Instructor solution manual to accompany physical chemistry 7th ed by peter atkins chap11

Solutions to exercises Discussion questions E11.1b A successful theory of black-body radiation must be able to explain the energy density distribution of the radiation as a function of w

Solutions to exercises

Discussion questions

E11.1(b) A successful theory of black-body radiation must be able to explain the energy density distribution of

the radiation as a function of wavelength, in particular, the observed drop to zero asλ → 0 Classical

theory predicts the opposite However, if we assume, as did Planck, that the energy of the oscillators that constitute electromagnetic radiation are quantized according to the relationE = nhν = nhc/λ,

we see that at short wavelengths the energy of the oscillators is very large This energy is too large for the walls to supply it, so the short-wavelength oscillators remain unexcited The effect of quantization

is to reduce the contribution to the total energy emitted by the black-body from the high-energy short-wavelength oscillators, for they cannot be sufficiently excited with the energy available

E11.2(b) In quantum mechanics all dynamical properties of a physical system have associated with them a

corresponding operator The system itself is described by a wavefunction The observable properties

of the system can be obtained in one of two ways from the wavefunction depending upon whether or not the wavefunction is an eigenfunction of the operator

When the function representing the state of the system is an eigenfunction of the operator, we

solve the eigenvalue equation (eqn 11.30)

 = ω

in order to obtain the observable values,ω, of the dynamical properties.

When the function is not an eigenfunction of, we can only find the average or expectation value

of dynamical properties by performing the integration shown in eqn 11.39

 =

∗ dτ.

E11.3(b) No answer

Numerical exercises

E11.4(b) The power is equal to the excitanceM times the emitting area

P = MA = σT4(2πrl)

= (5.67 × 10−8W m−2K−4) × (3300 K)4× (2π) × (0.12 × 10−3m) × (5.0 × 10−2m)

= 2.5 × 102W

Comment This could be a 250 W incandescent light bulb.

E11.5(b) Wien’s displacement law is

T λmax= c2/5 so λmax= c2

5T =

1.44 × 10−2m K

5(2500 K) = 1.15 × 10−6m= 1.15 µm

E11.6(b) The de Broglie relation is

h

h

6.626 × 10−34J s

(1.675 × 10−27kg) × (3.0 × 10−2m)

v = 1.3 × 10−5m s−1

E11.7(b) The de Broglie relation is

h

h

6.626 × 10−34J s

(9.11 × 10−31kg) × (0.45 × 10−9m)

v = 1.6 × 106m s−1

E11.8(b) The momentum of a photon is

p = h

λ =

6.626 × 10−34J s

350× 10−9m = 1.89 × 10−27kg m s−1

The momentum of a particle is

1.89 × 10−27kg m s−1

2(1.0078 × 10−3kg mol−1/6.022 × 1023mol−1)

v = 0.565 m s−1

E11.9(b) The energy of the photon is equal to the ionization energy plus the kinetic energy of the ejected

electron

Ephoton= Eionize+ Eelectron so hc

λ = Eionize+1

2mv2

andλ = hc

Eionize+1

2mv2 = (6.626 × 10−34J s) × (2.998 × 108m s−1)

5.12 × 10−18J+1

2(9.11 × 10−31kg) × (345 × 103m s−1)2

= 3.48 × 10−8m= 38.4 nm

E11.10(b) The uncertainty principle is

px ≥ 1

2h¯

so the minimum uncertainty in position is

x = h¯

2p =

¯

h

2mv =

1.0546 × 10−34J s

2(9.11 × 10−31kg) × (0.000 010) × (995 × 103m s−1)

= 5.8 × 10−6m

E11.11(b) E = hν = hc λ; E(per mole) = NAE = NAhc

λ

hc = (6.62608 × 10−34J s) × (2.99792 × 108m s−1) = 1.986 × 10−25J m

NAhc = (6.02214 × 1023mol−1) × (1.986 × 10−25J m) = 0.1196 J m mol−1

Thus,E = 1.986 × 10−25J m

λ ; E(per mole) =

0.1196 J m mol−1 λ

We can therefore draw up the following table

λ E/J E/(kJ mol−1)

(a) 200 nm 9.93 × 10−19 598

(b) 150 pm 1.32 × 10−15 7.98 × 105

(c) 1.00 cm 1.99 × 10−23 0.012

E11.12(b) Assuming that the4He atom is free and stationary, if a photon is absorbed, the atom acquires its

momentump, achieving a speed v such that p = mv.

