Unit 05 Photoelectric Effect Prob - pdf

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PHOTOELECTRIC EFFECT
SUPPLEMENTARY QUESTIONS - UNIT 05
______________________________________________________________
Answer the following questions.
Be sure to show all work, set-ups, units, etc. neatly and completely.
1. A metal had a threshold frequency of 5.8 x 1014 sec−1 for the photoelectric effect. Predict the effect (increase,
decrease, or remain the same) of each of the following changes on (i) the number of electrons emitted and (ii)
the kinetic energy of each emitted electron.
a. The frequency of radiation striking the metal increases from 5.2 x 1014 to 5.6 x 1014 sec−1. Assume
constant intensity.
(i) same - it is ZERO
(ii) same - it is ZERO
b. The frequency of radiation striking the metal is decreased from 6.2 x 1014 to 5.9 x 1014 sec−1. Assume
constant intensity.
(i) same - same intensity
(ii) decreases - less energetic light
c. The metal is moved closer to a source of radiation of frequency 6.2 x 1014 sec−1.
(i) increase - more intensity
(ii) same - same energy light
2. The threshold frequency of a metal is 1.23 x 1015 sec−1. Find the binding energy of an electron in the metal.
(8.15x10−19J)
E = hv = (6.6262x10-34 J-s)(1.23x1015Hz) = 8.15x10-19J
3. When radiation of frequency 9.2 x 1014 sec-1 strikes a metal, the metal emits electrons with kinetic energy 3.1
x 10−19 Joule. Find the threshold frequency of this metal. (4.5x1014 Hz)
hv = KE + W
(6.6262x10−34 J-s)(9.2x1014Hz) = 3.1x10−19J + W
W = 3.0x10−19J
W = work function = how tightly the electron is held
threshold is when KE = 0 and electron is just knocked off
so when KE = 0, W = hv
3.0x10−19J = (6.6262x10−34 J-s) v
v = 4.5x1014Hz
4. Find the threshold frequency of a metal whose binding energy is 259 kJ/mol. (4.30x10−19J)
5. Find the velocity of an electron emitted from a metal with a threshold frequency of 3.71 x 1014 sec−1, if the
metal is exposed to light of wavelength 422 nm. The mass of the electron is 9.11 x 10−31 kg.
Use threshold energy to find work function
W = hv = (6.6262x10−34 J-s)(3.71x1014Hz) = 2.46x10−19 J
(7.03x105m/s)
hv = KE + W
(6.6262x10−34J-s)(3.00x10^8 m/s/422 x 10^−9 m) = KE + 2.46x10−19J
KE = 2.25x10−19J
KE = 1/2 m u2 = 2.25x10−19J = 1/2 (9.11x10−31kg) u2
u = 7.03x105 m/s
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Revised - 2014-2015 - LCA from PK
PHOTOELECTRIC EFFECT
SUPPLEMENTARY QUESTIONS - UNIT 05
______________________________________________________________
6. The energy required to melt 1.00 gram of ice is 333J. Find the number of quanta of infrared radiation of
frequency 4.67x1013 sec−1 that must be absorbed in order to melt 1.00g of ice.
(1.08x1022quanta)
Blast from the past . . .
7. The atomic weight of antimony of natural origin is 121.75. It consists of only two isotopes, Sb-121, relative
atomic mass 120.90, and Sb -123, relative atomic mass 122.90. Find the natural abundance of the two
isotopes. (57.5% Sb-121 & 42.5% Sb-123)
x (120.90g) + (1 − x)(122.90g) = 121.75g
120.90x + 122.90 − 122.90x = 121.75
!
x = 57.5%
57.5% Sb − 121
42.5% Sb − 123
8. Naturally occurring iodine has an atomic weight of 126.9045. A 12.3849g sample of iodine is accidentally
contaminated with 1.00070g of 129I, a synthetic radioactive isotope of iodine used in treatment of certain
€ gland. The relative mass of 129I is 128.9050. Find the apparent “atomic weight” of
diseases of the thyroid
the contaminated iodine. (127.054 g/mol)
" 12.3849g %
" 1.00070g %
$$
''(126.9045g) + $$
''(128.9050g) = MoMa
!
# 13.3856g &
# 13.3856g &
117.417g + 9.63687 = 127.054g / mol = Moma
€
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Revised - 2014-2015 - LCA from PK

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