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“Atomic Structure”
Illustrative problem 1
Radio city broadcasts on a frequency of 5,090 KHz.What is the
wavelength of electromagnetic radiation emitted by the
transmitter?
c
λ=
ν
c is 3 × 10 8 m / s
3 × 108
λ=
=
5090 × 103
0.589 × 102
= 58.9 m
Illustrative Problem 2
The ratio of the energy of a photon of 2000 Å wavelength radiation to
that of 6000 Å
(a) ¼
(b) ½
radiation is
(b) 4
(d) 3
Solution:
hc
E = hν =
λ
hc
hc
=
E1 =
E2
λ1
λ2
0
6000 A
E1 λ 2
3
∴
== =
0
E2 λ1
2000 A
Hence, answer is (d).
Illustrative Problem 3
The energy of the electron in the second and third Bohr orbits of the
hydrogen atom is -5.42 X 10-12 and –2.41 X 10-12
respectively. Calculate the wavelength of the emitted radiation, when
the electron drops from third to second orbit.
Solution
∆E = E2 − E1
=−5.42 × 10−12 − (−2.41 × 10−12 )ergs
=
−3.01 × 10−12 ergs.
According to Planck’s quantum theory
E = hυ
E =h
c
λ
=
h 6.6 × 10−27 ergs
Solution
c
∴λ =
h
E
=
6.62 × 10−27 × 3 × 108
3.01 × 10−12
=
λ 6.6 × 10−5 cm
= 6.6 × 103 A0
1 A0 = 10−8 cm
Class Exercise - 1
Which of the following fundamental
particles are present in the nucleus
of an atom?
(a) Alpha particles and protons
(b) Protons and neutrons
(c) Protons and electrons
(d) Electrons, protons and neutrons
Solution
The nucleus of an atom is positively charged and almost
the entire mass of the atom is concentrated in it. Hence,
it contains protons and neutrons.
Hence, answer is (b).
Class Exercise - 2
The mass of the proton is
(a) 1.672 × 10–24 g
(b) 1.672 × 10–25 g
(c) 1.672 × 1025 g
(d) 1.672 × 1026 g
Solution
The mass of the proton is 1.672 × 10–24 g
Hence, answer is (a).
Class Exercise - 3
Which of the following is not true in case of an electron?
(a) It is a fundamental particle
(b) It has wave nature
(c) Its motion is affected by magnetic field
(d) It emits energy while moving in orbits
Solution
An electron does not emit energy while moving in orbit.
This is so because if it would have done that it would
have eventually fallen into the nucleus and the atom would have
collapsed.
Hence, answer is (d).
Class Exercise - 4
Positive charge of an atom is
(a) concentrated in the nucleus
(b) revolves around the nucleus
(c) scattered all over the atom
(d) None of these
Solution
Positive charge of an atom is present entirely in the
nucleus.
Hence, answer is (a).
Class Exercise - 5
Why only very few a-particles are deflected back on hitting a
thin gold foil?
Solution
Due to the presence of a very small centre in which the entire
mass is concentrated.
Class Exercise - 6
Explain why cathode rays are produced only when the pressure in the
discharge tube is very low.
Solution
This is happened because at higher pressure no electric current
flows through the tube as gases are poor conductor of electricity.
Class Exercise - 7
If a neutron is introduced into the nucleus of an atom, it would
result
in the change of
(a) number of electrons
(b) atomic number
(c) atomic weight
(d) chemical nature of the atom
Solution
Neutrons contribute in a major way to the weight of the nucleus,
thus addition of neutron would result in increase in the atomic
weight.
Hence, answer is (c).
Class Exercise - 8
The concept of stationary orbits lies in the fact that
(a) Electrons are stationary
(b) No change in energy takes place in
stationary orbit
(c) Electrons gain kinetic energy
(d) Energy goes on increasing
Solution
When an electron revolves in a stationary orbit, no energy
change takes place. Energy is emitted or absorbed only when
the electron jumps from one stationary orbit to another.
Hence, answer is (b).
Class Exercise - 9
What is the energy possessed by 1 mole of photons of radiations
of frequency 10 × 1014 Hz?
Solution
E = hv
E = 6.6 × 10–34 × 10 × 1014
E = 66 × 10–20 = 6.6 × 10–19 joules
∴ energy of 1 mole of photons = 6.6 × 10–19 × 6.023 × 1023
= 39.7518 × 104
= 397.518 kJ/mol
The Atomic Mass Scale and Mass
Weights
atomic Mass = (0.691)(62.9 amu)