v = m p m = 4.00 × 1.6605 × 10−27kg= 6.64¯2 × 10−27kg

p = h λ

(a) p = 6.626 × 10−34J s

200× 10−9m = 3.31¯3 × 10−27kg m s−1

v = m p = 3.31¯3 × 10−27kg m s−1

6.642 × 10−27kg = 0.499 m s−1

(b) p = 6.626 × 10−34J s

150× 10−12m = 4.41¯7 × 10−24kg m s−1

v = m p = 4.41¯7 × 10−24kg m s−1

6.642 × 10−27kg = 665 m s−1

(c) p = 6.626 × 10−34J s

1.00 × 10−2m = 6.626 × 10−32kg m s−1

v = m p = 6.626 × 10−32kg m s−1

6.642 × 10−27kg = 9.98 × 10−6m s−1

E11.13(b) Each emitted photon increases the momentum of the rocket byh/λ The final momentum of the

rocket will beNh/λ, where N is the number of photons emitted, so the final speed will be λm Nh

rocket

The rate of photon emission is the power (rate of energy emission) divided by the energy per photon (hc/λ), so

N = tP λ hc and v =

tP λ

hc



×

 h

λmrocket



= cm tP

rocket

v = (10.0 yr) × (365 day yr−1) × (24 h day−1) × (3600 s h−1) × (1.50 × 103W)

(2.998 × 108m s−1) × (10.0 kg)

= 158 m s−1

E11.14(b) Rate of photon emission is rate of energy emission (power) divided by energy per photon (hc/λ)

(a) rate=P λ

hc =

(0.10 W) × (700 × 10−9m) (6.626 × 10−34J s) × (2.998 × 108m s−1) = 3.52 × 1017s−1

(b) rate= (1.0 W) × (700 × 10−9m)

(6.626 × 10−34J s) × (2.998 × 108m s−1) = 3.52 × 1018s−1

E11.15(b) Wien’s displacement law is

T λmax= c2/5 so T = c2

5λmax = 1.44 × 10−2m K

5(1600 × 10−9m) = 1800 K

E11.16(b) Conservation of energy requires

Ephoton=  + EK= hν = hc/λ so EK= hc/λ − 

andEK=1

2mev2

sov =



2EK

me

1/2

(a) EK= (6.626 × 10−34J s) × (2.998 × 108m s−1)

650× 10−9m − (2.09 eV) × (1.60 × 10−19J eV−1)

But this expression is negative, which is unphysical There is no kinetic energy or velocity because the photon does not have enough energy to dislodge the electron

(b) EK= (6.626 × 10−34J s) × (2.998 × 108m s−1)

195× 10−9m − (2.09 eV) × (1.60 × 10−19J eV−1)

= 6.84 × 10−19J andv =



2(3.20 × 10−19J)

9.11 × 10−31kg

1/2

= 1.23 × 106m s−1

E11.17(b) E = hν = h/τ, so

(a) E = 6.626 × 10−34J s/2.50 × 10−15s= 2.65 × 10−19J= 160 kJ mol−1

(b) E = 6.626 × 10−34J s/2.21 × 10−15s= 3.00 × 10−19J= 181 kJ mol−1

(c) E = 6.626 × 10−34J s/1.0 × 10−3s= 6.62 × 10−31J= 4.0 × 10−10kJ mol−1

E11.18(b) The de Broglie wavelength is

λ = h

p

The momentum is related to the kinetic energy by

EK= p2

2m so p = (2mEK)1/2

The kinetic energy of an electron accelerated through 1 V is 1 eV= 1.60 × 10−19J, so

(2mEK)1/2

(a) λ = 6.626 × 10−34J s

(2(9.11 × 10−31kg) × (100 eV) × (1.60 × 10−19J eV−1))1/2

= 1.23 × 10−10m

(2(9.11 × 10−31kg) × (1.0 × 103eV) × (1.60 × 10−19J eV−1))1/2

= 3.9 × 10−11m

(2(9.11 × 10−31kg) × (100 × 103eV) × (1.60 × 10−19J eV−1))1/2

= 3.88 × 10−12m

E11.19(b) The minimum uncertainty in position is 100 pm Therefore, sincexp ≥ 1

2¯h

p ≥ ¯h

2x =

1.0546 × 10−34J s

2(100 × 10−12m) = 5.3 × 10−25kg m s−1

5.3 × 10−25kg m s−1

9.11 × 10−31kg = 5.8 × 10−5m s−1

E11.20(b) Conservation of energy requires

Ephoton= Ebinding+1

2mev2= hν = hc/λ so Ebinding= hc/λ −1

2mev2

andEbinding=(6.626 × 10−34J s) × (2.998 × 108m s−1)

