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9 | 2139-2142 | The
scattered alpha-particles on striking
the screen produced brief light flashes
or scintillations These flashes may be
viewed through a microscope and the
distribution of the number of scattered
particles may be studied as a function
of angle of scattering FIGURE 12 2 Schematic arrangement of the Geiger-Marsden experiment |
9 | 2140-2143 | These flashes may be
viewed through a microscope and the
distribution of the number of scattered
particles may be studied as a function
of angle of scattering FIGURE 12 2 Schematic arrangement of the Geiger-Marsden experiment A typical graph of the total number of a-particles scattered at different
angles, in a given interval of time, is shown in Fig |
9 | 2141-2144 | FIGURE 12 2 Schematic arrangement of the Geiger-Marsden experiment A typical graph of the total number of a-particles scattered at different
angles, in a given interval of time, is shown in Fig 12 |
9 | 2142-2145 | 2 Schematic arrangement of the Geiger-Marsden experiment A typical graph of the total number of a-particles scattered at different
angles, in a given interval of time, is shown in Fig 12 3 |
9 | 2143-2146 | A typical graph of the total number of a-particles scattered at different
angles, in a given interval of time, is shown in Fig 12 3 The dots in this
figure represent the data points and the solid curve is the theoretical
prediction based on the assumption that the target atom has a small,
dense, positively charged nucleus |
9 | 2144-2147 | 12 3 The dots in this
figure represent the data points and the solid curve is the theoretical
prediction based on the assumption that the target atom has a small,
dense, positively charged nucleus Many of the a-particles pass through
the foil |
9 | 2145-2148 | 3 The dots in this
figure represent the data points and the solid curve is the theoretical
prediction based on the assumption that the target atom has a small,
dense, positively charged nucleus Many of the a-particles pass through
the foil It means that they do not suffer any collisions |
9 | 2146-2149 | The dots in this
figure represent the data points and the solid curve is the theoretical
prediction based on the assumption that the target atom has a small,
dense, positively charged nucleus Many of the a-particles pass through
the foil It means that they do not suffer any collisions Only about 0 |
9 | 2147-2150 | Many of the a-particles pass through
the foil It means that they do not suffer any collisions Only about 0 14%
of the incident a-particles scatter by more than 1°; and about 1 in 8000
deflect by more than 90° |
9 | 2148-2151 | It means that they do not suffer any collisions Only about 0 14%
of the incident a-particles scatter by more than 1°; and about 1 in 8000
deflect by more than 90° Rutherford argued that, to deflect the a-particle
backwards, it must experience a large repulsive force |
9 | 2149-2152 | Only about 0 14%
of the incident a-particles scatter by more than 1°; and about 1 in 8000
deflect by more than 90° Rutherford argued that, to deflect the a-particle
backwards, it must experience a large repulsive force This force could
FIGURE 12 |
9 | 2150-2153 | 14%
of the incident a-particles scatter by more than 1°; and about 1 in 8000
deflect by more than 90° Rutherford argued that, to deflect the a-particle
backwards, it must experience a large repulsive force This force could
FIGURE 12 1 Geiger-Marsden scattering experiment |
9 | 2151-2154 | Rutherford argued that, to deflect the a-particle
backwards, it must experience a large repulsive force This force could
FIGURE 12 1 Geiger-Marsden scattering experiment The entire apparatus is placed in a vacuum chamber
(not shown in this figure) |
9 | 2152-2155 | This force could
FIGURE 12 1 Geiger-Marsden scattering experiment The entire apparatus is placed in a vacuum chamber
(not shown in this figure) Rationalised 2023-24
293
Atoms
be provided if the greater part of the
mass of the atom and its positive charge
were concentrated tightly at its centre |
9 | 2153-2156 | 1 Geiger-Marsden scattering experiment The entire apparatus is placed in a vacuum chamber