63
Cu isotope
16
The Atomic Weight Scale and
Atomic Weights
atomic weight = (0.691)(62.9 amu) + (0.309)(64.9 amu)


 


63
Cu isotope
17
65
Cu isotope
The Atomic Weight Scale and
Atomic Weights
atomic weight = (0.691)(62.9 amu) + (0.309)(64.9 amu)


 


63
Cu isotope
atomic weight = 63.5 amu for copper
18
65
Cu isotope
Atomic Number
Atomic number (Z) of an element is
the number of protons in the nucleus
of each atom of that element.
Element
# of protons
Atomic # (Z)
Carbon
6
6
Phosphorus
15
15
Gold
79
79
Question 2
The table on the next slide shows the five isotopes of germanium found in nature,
the abundance of each isotope, and the atomic mass of each isotope.
Calculate the atomic mass of germanium.
Answer
72.591 amu
The wavelength of the peak in the intensity graph is given by
Wien’s law (T must be in kelvin):
Wien’s
Displacement Law
EXAMPLE: Finding peak
wavelengths
QUESTIONS:
EXAMPLE :Finding peak
wavelengths
EXAMPLE : Finding peak
wavelengths
EXAMPLE :Finding peak
wavelengths
The Electron Volt
• Consider an electron
accelerating (in a vacuum)
from rest across a parallel
plate capacitor with a 1.0 V
potential difference.
• The electron’s kinetic energy
when it reaches the positive
plate is 1.60 x 10−19 J.
• Let us define a new unit of
energy, called the electron
volt, as 1 eV = 1.60 x 10−19 J.
Energy of an electron
QUESTION:
EXAMPLE 38.5 Energy of an
electron
EXAMPLE 38.5 Energy of an
electron
EXAMPLE : Energy of an electron
Example 1
 Write out the electron configuration of the elements

(i) carbon (Z = 6)

(ii) magnesium (Z = 12)

(iii) argon (Z = 18)
Solution

(i) 1s2 2s2 2p2

(ii) 1s2 2s2 2p6 3s2

(iii) 1s2 2s2 2p6 3s2 3p6
Example 2
 Determine which of the following are in the excited state

(i) 1s2 2s2 2p5 3s1

(ii) 1s2 2s2 2p6 3s2 3p6 4s1

(iii) 1s2 2s2 2p3

(iv) 1s2 2s2 2p6 3s1 3p5
Solution

(i) and (iv) are in excited states
Mass Number
Mass number is the number of
protons and neutrons in the nucleus
of an isotope: Mass # = p+ + n0
p+
n0
e- Mass #
8
10
8
18
Arsenic - 75
33
42
33
75
Phosphorus - 31
15
16
15
31
Nuclide
Oxygen - 18
Information in the Periodic Table
The number at the bottom of
each box is the average
atomic mass of that
element.
This number is the weighted average mass of all the naturally occurring isotopes of
that element.
Question 1
How does the atomic number of an element differ from the element’s mass
number?
Answer
The atomic number of an element is the number of protons in the nucleus. The mass
number is the sum of the number of protons and neutrons.
The Atomic Weight Scale and Atomic Weights


