4 solutions

Chem 6, 9 Section Spring 2002

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Assignment 4 Solutions

1. The “red line” in the Balmer series is the photon that results when the H atom changesstate from n = 3 to n = 2. This photon has an energy

Ephot = E3 – E2 = –2.18 × 10–18 J

32 –

–2.18 × 10–18 J

22 = 3.03 × 10–19 J .

This is the energy that the electron in the box will gain when it absorbs the photon. Theelectron starts in its ground state (n = 1, where this n is the particle-in-a-box quantumnumber—the other n is the H atom principal quantum number), absorbs this energy, andends in n = 2. This energy change is related to the box length L through

∆E = h2

8meL2 22 – 12 or L = 3h2

8me∆E = 7.73 × 10–10 m = 7.73 Å .

2. Problem 77 makes it clear that the symmetry of the wavefunction ensures that theprobability of finding the particle between 0 and L/4 is 1/4, as the figure below emphasizes.

x/L

Ψ2 n = 2

The area under the red curve for Ψ2 over the interval 0 ≤ x ≤ L/4 (the green bar) is the sameas the area over L/4 ≤ x ≤ L/2 or L/2 ≤ x ≤ 3L/4 or 3L/4 ≤ x ≤ L. But for n = 1, thesymmetry changes:

x/L

Ψ2 n = 1


2

Here, you can see at a glance that the area under Ψ2 from 0 to L/4 is much less than thearea from L/4 to L/2. Thus, the probability of finding the particle anywhere between 0 andL/4 is much less than the probability of finding it between L/4 and L/2. (Note however thatsymmetry can still be invoked here for some situations. The probability between 0 and L/4is the same as from 3L/4 to L, for instance.) The probability’s numerical value in this caseis given by the expression below (and I don’t expect you to know how to do this integraloff the top of your head!):

Probability = Ψ(x)2 dx0

L/4

= 2L

sin2 πxL

dx0

L/4

= π – 2

4π = 0.0908 .

3. First, we use the expression in Equation [15.22] to calculate the size of a hydrogenatom (Z = 1) in the 3d (n = 3, l = 2) state:

r nl =

n2a0Z

1 + 12

1 – l (l + 1)

n2 = 10.5 a0 .

Then we make a table of r nl values for various n and l values for He+ (Z = 2):

n l r nl /a0

1 0 0.752 0 3.002 1 2.503 0 6.753 1 6.253 2 5.254 0 12.004 1 11.504 2 10.504 3 9.005 0 18.755 1 18.255 2 17.255 3 15.755 4 13.75

This table shows that the n = 4, l = 2 (or 4d) He+ ion has the same average size as the Hatom in the 3d state. The He+ 4s and 4p states and all n > 4 states are larger than H 3d.Note as well that the average radius does not depend on the m quantum number.

4. First, let’s review the Zeff calculation. We know the Na first ionization energy: IE =8.23 × 10–19 J, and we measured the sodium D line wavelength in lab to be about λ = 590nm. This wavelength corresponds to a photon energy of


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Ephot = hcλ

= 3.37 × 10–19 J .

We can now construct the energy level diagram below for Na:

E = 0

E3p = –IE + Ephot

E3s = –IE

Ephot

n =

3p

3s

0 J

–4.95 × 10–19 J

–8.23 × 10–19 J

From the general expression for the energy of an atomic level in terms of an effectivenuclear charge:

En,l = – e4me

8ε02h2

Zeff,n,l

2

n2 = –2.18 × 10–18 J

Zeff,n,l2

n2

we can write an expression for Zeff:

Zeff,n,l = n2En,l

–2.18 × 10–18 J

Our level energies for these two n = 3 levels give

Zeff 3s = 32E3s

–2.18 × 10–18 J =

32(–8.32 × 10–19 J)

–2.18 × 10–18 J = 1.85

Zeff 3p = 32E3p

–2.18 × 10–18 J =

32(–4.95 × 10–19 J)

–2.18 × 10–18 J = 1.43

Now we can use Equation [15.25] to calculate average radii for these two orbitals:


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r nl =

n2a0Zeff

1 + 12

1 – l(l + 1)

n2

r 3s =

32a01.85

1 + 12

1 – 0(0 + 1)

32 = 7.30 a0

r 3p =

32a01.43

1 + 12

1 – 1(1 + 1)

32 = 8.74 a0

Now we can move on to Cs. The analogous transition in Cs involves n = 6 instead of n =3 levels. The Cs atom changes configuration from [Xe]6p1 to [Xe]6s1. The 873.5 nmphoton that is emitted in this change has an energy Ephot = 2.27 × 10–19 J, and with a6.24 × 10–19 J ionization energy, our energy level diagram looks like this:

E = 0

E6p = –IE + Ephot

E6s = –IE

Ephot

n =

6p

6s

0 J

–3.97 × 10–19 J

–6.24 × 10–19 J

Repeating the effective nuclear charge calculation (with n = 6) gives Zeff = 3.21 for the 6slevel and 2.56 for the 6p level.

4 solutions

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