(not shown in this figure) Rationalised 2023-24
293
Atoms
be provided if the greater part of the
mass of the atom and its positive charge
were concentrated tightly at its centre Then the incoming a-particle could get
very close to the positive charge without
penetrating it, and such a close
encounter would result in a large
deflection |
9 | 2154-2157 | The entire apparatus is placed in a vacuum chamber
(not shown in this figure) Rationalised 2023-24
293
Atoms
be provided if the greater part of the
mass of the atom and its positive charge
were concentrated tightly at its centre Then the incoming a-particle could get
very close to the positive charge without
penetrating it, and such a close
encounter would result in a large
deflection This agreement supported
the hypothesis of the nuclear atom |
9 | 2155-2158 | Rationalised 2023-24
293
Atoms
be provided if the greater part of the
mass of the atom and its positive charge
were concentrated tightly at its centre Then the incoming a-particle could get
very close to the positive charge without
penetrating it, and such a close
encounter would result in a large
deflection This agreement supported
the hypothesis of the nuclear atom This
is why Rutherford is credited with the
discovery of the nucleus |
9 | 2156-2159 | Then the incoming a-particle could get
very close to the positive charge without
penetrating it, and such a close
encounter would result in a large
deflection This agreement supported
the hypothesis of the nuclear atom This
is why Rutherford is credited with the
discovery of the nucleus In Rutherford’s nuclear model of
the atom, the entire positive charge and
most of the mass of the atom are
concentrated in the nucleus with the
electrons some distance away |
9 | 2157-2160 | This agreement supported
the hypothesis of the nuclear atom This
is why Rutherford is credited with the
discovery of the nucleus In Rutherford’s nuclear model of
the atom, the entire positive charge and
most of the mass of the atom are
concentrated in the nucleus with the
electrons some distance away The
electrons would be moving in orbits
about the nucleus just as the planets
do around the sun |
9 | 2158-2161 | This
is why Rutherford is credited with the
discovery of the nucleus In Rutherford’s nuclear model of
the atom, the entire positive charge and
most of the mass of the atom are
concentrated in the nucleus with the
electrons some distance away The
electrons would be moving in orbits
about the nucleus just as the planets
do around the sun Rutherford’s
experiments suggested the size of
the nucleus to be about 10–15 m to
10–14 m |
9 | 2159-2162 | In Rutherford’s nuclear model of
the atom, the entire positive charge and
most of the mass of the atom are
concentrated in the nucleus with the
electrons some distance away The
electrons would be moving in orbits
about the nucleus just as the planets
do around the sun Rutherford’s
experiments suggested the size of
the nucleus to be about 10–15 m to
10–14 m From kinetic theory, the size
of an atom was known to be 10–10 m,
about 10,000 to 100,000 times larger
than the size of the nucleus (see Chapter 10, Section 10 |
9 | 2160-2163 | The
electrons would be moving in orbits
about the nucleus just as the planets
do around the sun Rutherford’s
experiments suggested the size of
the nucleus to be about 10–15 m to
10–14 m From kinetic theory, the size
of an atom was known to be 10–10 m,
about 10,000 to 100,000 times larger
than the size of the nucleus (see Chapter 10, Section 10 6 in Class XI
Physics textbook) |
9 | 2161-2164 | Rutherford’s
experiments suggested the size of
the nucleus to be about 10–15 m to
10–14 m From kinetic theory, the size
of an atom was known to be 10–10 m,
about 10,000 to 100,000 times larger
than the size of the nucleus (see Chapter 10, Section 10 6 in Class XI
Physics textbook) Thus, the electrons would seem to be at a distance
from the nucleus of about 10,000 to 100,000 times the size of the nucleus
itself |
9 | 2162-2165 | From kinetic theory, the size
of an atom was known to be 10–10 m,
about 10,000 to 100,000 times larger