The atomic weight of an element is the weighted
average of the masses of its stable isotopes
Example 5-2: Naturally occurring Cu consists of
2 isotopes. It is 69.1% 63Cu with a mass of 62.9
amu, and 30.9% 65Cu, which has a mass of 64.9
amu. Calculate the atomic weight of Cu to one
decimal place.
Calculating Atomic Mass
Calculating Atomic Mass
Copper exists as a mixture of two isotopes.
The lighter isotope (Cu-63), with 29 protons and 34 neutrons, makes up 69.17% of
copper atoms.
The heavier isotope (Cu-65), with 29 protons and 36 neutrons, constitutes the
remaining 30.83% of copper atoms.
Calculating Atomic Mass
The atomic mass of Cu-63 is 62.930 amu, and the atomic mass of Cu-65 is 64.928 amu.
Use the data above to compute the atomic mass of copper.
Calculating Atomic Mass
First, calculate the contribution of each isotope to the average atomic mass, being sure
to convert each percent to a fractional abundance.
Calculating Atomic Mass
The average atomic mass of the element is the sum of the mass contributions of each
isotope.
Can you tell me: how many atoms are there in 43.2 grams of iron?
Using unit analysis:
The unit of the answer is atoms.
The unit of the information given is grams.
But there is no conversion factor for grams ---> atoms.
Reason: the mass of an atom is different for every element.
Solution: use the mole concept.
6.02x10 23 atoms Fe
1 mol Fe
= 4.66x 10 23 atoms Fe
x
43.2g Fe x
1 mol Fe
55.85 g Fe
Electron Orbits
Consider the planetary model for electrons which move in a circle around the positive
nucleus. The figure below is for the hydrogen atom.
r
FC
-
e-
Coulomb’s law:
+
Nucleus
mv 2
e2
=
r
4πε 0 r 2
FC =
Centripetal FC:
e
2
4πε 0 r
2
Radius of Hydrogen atom
mv
FC = 2
r
r=
2
e
2
4πε 0 mv
2
Example 1: Use the Balmer equation to find the
wavelength of the first line (n = 3) in the Balmer
series. How can you find the energy?
 1 1 
R  2 + 2 ; n =
3 R = 1.097 x 107 m-1
=
λ
2 n 
1
1
 1 1
= R  2 + 2 = R(0.361); λ=
0.361R
λ
2 3 
1
λ = 656 nm
λ=
7
-1
0.361(1.097 x 10 m )
1
The frequency and the energy are found from:
c = fλ and E = hf
Example 2: Find the radius of the Hydrogen atom in
its most stable state (n = 1).
n ε 0h
rn =
π me 2
2
2
r=
m = 9.1 x 10-31 kg
2
(1) (8.85 x 10
e = 1.6 x 10-19 C
-12 Nm 2
C2
-31
π (9.1 x 10
r = 5.31 x 10-11 m
)(6.63 x 10−34 J ⋅ s) 2
kg)(1.6 x 10-19 C) 2
r = 53.1 pm
Total Energy of an Atom
The total energy at level n is the sum of the kinetic and potential energies at that
level.
E=
K +U;
But we recall that:
2
vn =
e
2ε 0 nh
2
1
K=
mv
;
2
n ε 0h
rn =
π me 2
Total energy of Hydrogen
atom for level n.
2
2
e2
U=
4πε 0 r
Substitution for v and r
gives expression for total
energy.
me 4
En = − 2 2 2
8ε 0 n h
Energy for a Particular State
It will be useful to simplify the energy formula for a particular state by substitution of
constants.
m = 9.1 x 10-31 kg
e = 1.6 x 10-19 C
εo = 8.85 x 10--12 C2/Nm2
h = 6.63 x 10-34 J s
me 4
(9.1 x 10-31kg)(1.6 x 10-19 C) 4
En =
− 2 2 2 =
−
-12 C2 2 2
8ε 0 n h
8(8.85 x 10 Nm2 ) n (6.63 x 10-34 Js) 2
2.17 x 10-18 J
En = −
2
n
Or
−13.6 eV
En =
n2
Balmer Revisited
Total energy of Hydrogen
atom for level n.
me 4
En = − 2 2 2
8ε 0 n h
Negative because outside energy to
raise n level.
When an electron moves from an initial state ni to a final state nf, energy involved
is:
1
=
λ
me 4
2 3
2
8ε 0 h cn f
 1
1 
me 4
 2 − 2  ; If R=
2 3
2
8ε 0 h cn f
 n f ni 
hc
=
E=
E0 − E f ;
λ
Balmer’s
Equation:
1  −me 4
me 4
=
 2 2 2 + 2 2 2
λ hc  8ε 0 h n0 8ε 0 h n f
1
 1
1 
=
1.097 x 107 m -1
R  2 − 2 ; R =
n

n
λ
0
f


1



Energy Levels
We can now visualize the hydrogen atom with an electron at many possible energy levels.
Emission
Absorption
The energy of the atom increases on absorption (nf > ni)
and de-creases on emission (nf < ni).
Energy of nth
level:
−13.6 eV
E=
n2
The change in energy of the atom can be given in terms of initial ni and final nf levels:
 1
1 
−13.6 eV  2 − 2 
E=
n

n
0 
 f
Spectral Series for an Atom
The Lyman series is for transitions to n = 1 level.
The Balmer series is for transitions to n = 2 level.
The Pashen series is for
transitions to n = 3 level.
The Brackett series is for
transitions to n = 4 level.
n =1
n =2
n =3
n =4
n =5
n =6
 1 1 
−13.6 eV  2 − 2 
E=
n

