is for the 2p states with m = #1 are numerical) of the 0 eq made difficult because th singularities at 0 = 0 an nominator makes the firs T. Indeed this is exactly able solutions for most v you have access to softwa tial equations numericall to the acceptable and (a) Write down the diffe m = 0 and l = 2. Solv boundary conditions (7 Plot your result for 0 solution looks perfectly a boundary conditions 0(a and explain why this sol ceptable. (b) Repeat p 1.75. Explain why bo unacceptable; that is, ther for these values of m and 21.t1= R2p(r)sin0 eti (b) Prove that the sum of these two wave functions is the 2p, wave function (times an uninteresting factor of 2) and that the difference is the 2p, function (times 2i). Hint: Rewrite ei as cos d t i sin , and remember the relations for x and y in terms of r,0, in Fig. 8.11. SECTION 8.9 (Shells) 8.48 Consider the radial probability density P(r) for the ground state of hydrogen, as given by Eq. (8.85). By finding where P(r) is maximum, find the most proba- ble radius for this state. 8.49 Using the wave function (8.108) given in Problem 8.40, write doWn the radial probability density for a hy- drogen atom in a state with / = n - 1. Find the most probable radius. Notice that in this case (with / equal to its maximum possible value, / = n- 1) the quan- tum mechanical answer agrees with the Bohr model. which we would fin comes from the factor 4mr in the volume of the spherical shell (8.82). means that when we discuss the probability of different distances opposed to different positions), large distances are more heavily weighted, jus because larger r corresponds to larger spherical shells, with more volume than same state). the electron Afn r those with small r. For the ground state of hydrogen, with wave function (8.81), the radial If we substit probability density is (8.85 Pis(r) = 4T A2re r/ap T This is plotted in Fig. 8.18.* Perhaps its most striking property is that is maximum is at r = ag. That is, the most probable distance between the ele tron and proton in the 1s state is the Bohr radius aB. Thus, although quantur The integral mechanics gives a very different picture of the hydrogen atom (with the ele 1s tron's probability density spread continuously through space), it agrees exact ) with the Bohr model as to the electron's most probable radius in the groun ity state. Note that th with the pot se of Armed with the radial density P(r), one can calculate several impe tant properties of the atom. Problems 8.37 and 8.41 to 8.43 contain s examples, and here is another The 2s Way
is for the 2p states with m = #1 are numerical) of the 0 eq made difficult because th singularities at 0 = 0 an nominator makes the firs T. Indeed this is exactly able solutions for most v you have access to softwa tial equations numericall to the acceptable and (a) Write down the diffe m = 0 and l = 2. Solv boundary conditions (7 Plot your result for 0 solution looks perfectly a boundary conditions 0(a and explain why this sol ceptable. (b) Repeat p 1.75. Explain why bo unacceptable; that is, ther for these values of m and 21.t1= R2p(r)sin0 eti (b) Prove that the sum of these two wave functions is the 2p, wave function (times an uninteresting factor of 2) and that the difference is the 2p, function (times 2i). Hint: Rewrite ei as cos d t i sin , and remember the relations for x and y in terms of r,0, in Fig. 8.11. SECTION 8.9 (Shells) 8.48 Consider the radial probability density P(r) for the ground state of hydrogen, as given by Eq. (8.85). By finding where P(r) is maximum, find the most proba- ble radius for this state. 8.49 Using the wave function (8.108) given in Problem 8.40, write doWn the radial probability density for a hy- drogen atom in a state with / = n - 1. Find the most probable radius. Notice that in this case (with / equal to its maximum possible value, / = n- 1) the quan- tum mechanical answer agrees with the Bohr model. which we would fin comes from the factor 4mr in the volume of the spherical shell (8.82). means that when we discuss the probability of different distances opposed to different positions), large distances are more heavily weighted, jus because larger r corresponds to larger spherical shells, with more volume than same state). the electron Afn r those with small r. For the ground state of hydrogen, with wave function (8.81), the radial If we substit probability density is (8.85 Pis(r) = 4T A2re r/ap T This is plotted in Fig. 8.18.* Perhaps its most striking property is that is maximum is at r = ag. That is, the most probable distance between the ele tron and proton in the 1s state is the Bohr radius aB. Thus, although quantur The integral mechanics gives a very different picture of the hydrogen atom (with the ele 1s tron's probability density spread continuously through space), it agrees exact ) with the Bohr model as to the electron's most probable radius in the groun ity state. Note that th with the pot se of Armed with the radial density P(r), one can calculate several impe tant properties of the atom. Problems 8.37 and 8.41 to 8.43 contain s examples, and here is another The 2s Way
Related questions
Question
For Problem 8.48, am I to find these two things via taking the derivatives?
Transcribed Image Text:is for the 2p states with m = #1 are
numerical) of the 0 eq
made difficult because th
singularities at 0 = 0 an
nominator makes the firs
T. Indeed this is exactly
able solutions for most v
you have access to softwa
tial equations numericall
to the acceptable and
(a) Write down the diffe
m = 0 and l = 2. Solv
boundary conditions (7
Plot your result for 0
solution looks perfectly a
boundary conditions 0(a
and explain why this sol
ceptable. (b) Repeat p
1.75. Explain why bo
unacceptable; that is, ther
for these values of m and
21.t1= R2p(r)sin0 eti
(b) Prove that the sum of these two wave functions is
the 2p, wave function (times an uninteresting factor of
2) and that the difference is the 2p, function (times 2i).
Hint: Rewrite ei as cos d t i sin , and remember
the relations for x and y in terms of r,0, in Fig. 8.11.
SECTION 8.9 (Shells)
8.48 Consider the radial probability density P(r) for the
ground state of hydrogen, as given by Eq. (8.85). By
finding where P(r) is maximum, find the most proba-
ble radius for this state.
8.49 Using the wave function (8.108) given in Problem
8.40, write doWn the radial probability density for a hy-
drogen atom in a state with / = n - 1. Find the most
probable radius. Notice that in this case (with / equal
to its maximum possible value, / = n- 1) the quan-
tum mechanical answer agrees with the Bohr model.
Transcribed Image Text:which
we would fin
comes from the factor 4mr in the volume of the spherical shell (8.82).
means that when we discuss the probability of different distances
opposed to different positions), large distances are more heavily weighted, jus
because larger r corresponds to larger spherical shells, with more volume than
same state).
the electron
Afn
r
those with small r.
For the ground state of hydrogen, with wave function (8.81), the radial
If we substit
probability density is
(8.85
Pis(r) = 4T A2re r/ap
T
This is plotted in Fig. 8.18.* Perhaps its most striking property is that is
maximum is at r = ag. That is, the most probable distance between the ele
tron and proton in the 1s state is the Bohr radius aB. Thus, although quantur
The integral
mechanics gives a very different picture of the hydrogen atom (with the ele
1s
tron's probability density spread continuously through space), it agrees exact
)
with the Bohr model as to the electron's most probable radius in the groun
ity
state.
Note that th
with the pot
se
of
Armed with the radial density P(r), one can calculate several impe
tant properties of the atom. Problems 8.37 and 8.41 to 8.43 contain s
examples, and here is another
The 2s Way
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