than the size of the nucleus (see Chapter 10, Section 10 6 in Class XI
Physics textbook) Thus, the electrons would seem to be at a distance
from the nucleus of about 10,000 to 100,000 times the size of the nucleus
itself Thus, most of an atom is empty space |
9 | 2163-2166 | 6 in Class XI
Physics textbook) Thus, the electrons would seem to be at a distance
from the nucleus of about 10,000 to 100,000 times the size of the nucleus
itself Thus, most of an atom is empty space With the atom being largely
empty space, it is easy to see why most a-particles go right through a
thin metal foil |
9 | 2164-2167 | Thus, the electrons would seem to be at a distance
from the nucleus of about 10,000 to 100,000 times the size of the nucleus
itself Thus, most of an atom is empty space With the atom being largely
empty space, it is easy to see why most a-particles go right through a
thin metal foil However, when a-particle happens to come near a nucleus,
the intense electric field there scatters it through a large angle |
9 | 2165-2168 | Thus, most of an atom is empty space With the atom being largely
empty space, it is easy to see why most a-particles go right through a
thin metal foil However, when a-particle happens to come near a nucleus,
the intense electric field there scatters it through a large angle The atomic
electrons, being so light, do not appreciably affect the a-particles |
9 | 2166-2169 | With the atom being largely
empty space, it is easy to see why most a-particles go right through a
thin metal foil However, when a-particle happens to come near a nucleus,
the intense electric field there scatters it through a large angle The atomic
electrons, being so light, do not appreciably affect the a-particles The scattering data shown in Fig |
9 | 2167-2170 | However, when a-particle happens to come near a nucleus,
the intense electric field there scatters it through a large angle The atomic
electrons, being so light, do not appreciably affect the a-particles The scattering data shown in Fig 12 |
9 | 2168-2171 | The atomic
electrons, being so light, do not appreciably affect the a-particles The scattering data shown in Fig 12 3 can be analysed by employing
Rutherford’s nuclear model of the atom |
9 | 2169-2172 | The scattering data shown in Fig 12 3 can be analysed by employing
Rutherford’s nuclear model of the atom As the gold foil is very thin, it
can be assumed that a-particles will suffer not more than one scattering
during their passage through it |
9 | 2170-2173 | 12 3 can be analysed by employing
Rutherford’s nuclear model of the atom As the gold foil is very thin, it
can be assumed that a-particles will suffer not more than one scattering
during their passage through it Therefore, computation of the trajectory
of an alpha-particle scattered by a single nucleus is enough |
9 | 2171-2174 | 3 can be analysed by employing
Rutherford’s nuclear model of the atom As the gold foil is very thin, it
can be assumed that a-particles will suffer not more than one scattering
during their passage through it Therefore, computation of the trajectory
of an alpha-particle scattered by a single nucleus is enough Alpha-
particles are nuclei of helium atoms and, therefore, carry two units, 2e,
of positive charge and have the mass of the helium atom |
9 | 2172-2175 | As the gold foil is very thin, it
can be assumed that a-particles will suffer not more than one scattering
during their passage through it Therefore, computation of the trajectory
of an alpha-particle scattered by a single nucleus is enough Alpha-
particles are nuclei of helium atoms and, therefore, carry two units, 2e,
of positive charge and have the mass of the helium atom The charge of
the gold nucleus is Ze, where Z is the atomic number of the atom; for
gold Z = 79 |
9 | 2173-2176 | Therefore, computation of the trajectory
of an alpha-particle scattered by a single nucleus is enough Alpha-
particles are nuclei of helium atoms and, therefore, carry two units, 2e,
of positive charge and have the mass of the helium atom The charge of
the gold nucleus is Ze, where Z is the atomic number of the atom; for