 f n0 
Example 3: What is the energy of an emitted
photon if an electron drops from the n = 3 level
to the n = 1 level for the hydrogen atom?
 1
1 
−13.6 eV  2 − 2 
E=
n

n
0 
 f
Change in energy of the
atom.
1 1
E=
−13.6 eV  2 − 2  =
−12.1 eV
1 3 
∆E = -12.1 eV
The energy of the atom decreases by 12.1 eV as a photon of that energy is emitted.
You should show that 13.6 eV is required to move an electron from n = 1 to
n = ∞.
Summary (Cont.)
Spectrum for nf = 2 (Balmer)
434 nm
n=3
n=4
Emission spectrum
653 nm
n
6
n=5
486 nm
410 nm
The general equation for a change from one level to another:
Balmer’s
Equation:
 1
1 
=
1.097 x 107 m -1
R  2 − 2 ; R =
n

n
λ
0
f


1
Questions
1. a. Sketch a cathode-ray tube and label it with
the following parts: partially evacuated glass
vessel, electrically charged plates, cathode,
anode, fluorescent screen, high voltage
source.
b. Explain how it works.
Questions
2. Do you think the hypothesis inferring from the
observations are reasonable? Why or why not?
Explain in each hypothesis
Questions
3. How can the observations from the cathode
ray tube experiment lead to the new atomic
model “the plum pudding model”
Plum Pudding Model
• in 1904 J.J. Thomson proposed the plum pudding model of
atom.
• According to this model the electrons were randomly stuck
into a ball of positively charge matter.
positively charge
matter
electrons
© 2004 by Pearson Education Inc.
• A more modern name for this model might be
chocolate-chip muffin model.
Discovering the Electron
57
The Discovery of the Electron
Thomson found q/m for cathode ray particles to
be 1.76 x 1011 C/kg
This is close to 2000 x q/m for a hydrogen ion
Thomson’s conclusion: cathode-ray particles
(today called electrons) are approximately
1/2000 the mass of a hydrogen ion (today called
protons)
The Discovery of the Electron
Since Thomson didn’t know the charge or mass
of these particles couldn’t use
qV =
1
2
mv2
to determine their speed


He balanced Fe versus Fm
was undeflected
so that particle path
The Discovery of the Electron
Then,
 
Fe = Fm

q E = qv B

E
v= 
B
Using only a magnetic field,


Fm = Fc
v2
qv B = m
r
q v
=
m Br
The Discovery of the Electron
Examples: Practice Problem 1,
=
v

E
6.0 × 10 4 N C
4

=
= 2.4 × 10
2.50 T
B
m
s
Practice Problem 1,
1.0 × 106 m s
q
v
= =
= 1.0 × 108 C kg
m B r 1.0 T × 0.0100 m
√
The potential energy
of electrons is converted
to kinetic energy
Since change in energy is the
Electrons
are times
attracted
positively charged
voltage
the to
charge
plate. then
They½mv²=qV
accelerate towards it and small
percentage
escape
plate through small
Therefore
v=the
√(2qV/m)
hole, creating electron beam.
120 Volts
Plate is heated and
electrons boil off.
Velocity= 0
Potential Energy= ½ mv^2
6.3 Volts
We now know that v= √(2qV/m), so we can now easily find the
velocity of our beam of electrons.
q(charge) of an electron= -1.6•10^-19
V(volts)=120
m(mass) of an electron=9.11•10^-31 kg
Therefore:
v=√(2)(-1.6•10^-19)(120)/(9.11•10^-31)
v=√4.215•10^13
v=649•10^6 m/s
In order to predict
the angle at which
the electrons are
deflected, we must
first measure
the force that the
magnetic field inserts
upon the beam
To do this, we use the equation:
F=qvB
Magnetic field
The force is always
Perpendicular to the magnetic field
And the velocity of the electrons
Electrons
Like Solar Wind,
the electrons in the
CRT beam are
deflected
when entering a
magnetic field,
therefore the
electron
beam “bends.”
In order to find the force of the magnetic field, we must first calculate its strenghth.
Since F=qvB and, according to Newton’s second law, F=m•v²/r, we can deduce that
qvB=m•v²/r
Or
B=mv/qr
mass= 9.11•10^-31 kg
velocity= 6.492•10^6 m/s
Charge= 1.6•10^-19 C
And if the measured distance of the electron beam from the magnets
Is .075 meters
Then B= (9.11•10^-31)(6.492•10^6)/(1.6•10^-19)(.075)
B=2.772•10^-6 tesla
Now that we know the strength of the magnetic field at the electron beam, we can
Calculate the force which the field exerts upon the electrons.
F=qvB
F=(6.49•10^6) (1.6•10^-19)(2.772•10^-6
F=2.879•10^-18 N
Examples
1. In a Cathode ray tube an electron travels through an electrical
field of 5.80 x 103 N/C and a magnetic field of 6.60 x 10-3 T
undeflected.
 a.
what is the velocity of the electron?
b. If the electric field is turned off the magnetic field causes the
electron to travel in a circular path, what is the radius of the
circular path of the electron?
2. An electron travels through a CRT and a magnetic field of 4.20 x 10-3
T, the electron travels a circular path with a radius of 4.90 x 10-3 m.
a. What is the velocity of the electron?
b. What was the potential difference in the CRT?
3. An unknown particle travels in a CRT tube through a magnetic