121× 10−12m

−1

2(9.11 × 10−31kg) × (5.69 × 107m s−1)2

= 1.67 × 10−16J

Comment This calculation uses the non-relativistic kinetic energy, which is only about 3 per cent

less than the accurate (relativistic) value of 1.52 × 10−15J In this exercise, however,Ebindingis a small difference of two larger numbers, so a small error in the kinetic energy results in a larger error

inEbinding: the accurate value isEbinding= 1.26 × 10−16J

Solutions to problems

Solutions to numerical problems

k , [θE]=

J s× s−1

J K−1 = K

In terms ofθEthe Einstein equation [11.9] for the heat capacity of solids is

C V = 3R



θE

T

2

×



eθE/2T

eθE/T − 1

2

, classical value = 3R

It reverts to the classical value when T θE or when hν

kT

(Section 11.1) The criterion for classical behaviour is therefore that T θE

θE= hν k = (6.626 × 10−34J Hz−1) × ν

1.381 × 10−23J K−1 = 4.798 × 10−11(ν/Hz)K

(a) Forν = 4.65 × 1013Hz, θE= (4.798 × 10−11) × (4.65 × 1013K) = 2231 K

(b) Forν = 7.15 × 1012Hz, θE= (4.798 × 10−11) × (7.15 × 1012K) = 343 K

Hence

(a) C V

3R =



2231 K

298 K

2

×



e223¯1/(2×298)

e2231/298− 1

2

= 0.031

(b) C V

3R =



343 K

298 K

2

×



e343/(2×298)

e343/298− 1

2

= 0.897

Comment For many metals the classical value is approached at room temperature; consequently,

the failure of classical theory became apparent only after methods for achieving temperatures well below 25◦C were developed in the latter part of the nineteenth century.

P11.5 The hydrogen atom wavefunctions are obtained from the solution of the Schr¨odinger equation in

Chapter 13 Here we need only the wavefunction which is provided It is the square of the wavefunction that is related to the probability (Section 11.4)

ψ2= 1

πa3 0

e−2r/a0, δτ = 4

3πr3

0, r0= 1.0 pm

If we assume that the volumeδτ is so small that ψ does not vary within it, the probability is given by

ψ2δτ = 4r03

3a3 0

e−2r/a0 =4

3 ×



1.0

53

3

e−2r/a0

(a) r = 0 : ψ2δτ = 4

3



1.0

53

3

= 9.0 × 10−6

(b) r = a0: ψ2δτ = 4

3



1.0

53

3

e−2 = 1.2 × 10−6

Question If there is a nonzero probability that the electron can be found atr = 0 how does it avoid

destruction at the nucleus? (Hint See Chapter 13 for part of the solution to this difficult question.)

P11.7 According to the uncertainty principle,

pq ≥ 1

2h,¯ whereq and p are root-mean-square deviations:

q = (x2 − x2)1/2 and p = (p2 − p2)1/2

To verify whether the relationship holds for the particle in a state whose wavefunction is

= (2a/π)1/4e−ax2

,

We need the quantum-mechanical averagesx, x2, p, and p2

x =

∗x2 dτ =

∞

∞−



2a π

1/4

e−ax2

x



2a π

1/4

e−ax2

dx,

x =



2a π

1/2 ∞

−∞

xe −2ax2

dx = 0;

x2 =

∞

−∞



2a π

1/4

e−ax2

x2



2a π

1/4

e−ax2

dx =



2a π

1/2 ∞

∞−

x2e−2ax2

dx,

x2 =



2a π

1/2

π1/2

2(2a)3/2 = 1

4a;

so q = 1

2a1/2

p =

∞

−∞

∗



¯

h

i

d

dx



dx and p2 =

∞

−∞

∗



−¯h2d2

dx2



dx.

We need to evaluate the derivatives:

d

dx =



2a π

1/4 (−2ax)e −ax2

and d

2

dx2 =



2a π

1/4 [(−2ax)2e−ax2

+ (−2a)e −ax2]=



2a π

1/4

(4a2x2− 2a)e −ax2.

So p =

∞

−∞



2a π

1/4

e−ax2

¯

h

i

 

2a π

1/4

(−2ax)e −ax2 dx

= −2¯h i



2a π

1/2 ∞

−∞

xe −2ax2dx = 0;

p2 =

∞

−∞



2a π

1/4

e−ax2

(−¯h2)



2a π

1/4

(4a2x2− 2a)e −ax2

dx,

p2 = (−2a¯h2)



2a π

1/2 ∞

−∞

(2ax2− 1)e −2ax2

dx,

p2 = (−2a¯h2)



2a π

1/2

2a π1/2

2(2a)3/2− π1/2

(2a)1/2



= a¯h2; and p = a1/2¯h.