gold Z = 79 Since the nucleus of gold is about 50 times heavier than an
a-particle, it is reasonable to assume that it remains stationary
throughout the scattering process |
9 | 2174-2177 | Alpha-
particles are nuclei of helium atoms and, therefore, carry two units, 2e,
of positive charge and have the mass of the helium atom The charge of
the gold nucleus is Ze, where Z is the atomic number of the atom; for
gold Z = 79 Since the nucleus of gold is about 50 times heavier than an
a-particle, it is reasonable to assume that it remains stationary
throughout the scattering process Under these assumptions, the
trajectory of an alpha-particle can be computed employing Newton’s
second law of motion and the Coulomb’s law for electrostatic
force of repulsion between the alpha-particle and the positively
charged nucleus |
9 | 2175-2178 | The charge of
the gold nucleus is Ze, where Z is the atomic number of the atom; for
gold Z = 79 Since the nucleus of gold is about 50 times heavier than an
a-particle, it is reasonable to assume that it remains stationary
throughout the scattering process Under these assumptions, the
trajectory of an alpha-particle can be computed employing Newton’s
second law of motion and the Coulomb’s law for electrostatic
force of repulsion between the alpha-particle and the positively
charged nucleus FIGURE 12 |
9 | 2176-2179 | Since the nucleus of gold is about 50 times heavier than an
a-particle, it is reasonable to assume that it remains stationary
throughout the scattering process Under these assumptions, the
trajectory of an alpha-particle can be computed employing Newton’s
second law of motion and the Coulomb’s law for electrostatic
force of repulsion between the alpha-particle and the positively
charged nucleus FIGURE 12 3 Experimental data points (shown by
dots) on scattering of a-particles by a thin foil at
different angles obtained by Geiger and Marsden
using the setup shown in Figs |
9 | 2177-2180 | Under these assumptions, the
trajectory of an alpha-particle can be computed employing Newton’s
second law of motion and the Coulomb’s law for electrostatic
force of repulsion between the alpha-particle and the positively
charged nucleus FIGURE 12 3 Experimental data points (shown by
dots) on scattering of a-particles by a thin foil at
different angles obtained by Geiger and Marsden
using the setup shown in Figs 12 |
9 | 2178-2181 | FIGURE 12 3 Experimental data points (shown by
dots) on scattering of a-particles by a thin foil at
different angles obtained by Geiger and Marsden
using the setup shown in Figs 12 1 and
12 |
9 | 2179-2182 | 3 Experimental data points (shown by
dots) on scattering of a-particles by a thin foil at
different angles obtained by Geiger and Marsden
using the setup shown in Figs 12 1 and
12 2 |
9 | 2180-2183 | 12 1 and
12 2 Rutherford’s nuclear model predicts the solid
curve which is seen to be in good agreement with
experiment |
9 | 2181-2184 | 1 and
12 2 Rutherford’s nuclear model predicts the solid
curve which is seen to be in good agreement with
experiment Rationalised 2023-24
Physics
294
EXAMPLE 12 |
9 | 2182-2185 | 2 Rutherford’s nuclear model predicts the solid
curve which is seen to be in good agreement with
experiment Rationalised 2023-24
Physics
294
EXAMPLE 12 1
The magnitude of this force is
2
0
(2 )(
)
1
4
e
Ze
F
r
ε
=
π
(12 |
9 | 2183-2186 | Rutherford’s nuclear model predicts the solid
curve which is seen to be in good agreement with
experiment Rationalised 2023-24
Physics
294
EXAMPLE 12 1
The magnitude of this force is
2
0
(2 )(
)
1
4
e
Ze
F
r
ε
=
π
(12 1)
where r is the distance between the a-particle and the nucleus |
9 | 2184-2187 | Rationalised 2023-24
Physics
294
EXAMPLE 12 1
The magnitude of this force is
2
0
(2 )(
)
1
4
e
Ze
F
r
ε
=
π
(12 1)
where r is the distance between the a-particle and the nucleus The force
is directed along the line joining the a-particle and the nucleus |
9 | 2185-2188 | 1
The magnitude of this force is
2
0
(2 )(
)
1
4
e
Ze
F
r
ε
=
π
(12 1)
where r is the distance between the a-particle and the nucleus The force
is directed along the line joining the a-particle and the nucleus The