field of 4.60 x 10-2 T at a velocity of 5.20 x 105 m/s, the magnetic
field causes the particle to travel in a circular path with a radius
of 0.235 m.
What is the charge to mass ratio of this particle?
Deflection of Electrons in a
Uniform Electric Field (1)


Consider an electron beam directed between two
oppositely charged parallel plates as shown below.
With a constant potential difference between the two
deflecting plates, the trace is curved towards the
positive plate.
+
d
-
Deflection of Electrons in a
Uniform Electric Field (2)

The force acting on each electron in the field is
given by
eVP
F = eE =
d
where E = electric field strength,
Vp = p.d. between plates,
d = plate spacing.
Deflection of Electrons in a
Uniform Electric Field (3)

The vertical displacement y is given by
1 2 1 eV p 2
y = at = (
)t
2
2 md
1 eV p x 2
= (
) 2
2 md v
This is the equation for a parabola.
Deflection of electrons in a
uniform magnetic field
Deflection of Electrons in a Uniform Magnetic
Field

The force F acting on an electron in a uniform
magnetic field is given by
F = Bev
Since the magnetic force F is at right angles to
the velocity direction, the electron moves round
a circular path.
Deflection of Electrons in a Uniform Magnetic
Field

The centripetal acceleration of the electrons is
Bev
a=
m
v 2 Bev
Hence a =
=
r
m
mv
r=
eB
which gives
Determination of Specific Charge Using a
Fine Beam Tube
Ber
Since v =
m
(For an electron moving in a
uniform magnetic field)
and the kinetic energy of the electron provided
by the electron gun is
1 2
mv = eV
2
Where V is the anode voltage.
Determination of Specific Charge Using
a Fine Beam Tube (3)
1 Ber 2
So
) = eV
m(
2
m
Rearrange the equation gives
e
2V
= 2 2
m B r
The value of the specific charge of an electron
is now known accurately to be
(1.758803 ± 0.000003) × 1011 C/kg
Thomson’s e/m Experiment (1)
Thomson’s apparatus for measuring the ratio e/m
+
×
×
v
×
×
× × × × × × × ×
× × × × × × × ×
× × × × × × × ×
× × × × × × × ×
-
Thomson’s e/m Experiment (2)



A beam of electron is produced by an electron
gun with an accelerating voltage V.
The electron beam is arranged to travel through
an electric field and a magnetic field which are
perpendicular to each other.
The apparatus is set-up so that an electron from
the gun is undeflected.
Thomson’s e/m experiment (3)

As the electron from the gun is undeflected, this gives
FE = FB
i.e.
eE = Bev
On the other hand,
eE
E
⇒v=
B
v
Bev
1
mv 2 = eV
2
Combining the equations, we get
2
e
E
=
m 2VB 2
A
F
B
+Q
+V
d
To move the charge from
A to B a force equal to F would
be needed. The work done
moving Q from A to B is W.
W = Fd
But
E = F/Q or F = EQ
So, W = EQd
V
Potential difference is defined as
work done per unit charge:
V = W/Q
d
position
So:
V = W/Q = EQd /Q
V = Ed
E = V/d
(Uniform fields)
Acceleration: gravitational and electrical
An accelerating charge
A falling ball
starts at high
gravitational
potential energy
gravitational
field g
starts at high
electrical
potential
energy
charge q
mass m
height h
electric
field
+
gains
kinetic
energy
potential
difference V
gains
kinetic
energy
ends at low
gravitational
potential energy
+
–
ends at low
electrical
potential energy
+
electric
potential
gravitational
potential
ball falls down gravitational
potential hill
ΔEp=mgΔh
gain of kinetic energy = loss of potential energy = mgh
charge ‘falls down’ electrical
potential hill
ΔW = QΔV
gain of kinetic energy = loss of potential energy = qV
An electric field accelerates a charge as a gravitational field accelerates a mass
E
Can you come up with an expression
for the charge on the droplet?
Identify the forces on a charge in an electric field
Draw the forces acting on the drop
F= Eq
mg
q=
E
Can you come up with an expression
for the charge on the droplet?
F= mg
Identify the forces on a charge in an electric field
Draw the forces acting on the drop
electric force upward = weight downward
 EQ = mg
 Q = mg/E = 1.8 x 10-15 x 9.81 / 1.9 x 104 =
9.4 x 10-19 C
1)
ΔEp=m
ΔW = Q
gΔh
ΔV
Quantization of Charge
Example: Practice Problem 1, page 763 (Pearson


