Download Recitation on atomic structure Solution

Modern Physics: PHY-104
Spring Semester 2013
Recitation on atomic structure
Solution
1. In the Bohr model, compare the magnitudes of the electron’s kinetic and potential
energies in orbit. What about their signs? What does this mean?
Answer 1:
When an electron moves in an orbit, the Coulomb force acting on it balances the
centripetal acceleration, which keeps the electron in its orbit. i.e.,
1 Ze2
mv 2
=
4πε0 r2
r
1 Ze2
= mv 2 ,
4πε0 r
(1)
where v is the speed of electron and r is the radius of the orbit. The kinetic energy of
this electron is,
1 2
mv
2(
)
1 Ze2
1
=
2 4πε0 r
1
= − (V )
2
1
|K| = V.
2
K =
(using equation1)
2
Ze
where V = − 4πε
is the potential energy of this electron. This implies that the
0r
magnitude of the kinetic energy of an electron is half the magnitude of potential
energy. Now, the total energy of this electron is,
E = K +V
1
= − V +V
2
1
= V,
2
which immediately implies that if the potential energy is negative, the total energy
will also be negative, which is exactly the case in hydrogen atom. Negative energies
correspond to electrons bound to the nucleus, while positive energies imply that the
electron is free from the attractive pull of the nucleus, the atom is ionized.
Date: 4 March, 2013
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Modern Physics: PHY-104
Spring Semester 2013
2. In stars, the Pickering series is found in the He+ spectrum. It is emitted when the
electron in He+ jumps from higher levels into the level with n = 4.
(a) State the exact formula for the wavelength of lines belonging to this series.
(b) In what region of the spectrum is the series?
(c) Find the wavelength of the series limit.
(d) Find the ionization potential, if He+ is in the ground state, in electron volts.
Answer 2:
When an electron makes a transition from an initial quantum state (ni ) to a final
quantum state (nf ), the change in internal energy is,
∆E = Ei − Ef = −
E◦ E◦
+ 2,
n2i
nf
where
E◦ = +
me4 Z 2
= +13.6Z eV.
32π 2 ε2◦ k 2
In the case of Pickering series found in He+ spectrum, the electron jumps from higher
levels into nf = 4. Therefore,
(
)
1
1
1
2
= R∞ Z
−
λ
16 n2i
( 2
)
2 ni − 16
= R∞ Z
16n2i
16n2
1
λ = 2 i
·
ni − 16 R∞ Z 2
(b) If we take ni = 5, i.e., the transition is taking place from n = 5 to n = 4 energy
level, the corresponding wavelength is,
(
)
16n2i
1
λ =
2
n − 16 R∞ Z 2
( i
)
16
1
=
16
1 − n2 10973731.6 × 4
i
= 1012 nm,
which is in the infrared region.
(c) The series limit occurs when an electron makes a transition from the continuum,
ni → ∞ to nf → 4. Therefore, the wavelength of series limit corresponds to the
Date: 4 March, 2013
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Modern Physics: PHY-104
Spring Semester 2013
transition from n = 9 to n = 4, for which,
(
)
16n2i
1
λ =
2
n − 16 R∞ Z 2
( i
)
16 × 81
1
=
81 − 16 R∞ Z 2
16
=
R∞ Z 2
= 454.2 nm,
which is in the ultraviolet region.
(d) The ionization potential of helium (He+ having Z = 2) is the energy required
to completely remove the electron, i.e., to take it from the ground state ni = 1 to
continuum nf → ∞.
)
(
1
1
1
2
= R∞ Z
− 2
λ
n2i
nf
(
)
1
2
= R∞ Z 1 −
∞
1
⇒λ =
R∞ Z 2
= 22.8 nm.
This corresponds to the ionization energy,
hc
6.67 × 10−34 × 3 × 108
=
= 8.8 × 10−18 J ≈ 55 eV.
λ
22.8 × 10−9
3. A muon is a particle with a charge equal to that of an electron and a mass equal to 207
times the mass of an electron. Muonic lead is formed when
208
Pb captures a muon to
replace an electron. Assume that the muon moves in such a small orbit that it “sees”
a nuclear charge of Z = 82. According to the Bohr theory, what are the radius and
energy of the ground state of muonic lead?.
Answer 3:
The effective mass of muon, when it moves in a small orbit and “sees” a nuclear charge
of Z = 82, is given by the following expression,
µ =
Date: 4 March, 2013
mµ M
,
mµ + M
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Modern Physics: PHY-104
Spring Semester 2013
where mµ = 207me is the mass of muon and M = 82mp is the mass of nucleus that it
sees. Therefore,
(207 me )(82 mp )
207 me + 82 mp
(207 × 9.11 × 10−31 kg)(82 × 1.6 × 10−27 kg)
=
207 × 9.11 × 10−31 kg + (82 × 1.6 × 10−27 kg)
= 1.87 × 10−28 kg.
µ =
According to Bohr theory, the radius is,
r = 4πε0
n2 ~2
µZe2
1
12 (1.054 × 10−34 )2
9 × 109 (1.87 × 10−28 )82(1.6 × 10−19 )2
= 3.1 × 10−15 m.
=
The energy of the ground state of muonic lead is,
me4 Z 2
(4πε0 )2 2~2 n2
(1.87 × 10−27 )82(1.6 × 10−19 )4
= −
(1.1 × 10−10 )2 2(1.054 × 10−34 )2 (1)2
= −18.5 MeV.