magnitude and direction of the force on an a-particle continuously
changes as it approaches the nucleus and recedes away from it |
9 | 2186-2189 | 1)
where r is the distance between the a-particle and the nucleus The force
is directed along the line joining the a-particle and the nucleus The
magnitude and direction of the force on an a-particle continuously
changes as it approaches the nucleus and recedes away from it 12 |
9 | 2187-2190 | The force
is directed along the line joining the a-particle and the nucleus The
magnitude and direction of the force on an a-particle continuously
changes as it approaches the nucleus and recedes away from it 12 2 |
9 | 2188-2191 | The
magnitude and direction of the force on an a-particle continuously
changes as it approaches the nucleus and recedes away from it 12 2 1 Alpha-particle trajectory
The trajectory traced by an a-particle depends on the impact parameter,
b of collision |
9 | 2189-2192 | 12 2 1 Alpha-particle trajectory
The trajectory traced by an a-particle depends on the impact parameter,
b of collision The impact parameter is the perpendicular distance of the
initial velocity vector of the a-particle from the centre of the nucleus (Fig |
9 | 2190-2193 | 2 1 Alpha-particle trajectory
The trajectory traced by an a-particle depends on the impact parameter,
b of collision The impact parameter is the perpendicular distance of the
initial velocity vector of the a-particle from the centre of the nucleus (Fig 12 |
9 | 2191-2194 | 1 Alpha-particle trajectory
The trajectory traced by an a-particle depends on the impact parameter,
b of collision The impact parameter is the perpendicular distance of the
initial velocity vector of the a-particle from the centre of the nucleus (Fig 12 4) |
9 | 2192-2195 | The impact parameter is the perpendicular distance of the
initial velocity vector of the a-particle from the centre of the nucleus (Fig 12 4) A given beam of a-particles has a
distribution of impact parameters b, so that
the beam is scattered in various directions
with different probabilities (Fig |
9 | 2193-2196 | 12 4) A given beam of a-particles has a
distribution of impact parameters b, so that
the beam is scattered in various directions
with different probabilities (Fig 12 |
9 | 2194-2197 | 4) A given beam of a-particles has a
distribution of impact parameters b, so that
the beam is scattered in various directions
with different probabilities (Fig 12 4) |
9 | 2195-2198 | A given beam of a-particles has a
distribution of impact parameters b, so that
the beam is scattered in various directions
with different probabilities (Fig 12 4) (In
a beam, all particles have nearly same
kinetic energy |
9 | 2196-2199 | 12 4) (In
a beam, all particles have nearly same
kinetic energy ) It is seen that an a-particle
close to the nucleus (small impact
parameter) suffers large scattering |
9 | 2197-2200 | 4) (In
a beam, all particles have nearly same
kinetic energy ) It is seen that an a-particle
close to the nucleus (small impact
parameter) suffers large scattering In case
of head-on collision, the impact parameter
is minimum and the a-particle rebounds
back (q @ p) |
9 | 2198-2201 | (In
a beam, all particles have nearly same
kinetic energy ) It is seen that an a-particle
close to the nucleus (small impact
parameter) suffers large scattering In case
of head-on collision, the impact parameter
is minimum and the a-particle rebounds
back (q @ p) For a large impact parameter,
the a-particle goes nearly undeviated and
has a small deflection (q @ 0) |
9 | 2199-2202 | ) It is seen that an a-particle
close to the nucleus (small impact
parameter) suffers large scattering In case
of head-on collision, the impact parameter
is minimum and the a-particle rebounds
back (q @ p) For a large impact parameter,
the a-particle goes nearly undeviated and
has a small deflection (q @ 0) The fact that only a small fraction of the
number of incident particles rebound back
indicates that the number of a-particles
undergoing head on collision is small |
9 | 2200-2203 | In case
of head-on collision, the impact parameter
is minimum and the a-particle rebounds
back (q @ p) For a large impact parameter,