Finally, qp = 1

2a1/2 × a1/2¯h = 1/2¯h,

which is the minimum product consistent with the uncertainty principle

Solutions to theoretical problems

P11.9 We look for the value ofλ at which ρ is a maximum, using (as appropriate) the short-wavelength

(high-frequency) approximation

ρ = 8πhc

λ5

 1

ehc/λkT − 1

 [11.5]

dρ

dλ = −

5

λ ρ +

hc

λ2kT



ehc/λkT

ehc/λkT − 1



ρ = 0 at λ = λmax

Then,−5 + hc

ehc/λkT

ehc/λkT − 1 = 0 Hence, 5− 5ehc/λkT + hc

λkTehc/λkT = 0

If hc

λkT 1 [short wavelengths, high frequencies], this expression simplifies We neglect the initial 5,

cancel the two exponents, and obtain

hc = 5λkT for λ = λmax and hc

λkT 1

or λmaxT = hc

5k = 2.88 mm K , in accord with observation.

Comment Most experimental studies of black-body radiation have been done over a wavelength

range of a factor of 10 to 100 of the wavelength of visible light and over a temperature range of 300 K

to 10 000 K.

Question Does the short-wavelength approximation apply over all of these ranges? Would it apply

to the cosmic background radiation of the universe at 2.7 K where λmax≈ 0.2 cm?

P11.10 ρ = 8πhc

λ5

 1

ehc/λkT − 1

 [11.5]

Asλ increases, hc

λkT decreases, and at very long wavelength

the exponential in a power series Letx = hc/λkT , then

ex = 1 + x + 1

2!x2+ 1 3!x3+ · · ·

ρ = 8πhc

λ5



1

1+ x + 1

2!x2+ 1 3!x3+ · · · − 1



lim

λ→∞ ρ = 8πhc

λ5

 1

1+ x − 1

= 8πhc

λ5

 1

hc/λkT



= 8πkT

λ4

This is the Rayleigh–Jeans law [11.3]

P11.12 ρ = 8πhc

ehc/λkT − 1 [11.5]

∂ρ

40πhc

 1

ehc/λkT − 1



−



8πhc

λ5



×e

hc/λkT × −λ hc2kT

ehc/λkT − 1 2

= 8πhc

λ5 ×

 1

ehc/λkT − 1



×



−λ5+ hc

λ2kT

ehc/λkT

ehc/λkT − 1



=



−5

λ



× ρ



1− hc

5λkT

1

1− e−hc/λkT

∂ρ

∂λ = 0 when λ = λmax and

hc

5λmaxkT



1

1− e−hc/λmaxkT

= 1

5λmaxkT

hc 1− e−hc/λmaxkT

= 1

Letx = hc

λmaxkT; then

5

x

1− e−x = 1 or 5

x =

1

1− e−x

The solution of this equation isx = 4.965.

Thenh = 4.965λmaxkT

However

M = σ T4=



2π5k4

15c2h3



Substituting (1) into (2) yields

M ≈



2π5k4

15c2



×



c

4.965λmaxkT

3

T4

≈ 2π5ckT

1835.9λ3 max

k ≈ 1835.9λ3maxM

2π5cT

≈ 1835.9(1.451 × 10−6m)3× (904.48 × 103W)

2π5(2.998 × 108m s−1) × (2000 K) × (1.000 m2)

Substituting (3) into (1)

h ≈ 5(1.451 × 10−6m) × (1.382 × 10−23J K−1) × (2000 K)

2.998 × 108m s−1

h ≈ 6.69 × 10−34J s

Comment These calculated values are very close to the currently accepted values for these constants P11.14 In each case formNψ; integrate

(Nψ)∗(Nψ) dτ

set the integral equal to 1 and solve forN.



