Transcript Atomic Physics

Atomic Physics
Question 1
Calculate the Kinetic Energy
gained by an electron when it is
accelerated through a P.D. of
50kV in an X-ray tube
• Firstly we know that KE is measured in
Joules.
• JDeV…..
• eV = ½mv2(KE)
• eV = (1.6 X 10-19)(50000) = 8 X 10-15 Joules = KE
Calculate the minimum wavelength
of an X-ray emitted from the anode.
(Planck’s = 6.6 X 10-34Js; c = 3 X 10-8ms-1;
Charge on electron = 1.6 X 10-19C)
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What formulae do we know with Planck?
E = hf
E = hc/λ
We know E from last question = 8 X 10-15 J
Calculate the minimum wavelength
of an X-ray emitted from the anode.
(Planck’s = 6.6 X 10-34Js; c = 3 X 10-8ms-1;
Charge on electron = 1.6 X 10-19C)
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E = hc/λ
8 X 10-15 J = (6.6 X 10-34)(3 X 10-8ms-1)/λ
λ = (6.6 X 10-34)(3 X 10-8ms-1)/ 8 X 10-15
λ = 2.475 X 10-27 m
Question 2
The work function of zinc is 6.9 X 1019J. What is the minimum frequency of
UV radiation that will cause the
photoelectric effect to occur in zinc.
(Planck’s = 6.6 X 10-34Js)
• The main equation that contains work
function (and the most important is
E = Φ + ½ mv2
• This is also written as:
• hf = hfo + ½ mv2
The work function of zinc is 6.9 X 10-19J.
What is the minimum frequency of UV
radiation that will cause the photoelectric
effect to occur in zinc. (Planck’s = 6.6 X
10-34Js)
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E = Φ + ½ mv2
hf = hfo + ½ mv2
Hence Φ = hfo
fo = Φ/h = 6.9 X 10-19J/ 6.6 X 10-34Js
fo = 1.05 X 10-15 Hz
Question 3
Zinc is illuminated with UV light of
wavelength 240nm. The work function is 4.3
eV. Calculate the threshold frequency of zinc
and Max KE of emitted electrons.
(Planck’s = 6.6 X 10-34Js; c = 3 X 10-8ms-1;
Charge on electron = 1.6 X 10-19C)
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What is the main formula we use?
What constants might give us a hint?)
E = Φ + ½ mv2
hf = hfo + ½ mv2
Hence Φ = hfo
Zinc is illuminated with UV light of
wavelength 240nm. The work function is 4.3
eV. Calculate the threshold frequency of zinc
and Max KE of emitted electrons
• Data above has to be converted into forms that
we can use.
• 240 nm…………..240 X 10-9 m
• The work function 4.3 eV needs to be converted
into Joules……….JDeV………..4.3 eV becomes
4.3 X (1.6 X 10-19) = 6.4 X 10-19 Joules
Zinc is illuminated with UV light of
wavelength 240nm. The work function is 4.3
eV. Calculate the threshold frequency of zinc
and Max KE of emitted electrons
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E = Φ + ½ mv2
hf = hfo + ½ mv2
Hence Φ = hfo
fo = Φ/h = 6.4 X 10-19J/ 6.6 X 10-34Js
fo = 0.97 X 10-15 Hz
Zinc is illuminated with UV light of
wavelength 240nm. The work function is 4.3
eV. Calculate the threshold frequency of zinc
and Max KE of emitted electrons
• E = Φ + ½ mv2
• hc/λ = hfo + ½ mv2
• (6.6 X 10-34)(3 X 10-8) = 6.4 X 10-19 + ½ mv2
240 X 10-9
• (6.6 X 10-34)(3 X 10-8)
= ½ mv2 (KE)
(240 X 10-9)(6.4 X 10-19)
Question 4
An electron accelerates from the cathode
through a P.D. of 4kV in a CRT.
How much energy does the electron gain?
What is the speed of the electron at the
anode?
• As usual we see that they are looking for
the energy gained
• eV = ½mv2
• eV = (1.6 X 10-19)(4000) = 6.4 X 10-16 Joules = KE
An electron accelerates from the cathode
through a P.D. of 4kV in a CRT.
How much energy does the electron gain?
What is the speed of the electron at the
anode?
• eV = ½mv2
• 6.4 X 10-16 Joules = ½mv2
• 6.4 X 10-16 Joules = ½(9.1 X 10-31)v2
• (6.4 X 10-16)(2) = v2
(9.1 X 10-31)
• V = 1.18 X 108 ms-1
After leaving the anode, the electron
travels at a constant speed and enters
a magnetic field at right angles where it
is reflected. The flux density is 5 X 10-2
T. Calculate the force acting on the
electron?
• F = qvB
• F = (1.6 X 10-19)(1.18 X 108)(5 X 10-2)
• F = 9.44 X 10-13 N
After leaving the anode, the electron travels
at a constant speed and enters a magnetic
field at right angles where it is reflected. The
flux density is 5 X 10-2 T. Calculate the
radius of the circular path followed by the
electron in the field?
• F = qvB
F = mv2/r
• qvB = mv2/r
• qvBr = mv2
• qBr = mv
• r
= mv/qB
After leaving the anode, the electron travels
at a constant speed and enters a magnetic
field at right angles where it is reflected. The
flux density is 5 X 10-2 T. Calculate the
radius of the circular path followed by the
electron in the field?