the a-particle goes nearly undeviated and
has a small deflection (q @ 0) The fact that only a small fraction of the
number of incident particles rebound back
indicates that the number of a-particles
undergoing head on collision is small This,
in turn, implies that the mass and positive charge of the atom is
concentrated in a small volume |
9 | 2201-2204 | For a large impact parameter,
the a-particle goes nearly undeviated and
has a small deflection (q @ 0) The fact that only a small fraction of the
number of incident particles rebound back
indicates that the number of a-particles
undergoing head on collision is small This,
in turn, implies that the mass and positive charge of the atom is
concentrated in a small volume Rutherford scattering therefore, is a
powerful way to determine an upper limit to the size of the nucleus |
9 | 2202-2205 | The fact that only a small fraction of the
number of incident particles rebound back
indicates that the number of a-particles
undergoing head on collision is small This,
in turn, implies that the mass and positive charge of the atom is
concentrated in a small volume Rutherford scattering therefore, is a
powerful way to determine an upper limit to the size of the nucleus FIGURE 12 |
9 | 2203-2206 | This,
in turn, implies that the mass and positive charge of the atom is
concentrated in a small volume Rutherford scattering therefore, is a
powerful way to determine an upper limit to the size of the nucleus FIGURE 12 4 Trajectory of a-particles in the
coulomb field of a target nucleus |
9 | 2204-2207 | Rutherford scattering therefore, is a
powerful way to determine an upper limit to the size of the nucleus FIGURE 12 4 Trajectory of a-particles in the
coulomb field of a target nucleus The impact
parameter, b and scattering angle q
are also depicted |
9 | 2205-2208 | FIGURE 12 4 Trajectory of a-particles in the
coulomb field of a target nucleus The impact
parameter, b and scattering angle q
are also depicted Example 12 |
9 | 2206-2209 | 4 Trajectory of a-particles in the
coulomb field of a target nucleus The impact
parameter, b and scattering angle q
are also depicted Example 12 1 In the Rutherford’s nuclear model of the atom, the
nucleus (radius about 10–15 m) is analogous to the sun about which
the electron move in orbit (radius » 10–10 m) like the earth orbits
around the sun |
9 | 2207-2210 | The impact
parameter, b and scattering angle q
are also depicted Example 12 1 In the Rutherford’s nuclear model of the atom, the
nucleus (radius about 10–15 m) is analogous to the sun about which
the electron move in orbit (radius » 10–10 m) like the earth orbits
around the sun If the dimensions of the solar system had the same
proportions as those of the atom, would the earth be closer to or
farther away from the sun than actually it is |
9 | 2208-2211 | Example 12 1 In the Rutherford’s nuclear model of the atom, the
nucleus (radius about 10–15 m) is analogous to the sun about which
the electron move in orbit (radius » 10–10 m) like the earth orbits
around the sun If the dimensions of the solar system had the same
proportions as those of the atom, would the earth be closer to or
farther away from the sun than actually it is The radius of earth’s
orbit is about 1 |
9 | 2209-2212 | 1 In the Rutherford’s nuclear model of the atom, the
nucleus (radius about 10–15 m) is analogous to the sun about which
the electron move in orbit (radius » 10–10 m) like the earth orbits
around the sun If the dimensions of the solar system had the same
proportions as those of the atom, would the earth be closer to or
farther away from the sun than actually it is The radius of earth’s
orbit is about 1 5 ´ 1011 m |
9 | 2210-2213 | If the dimensions of the solar system had the same
proportions as those of the atom, would the earth be closer to or
farther away from the sun than actually it is The radius of earth’s
orbit is about 1 5 ´ 1011 m The radius of sun is taken as 7 ´ 108 m |
9 | 2211-2214 | The radius of earth’s
orbit is about 1 5 ´ 1011 m The radius of sun is taken as 7 ´ 108 m Solution The ratio of the radius of electron’s orbit to the radius of