Physics)
Fe = Fg


qE =mg
q × 5.0 × 105 N C = 2.4 × 10 −14 kg × 9.81m s 2
q 4.7 × 10 −19 C
=
4.7 × 10 −19 C
−
e
=
3
1.60 × 10 −19 C e−
Example



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A beam of electrons travels an undeflected path in a cathode ray
tube. E is 7.0 x 103 N/C. B is 3.5 x 10-2 T. What is the speed of the
electrons as they travel through the tube?
What we know:
 E = 7.0 x 103 N/C
B=3.5 x 10-2 T
Equation:
 v = E/B
Substitute:
 v = (7.0 x 103N/A s) / (3.5 x 10-2 N/A m)
Solve!
 v = 2.0 x 105 m/s
Example
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An electron of mass 9.11 x 10-31 kg moves with a speed of 2.0 x
105 m/s across a magnetic field. The magnetic induction is 8.0
x 10-4 T. What is the radius of the circular path followed by the
electrons while in the field?
What we know:
 M = 9.11 x 10-31 kg
B=8.0 x 10-4 T
v=2.0 x 105 m/s
Equation:
 Bqv = mv2/r so r = (mv) / (Bq)
Substitute:
 R = (9.11x10-31kg)(2.0x105 m/s)
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(8.0x10-4N/Am)(1.6x10-19As)
Solve! r = 1.4 x 10-3 m
Modifications of the Bohr Theory
– Elliptical Orbits
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Sommerfeld extended the results to include elliptical
orbits
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Retained the principle quantum number, n
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Added the orbital quantum number, ℓ
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Determines the energy of the allowed states
ℓ ranges from 0 to n-1 in integer steps
All states with the same principle quantum number are said
to form a shell
The states with given values of n and ℓ are said to form a
subshell
Example
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A negatively charged polystyrene sphere of mass 3.3 x 10-15 kg is held at rest
between 2 parallel plates separated by 5.0 mm when the potentialdifference
between them is 170 V. How many excess electrons are on the sphere?
Solution:
Using qV/d = mg gives q = 9.5 x 10-19 C
Taking e to be 1.6 x 10-19, q corresponds to 5.9 excess electrons
Because we cannot have 0.9 of an electron, the answer must be 6 electrons,
the discrepancy must be due to experimental inaccuray or rounding errors
Example
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Electrons are accelerated to a speed of 8.4 x 106 m/s. they then pass into a
region of uniform magnetic flux density 0.50 mT. The path of the electrons in
the field is a circle with a radius 9.6 cm. Calculate:
a) the specific charge of the electron
b) the mass of the electron, assuming the charge on the electron is 1.6 x
10-19 C
Solution
a) e/m = v/Br = 1.8 x 1011 C/kg
b) using e/m, m = 9.1 x 10-31 kg
Example
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It is required to select charged ions which have a speed of 4.2 x 106 m s-1. the
electric field strength in the velocity selector is 3.2 x 104 V m-1. Calculate the
magnetic field strength required.
Solution
v = E/B = 7.6 x 10-3 T
Example
A charged particle has mass 6.7 x 10-27 kg and charge +3.2 x 10-19 C. It
is travelling at speed 2.5 x 108 m s-1 when it enters a region of space
where there is a uniform magnetic field of flux density 1.6 T at right
angles to its direction of motion. Calculate
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a) the force on the particle due to the magnetic field
b) the radius of its orbit in the field
Solution
a)
F = Bqv sin θ = 1.3 x 10-10 N
b)
centripetal force is provided by the electromotive force
mv2/r = Bqv
r = 3.2 m