En = −
4. One of the earliest experiments to show that p = γmv (rather than p = mv) was that
of Neumann. [G. Neumann, Ann. Physik 45:529 (1914)].
V
+ d
Radium
source
e
E
l
B
B
r
Photo plate
detector
y
Electron
impact point
Center of curvature
The Neumann apparatus
The apparatus shown in Figure is identical to Thomsons except that the source of highspeed electrons is a radioactive radium source and the magnetic field B is arranged to
Date: 4 March, 2013
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Modern Physics: PHY-104
Spring Semester 2013
act on the electron over its entire trajectory from source to detector. The combined
electric and magnetic fields act as a velocity selector, only passing electrons with
speed v, where v = V /Bd, while in the region where there is only a magnetic field
the electron moves in a circle of radius r, with r given by p = Bre. This latter region
(E = 0, B =constant) acts as a momentum selector because electrons with larger
momenta have paths with larger radii.
(a) Show that the radius of the circle described by the electron is given by r = (ℓ2 +
y 2 )/2y.
(b) Typical values for the Neumann experiment were d = 2.51 × 10−4 m, B = 0.0177
T, and ℓ = 0.0247 m. For V = 1060 V, y, the most critical value, was measured to
be 0.0024 ± 0.0005 m. Show that these values disagree with the y value calculated
from p = mv but agree with the y value calculated from p = γmv within experimental
error. (Hint: Find v from Equation 4.6, use mv = Bre or γmv = Bre to find r, and
use r to find y.)
Answer 4:
(a) Refer to the diagram shown below,
+ d
V
Radium
source
e
E
l
B
Photo plate
detector
Photo plate
detector
l
P
y
B
r
Center of curvature
y
r
l
r-y
θ
C
Electron
impact point Center of
curvature
r
Q
Electron
impact point
Suppose the electron reaches from point P to point Q in τ seconds, the angle θ is
shown. From the geometrical construction,
r−y
, and
r
ℓ
sin θ = .
r
cos θ =
Date: 4 March, 2013
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Modern Physics: PHY-104
Spring Semester 2013
Since cos2 θ + sin2 θ = 1,
(
)2 ( )2
r−y
ℓ
⇒
+
= 1
r
r
( 2
) ( 2)
r + y 2 − 2ry
ℓ
+
= 1
r2
r2
r2 + y 2 − 2ry + ℓ2 = r2
y 2 − 2ry + ℓ2 = 1
(2)
2ry = ℓ2 + y 2
ℓ2 + y 2
r =
,
2y
as required.
(b) We are given that,
d = 2.51 × 10−4 m
B = 0.0177 T
ℓ = 0.0247 m
V = 1060 V
ymeasured = (0.0024 ± 0.0005)m.
Using non-relativistic expressions:
⃗ + B)
⃗ field regions,
In the (E
Electric force = Magnetic force
eE = evB
E
V
v =
=
,
B
Bd
(3)
⃗ only region,
whereas in the B
p = mv = eBr
mv
⇒r =
eB )(
(
)
m
V
=
eB
Bd
mV
.
r =
eB 2 d
(4)
from (3)
Substitution of the given values yields,
r=
Date: 4 March, 2013
9.1 × 10−31 kg × 1060 V
,
1.6 × 10−19 C × (0.0177 T)2 × 2.51 × 10−4 m
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Modern Physics: PHY-104
Spring Semester 2013
where 1Tesla = V s/m2 .
9.1 × 10−31 kg × 1060 V
1.6 × 10−19 C × (0.0177 Vs/m2 )2 × 2.51 × 10−4 m
kgm3
= 0.0767 ·
,
CVs2
r =
where kgm2 /s2 = 1 J and 1 V = J/C.
⇒ r = 0.0767 m.
Now we have the equation (2),
2yr = ℓ2 + y 2
(5)
y 2 − 2ry + ℓ2 = 0,
whose solution is,
√
(−2r)2 − 4(1)(ℓ2 )
2(1)
√
2r ± 4r2 − 4ℓ2
=
√ 2
= r ± r2 − ℓ2 .
y =
−(−2r) ±
(6)
Substitution of the values of r and ℓ yields,
y = 0.0767 m ±
√
(0.0767 m)2 − (0.0247 m)2
= 0.0767 m ± 0.0726 m
= (0.1493, 0.0041) m.
None of these values match the experimental results.
Using relativistic expressions:
From equation (3),
v=
V
.
Bd
By substituting given values we obtain,
v=
Date: 4 March, 2013
1060 V
= 2.3859 × 108 m/s.
0.0177 T × 2.51 × 10−4 m
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Modern Physics: PHY-104
Spring Semester 2013
This is 80% of the speed of light and the relativistic expression for the momentum
must be used. Using γ = eBr instead of equation (4), we obtain,
r = γ × 0.0767 m,
where relativistic mass mr in terms of rest mass m◦ is given by,
mr = γm◦ = √
⇒γ = √
m◦
1 − v 2 /c2
1
1 − v 2 /c2
1
= √
1 − [(2.3859 × 108 m/s)2 /(3 × 108 m/s)2 ]
= 1.64957688
⇒ r = 1.64957688 × 0.0767 m = 0.1266 m.
Inserting this value of r and given value of ℓ in equation (6), yields,
y = r±
√
r2 − ℓ2
√
= 0.1266 m ± (0.1266 m)2 − (0.0247 m)2
= (0.25675, 0.0025) m.
The latter value matches the experimental result to within experimental accuracy.
Date: 4 March, 2013
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