nucleus is (10–10 m)/(10–15 m) = 105, that is, the radius of the electron’s
orbit is 105 times larger than the radius of nucleus |
9 | 2212-2215 | 5 ´ 1011 m The radius of sun is taken as 7 ´ 108 m Solution The ratio of the radius of electron’s orbit to the radius of
nucleus is (10–10 m)/(10–15 m) = 105, that is, the radius of the electron’s
orbit is 105 times larger than the radius of nucleus If the radius of
the earth’s orbit around the sun were 105 times larger than the radius
of the sun, the radius of the earth’s orbit would be 105 ´ 7 ´ 108 m =
7 ´ 1013 m |
9 | 2213-2216 | The radius of sun is taken as 7 ´ 108 m Solution The ratio of the radius of electron’s orbit to the radius of
nucleus is (10–10 m)/(10–15 m) = 105, that is, the radius of the electron’s
orbit is 105 times larger than the radius of nucleus If the radius of
the earth’s orbit around the sun were 105 times larger than the radius
of the sun, the radius of the earth’s orbit would be 105 ´ 7 ´ 108 m =
7 ´ 1013 m This is more than 100 times greater than the actual
orbital radius of earth |
9 | 2214-2217 | Solution The ratio of the radius of electron’s orbit to the radius of
nucleus is (10–10 m)/(10–15 m) = 105, that is, the radius of the electron’s
orbit is 105 times larger than the radius of nucleus If the radius of
the earth’s orbit around the sun were 105 times larger than the radius
of the sun, the radius of the earth’s orbit would be 105 ´ 7 ´ 108 m =
7 ´ 1013 m This is more than 100 times greater than the actual
orbital radius of earth Thus, the earth would be much farther away
from the sun |
9 | 2215-2218 | If the radius of
the earth’s orbit around the sun were 105 times larger than the radius
of the sun, the radius of the earth’s orbit would be 105 ´ 7 ´ 108 m =
7 ´ 1013 m This is more than 100 times greater than the actual
orbital radius of earth Thus, the earth would be much farther away
from the sun It implies that an atom contains a much greater fraction of empty
space than our solar system does |
9 | 2216-2219 | This is more than 100 times greater than the actual
orbital radius of earth Thus, the earth would be much farther away
from the sun It implies that an atom contains a much greater fraction of empty
space than our solar system does Rationalised 2023-24
295
Atoms
EXAMPLE 12 |
9 | 2217-2220 | Thus, the earth would be much farther away
from the sun It implies that an atom contains a much greater fraction of empty
space than our solar system does Rationalised 2023-24
295
Atoms
EXAMPLE 12 2
Example 12 |
9 | 2218-2221 | It implies that an atom contains a much greater fraction of empty
space than our solar system does Rationalised 2023-24
295
Atoms
EXAMPLE 12 2
Example 12 2 In a Geiger-Marsden experiment, what is the distance
of closest approach to the nucleus of a 7 |
9 | 2219-2222 | Rationalised 2023-24
295
Atoms
EXAMPLE 12 2
Example 12 2 In a Geiger-Marsden experiment, what is the distance
of closest approach to the nucleus of a 7 7 MeV a-particle before it
comes momentarily to rest and reverses its direction |
9 | 2220-2223 | 2
Example 12 2 In a Geiger-Marsden experiment, what is the distance
of closest approach to the nucleus of a 7 7 MeV a-particle before it
comes momentarily to rest and reverses its direction Solution The key idea here is that throughout the scattering process,
the total mechanical energy of the system consisting of an a-particle
and a gold nucleus is conserved |
9 | 2221-2224 | 2 In a Geiger-Marsden experiment, what is the distance
of closest approach to the nucleus of a 7 7 MeV a-particle before it
comes momentarily to rest and reverses its direction Solution The key idea here is that throughout the scattering process,
the total mechanical energy of the system consisting of an a-particle
and a gold nucleus is conserved The system’s initial mechanical
energy is Ei, before the particle and nucleus interact, and it is equal
to its mechanical energy Ef when the a-particle momentarily stops |
9 | 2222-2225 | 7 MeV a-particle before it
comes momentarily to rest and reverses its direction Solution The key idea here is that throughout the scattering process,