• r
= mv/qB
• r
= (9.1 X 10-31)(1.18 X 108)
(1.6 X 10-19)(5 X 10-2)
r = 1.34 X 10-2 m
Question 5
Radium-226 undergoes α-decay and
has a decay constant of 1.35 X 10-11s-1.
Calculate the number of α-particles
emitted per second by a 2μg sample of
this isotope.
(1 mol of radium-226 is 226 g; Avagadros constant
= 6.02 X 1021 mol-1)
• 1 mol = 226 g = 6.02 X 1021 atoms
• 1g =6.02 X 1021/226 = 2.66 X 1019 atoms
• 1μg = 2.66 X 1019 / 106= 2.66 X 1013 atoms
• 2μg = 5.32 X 1013 atoms
Radium-226 undergoes α-decay and
has a decay constant of 1.35 X 10-11s-1.
Calculate the number of α-particles
emitted per second by a 2μg sample of
this isotope.
(1 mol of radium-226 is 226 g; Avagadros constant
= 6.02 X 1021 mol-1)
• Number of α-particles = λN
• λN = (5.32 X 1013)(1.35 X 10-11)
• No. of α-particles = λN = 7.182 X 102 particles
Question 6
A detector records 1200 counts per
minute when the activity of a
radioactive sample is first measured.
Six minutes later the activity has fallen
to 150 counts per minute. Calculate the
half-life of the sample
• 1200
600
300
150
1 half-life 2 half-lives 3 half-lives
• 3 half lives in 6 min…Each half-life in 2 min
Question 7
An ancient wooden cup from an
archaeological site has an activity of
2.1 Bq. The corresponding for newly
cut wood is 8.4 Bq. If the half-life of
Carbon-14 is 5730 years, estimate the
age of the cup.
• 8.4Bq
4.2Bq
2.1Bq
1 half-life
2 half-lives
• 2 half lives each half-life being 5730 years
each…….11460 years
Question 8
The power generator in a nuclear
reactor is 150MW. Calculate the number
of fissions occurring per second in the
reactor, give that 180MeV of energy is
released per fission
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Firstly, get all units into SI units.
150 MW is 150 X 106 Watts
180MeV is 180 X 106 eV but this isn’t an SI
180 x 106 eV…….…..…JDeV……………..
(180 x 106)(1.6 X 10-19) = 2.88 X 10-11 Joules
The power generator in a nuclear reactor is
150MW. Calculate the number of fissions
occurring per second in the reactor, give that
180MeV of energy is released per fission
• Watts = 150 X 106 Watts
• Watts = Energy per Second
• Energy released per fission = 2.88 X 10-11 J
• Fissions per Sec = Energy/ Energy released
per fission
• Fissions per Sec = (150 X 106)/(2.88 X 10-11)
= 4.32 X 10-3 fissions per sec
Question 9
If a sample of radium contains 2.6 X
1021 radium-226 and is emitting 3.5 X
1010 particles per second, calculate the
decay constant
• Rate of decay = λN
• Rate of decay/N = λ
• 3.5 X 1010/ 2.6 X 1021 = λ
• λ = 1.35 X 10-11 s-1
If a sample of radium contains 2.6 X 1021
radium-226 and is emitting 3.5 X 1010
particles per second, calculate the half-life
if the rate of decay is 1.35 X 10-11 s-1
• T½ = 0.693/λ
• T½ = 0.693/1.35 X 10-11
• T½ = 5.15 X 1010 s
Question 10
A large number of solar panels are joined
together in series and cover an area of
20m2. The efficiency of the solar cells is
20%. If the solar cell constant is 1400
Wm-2, what is the max power generated
by the solar cells
• Because the efficiency is only 20% on
20m2, it’s like having only 4m2 solar cells.
• Solar constant is 1400 Wm-2. From the unit I
see that this is power/ area
A large number of solar panels are joined
together in series and cover an area of
20m2. The efficiency of the solar cells is
20%. If the solar cell constant is 1400
Wm-2, what is the max power generated
by the solar cells
• Solar constant is 1400 Wm-2. From the unit I
see that this is power/ area.
• 1400 = Power/ area
• 1400 X area = Power
• 1400 X 4 = Power = 5600W
Question 11
Cobalt-60 is a radioactive isotope
with a half-life of 5.26 years and
emits β-particles. Write an equation
to represent the decay of cobalt-60.
• Cobalt-60 ---------- β + substance-61
Cobalt-60 is a radioactive isotope
with a half-life of 5.26 years and
emits β-particles. Calculate the
decay constant of cobalt-60.
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T½
λ
λ
λ
= 0.693/λ
= 0.693/ T½
= 0.693/5.26
= 0.13175 y-1
Cobalt-60 is a radioactive isotope with
a half-life of 5.26 years and emits βparticles. Calculate the rate of decay of
cobalt-60 when it has 2.5 X 1021 atoms.
Rate of decay = λN
Rate of decay = λ X N
Rate of decay = 0.13175/ 2.5 X 1021
Rate of decay = 0.33 X 1021
Rate of decay = 3.3 X 1020 particles
Question 12
A neutral pion is unstable with a decay
constant of 2.5 X 1012 s-1. What is the
half-life of a neutral pion?
• T½ = 0.693/λ
• T½ = 0.693/2.5 X 1012
• T½ = 2.772 X 10-13 s