 × 10 −^34 J · s E = hf hf = KEmax + W 0 1 eV = 1. 602 × 10 −^19 J 1 λ
= R
( 1 n^2 f
n^2 i
) R = 1. 097 × 107 m−^1 En = −(13.6 eV)
Z^2
n^2
rn = (5. 29 × 10 −^11 m)
n^2 Z^2
A = Z+N mH = 1.0078 u mn = 1.0087 u (1 u)c^2 = 931.5 MeV Ebind = ∆mc^2