the total mechanical energy of the system consisting of an a-particle
and a gold nucleus is conserved The system’s initial mechanical
energy is Ei, before the particle and nucleus interact, and it is equal
to its mechanical energy Ef when the a-particle momentarily stops The initial energy Ei is just the kinetic energy K of the incoming
a- particle |
9 | 2223-2226 | Solution The key idea here is that throughout the scattering process,
the total mechanical energy of the system consisting of an a-particle
and a gold nucleus is conserved The system’s initial mechanical
energy is Ei, before the particle and nucleus interact, and it is equal
to its mechanical energy Ef when the a-particle momentarily stops The initial energy Ei is just the kinetic energy K of the incoming
a- particle The final energy Ef is just the electric potential energy U
of the system |
9 | 2224-2227 | The system’s initial mechanical
energy is Ei, before the particle and nucleus interact, and it is equal
to its mechanical energy Ef when the a-particle momentarily stops The initial energy Ei is just the kinetic energy K of the incoming
a- particle The final energy Ef is just the electric potential energy U
of the system The potential energy U can be calculated from
Eq |
9 | 2225-2228 | The initial energy Ei is just the kinetic energy K of the incoming
a- particle The final energy Ef is just the electric potential energy U
of the system The potential energy U can be calculated from
Eq (12 |
9 | 2226-2229 | The final energy Ef is just the electric potential energy U
of the system The potential energy U can be calculated from
Eq (12 1) |
9 | 2227-2230 | The potential energy U can be calculated from
Eq (12 1) Let d be the centre-to-centre distance between the a-particle and
the gold nucleus when the a-particle is at its stopping point |
9 | 2228-2231 | (12 1) Let d be the centre-to-centre distance between the a-particle and
the gold nucleus when the a-particle is at its stopping point Then
we can write the conservation of energy Ei = Ef as
2
0
0
1
(2 )(
)
2
4
4
e Ze
Ze
K
d
d
ε
ε
=
=
π
π
Thus the distance of closest approach d is given by
2
0
2
4
Ze
d
εK
=
π
The maximum kinetic energy found in a-particles of natural origin is
7 |
9 | 2229-2232 | 1) Let d be the centre-to-centre distance between the a-particle and
the gold nucleus when the a-particle is at its stopping point Then
we can write the conservation of energy Ei = Ef as
2
0
0
1
(2 )(
)
2
4
4
e Ze
Ze
K
d
d
ε
ε
=
=
π
π
Thus the distance of closest approach d is given by
2
0
2
4
Ze
d
εK
=
π
The maximum kinetic energy found in a-particles of natural origin is
7 7 MeV or 1 |
9 | 2230-2233 | Let d be the centre-to-centre distance between the a-particle and
the gold nucleus when the a-particle is at its stopping point Then
we can write the conservation of energy Ei = Ef as
2
0
0
1
(2 )(
)
2
4
4
e Ze
Ze
K
d
d
ε
ε
=
=
π
π
Thus the distance of closest approach d is given by
2
0
2
4
Ze
d
εK
=
π
The maximum kinetic energy found in a-particles of natural origin is
7 7 MeV or 1 2 × 10–12 J |
9 | 2231-2234 | Then
we can write the conservation of energy Ei = Ef as
2
0
0
1
(2 )(
)
2
4
4
e Ze
Ze
K
d
d
ε
ε
=
=
π
π
Thus the distance of closest approach d is given by
2
0
2
4
Ze
d
εK
=
π
The maximum kinetic energy found in a-particles of natural origin is
7 7 MeV or 1 2 × 10–12 J Since 1/4pe0 = 9 |
9 | 2232-2235 | 7 MeV or 1 2 × 10–12 J Since 1/4pe0 = 9 0 × 109 N m2/C2 |
9 | 2233-2236 | 2 × 10–12 J Since 1/4pe0 = 9 0 × 109 N m2/C2 Therefore
with e = 1 |
9 | 2234-2237 | Since 1/4pe0 = 9 0 × 109 N m2/C2 Therefore
with e = 1 6 × 10–19 C, we have,
9
2
2
–19
2
12
(2)(9 |
9 | 2235-2238 | 0 × 109 N m2/C2 Therefore
with e = 1 6 × 10–19 C, we have,
9
2
2
–19
2
12
(2)(9 0
10 Nm /
)(1 |
9 | 2236-2239 | Therefore
with e = 1 6 × 10–19 C, we have,
9
2
2
–19
2
12
(2)(9 0
10 Nm /
)(1 6
10
) Z
1 |
9 | 2237-2240 | 6 × 10–19 C, we have,
9
2
2
–19
2
12
(2)(9 0
10 Nm /
)(1 6
10
) Z
1 2
10
J
C
C
d
−
×
×
=
×
= 3 |
9 | 2238-2241 | 0
10 Nm /
)(1 6
10
) Z
1 2
10
J
C
C
d
−
×
×
=
×
= 3 84 × 10–16 Z m
The atomic number of foil material gold is Z = 79, so that
d (Au) = 3 |
Subsets and Splits