2−a r

0



e−r/2a0

ψ2= N2



2−a r

0

2

e−r/a0

ψ2dτ = N2

∞

0



4r2−4r3

a0 +r4

a2 0



e−r/a0dr

π

0

sinθ dθ

2π

0

dφ

= N2



4× 2a3

0− 4 ×6a04

a0 +24a

5 0

a2 0



× (2) × (2π) = 32πa3

0N2;

hence N =

 1

32πa3 0

1/2

where we have used

∞

0

x ne−axdx = n!

a n+1 [Problem 11.13]

(b) ψ = Nr sin θ cos φ e −r/(2a0)

ψ2dτ = N2

∞

0 r4e−r/a0dr

π

0

sin2θ sin θ dθ

2π

0

cos2φ dφ

= N24!a5

0

1

−1(1 − cos2θ) d cos θ × π

= N24!a5

0



2−2 3



π = 32πa5

0N2

0; hence N =

 1

32πa5 0

1/2

where we have used

π

0

cosn θ sin θ dθ = −

−1

1

cosn θ d cos θ =

1

−1x ndx

and the relations at the end of the solution to Problem 11.8 [See Student’s solutions manual.]

P11.16 Operate on each function with i; if the function is regenerated multiplied by a constant, it is an

eigenfunction ofi and the constant is the eigenvalue.

(a) f = x3− kx

i(x3− kx) = −x3+ kx = −f

Therefore,f is an eigenfunction with eigenvalue, −1

(b) f = cos kx

i cos kx = cos(−kx) = cos kx = f

Therefore,f is an eigenfunction with eigenvalue, +1

(c) f = x2+ 3x − 1

i(x2+ 3x − 1) = x2− 3x − 1 = constant × f

Therefore,f is not an eigenfunction of i.

P11.19 The kinetic energy operator, ˆT , is obtained from the operator analogue of the classical equation

EK= p2

2m

that is,

ˆT = ( ˆp)2

2m

ˆp x =h¯ i

d

dx [11.32]; hence ˆp2x = −¯h2

d2

dx2 and ˆT = − h¯2

2m

d2

dx2

Then

T  = N2

ψ∗



ˆp2

x

2m



ψ dτ =



ψ∗ 2ˆp m2ψ dτ



ψ∗ψ dτ



N2= 1

ψ∗ψ dτ

= −¯h

2

2m



ψ∗ ddx22(eikxcosχ + e −ikxsinχ) dτ



ψ∗ψ dτ

= −¯h

2

2m



ψ∗(−k2) × (eikxcosχ + e −ikxsinχ) dτ



¯

h2k2

ψ∗ψ dτ

2m ψ∗ψ dτ =

¯

h2k2

2m

P11.20 p x =h¯

i

d

dx [11.32]

p x  = N2

ψ∗ˆp x ψ dx; N2= 1

ψ∗ψ dτ

=



ψ∗ˆp x ψ dx



ψ∗ψ dx =

¯

h

i



ψ∗ dψ

dx

dx



ψ∗ψ dx

(a) ψ = eikx , dψ

dx = ikψ

Hence,

p x =

¯

h

i × ikψ∗ψ dx



ψ∗ψ dx = k¯h

(b) ψ = cos kx, dψ

dx = −k sin kx

∞

−∞ψ∗dψ

dx dx = −k

∞

−∞coskx sin kx dx = 0

Therefore,p x = 0

(c) ψ = e −αx2, dψ

dx = −2αxe −αx

2

∞

−∞ψ∗dψ

dx dx = −2α

∞

−∞xe −2αx2dx = 0 [by symmetry, since x is an odd function]

Therefore,p x = 0

P11.23 No solution

Solution to applications

P11.27 (a) Consider any infinitesimal volume element dx dy dz within the hemisphere (Figure 11.1) that

has a radius equal to the distance traveled by light in the time dt (c dt) The objective is to find the

total radiation flux perpendicular to the hemisphere face at its center Imagine an infinitesimal areaA at that point Let r be the distance from dx dy dz to A and imagine the infinitesimal

areaA perpendicular tor E is the total isotropic energy density in dx dy dz E dx dy dz is

the energy emitted in dt A/4πr2is the fraction of this radiation that passes throughA The radiation flux that originates from dx dy dz and passes through Ain dt is given by:

J A = ✁

A

4πr2

E dx dy dz

E dx dy dz

4πr2dt

The contribution ofJ A to the radiation flux throughA, J A, is given by the expressionJ A ×

(A cos θ)/A = J Acosθ The integration of this expression over the whole hemisphere gives an

c dt

c dt

A

J A

J A⬘

dx dy dz

A⬘



Figure 11.1

has a radius equal to the distance traveled by light in the time dt (c dt) The objective is to find the

total radiation flux perpendicular to the hemisphere face at its... Schrăodinger equation in

Chapter 13 Here we need only the wavefunction which is provided It is the square of the wavefunction that is related to the probability (Section 11.4)

ψ2=... ndx

and the relations at the end of the solution to Problem 11.8 [See Student’s solutions manual. ]