homework atomic

8/18/2019 homework atomic

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The distance rH is calculated from the bond distance rOH and half of the bond angle, θ = 52.25◦ as

rH = rOH sin θ

= 95.8 pm × sin(52.25◦)

= 75.75 pm

The moment of inertia for this system is, therefore,

I = 2 ×mH × r2H

= 2 ×1.0078 × 10−3 kg mol−1

6.022× 1023 mol−1 × (75.75 × 10−12 m)2

= 1.92056 × 10−47 kg m2

The rotational constant is, therefore,

eB =

h

8π2cI

= 6.626× 10−34 J s

8π× 2.998 × 1010 cm s−1 × I kg m2

= 14.57 cm−1.

The spacings in the rotational spectrum would, therefore, be equal to 2 eB or 29.14 cm−1.3. What is the absorption coefficient of a solute that absorbs 90% of a certain wavelength of light when the

beam is passed through a 0.1 m cell containing a 0.25 M solution?

Answer

Since 90% of the light is absorbed, I 0/I = 100/10 = 10. From the Lambert-Beer Law, we have

log10 ³ I 0I ´ = ²cl. Therefore,²=

1

cl log10

µI 0I

¶=

1

0.25 M × 0.1 m log10(10).

=40.0 M−1 m−1.

4. The RbH molecule has re =236.7 pm, ν 0 = 938.8 cm−1 and ν 0xe = 14.15 cm

−1.

(a) What is the value of the harmonic force constant kh for this molecule?

(b)

A more accurate model for the potential energy of a diatomic molecule is given by the Morse potential(See Eq. 13.146)

E p(r) = Deh

1− eα(r−re)i2.

Expand the potential function in a Taylor series about r = re, and obtain an expression for the harmonicforce constant kh in terms of the Morse parameters De and α.

(c) Using the results of parts (a) and (b), and the relationships

xe = α2h̄

2µω; where ω =

1

c

µkhµ

¶1/2.

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where c is the velocity of light in cm s−1, calculate the value of α.(d) Find the value of the classical bond dissociation energy De.(e) Plot the potential energy function from 50 pm to 800 pm.

Answer:

(a) The harmonic force constant is calculated from the relationship ω = 1c µkhµ ¶

1/2

or ν 0(cm−1

) = 12πc µkhµ ¶

1/2

.The reduced mass of RbH is

µ= mHmRbmH + mRb

'mH = 1.008× 10−3 kg mol−1

6.002 × 1023 mol−1

=1.6739 × 10−27 kg

The force constant is now calculated as

kh=4π2c2µν 0

=4π2(2.998× 1010 cm s−1)2 × 1.6739 × 10−27 kg × 938.82 cm−2

=52.348 N m−

1.

(b) The Taylor expansion of the potential about r = re yields

E p(r) ' E p(re) +

µ∂ E p∂ r

¶re

(r − re) + 1

2!

Ã∂ 2E p∂ r2

!re

(r− re)2 + ....

Using the Morse expression given above for E p(r), we get E p(re) = 0 and

µ∂ E p∂ r

¶re

= 0 (minimum of

the function). So,

E p(r) ' 12!Ã∂ 2V ∂ r2

!re

(r−re)2 , (1)

if we ignore terms higher than the second power. By explicit differentiation of the Morse function, we

get Ã∂ 2V

∂ r2

!re

= 2Deα2.

Therefore, from Eq. (1),

V (r) '1

2

³2Deα

2´

(r− re)2 . (2)

Comparing Eq. (2) to the Harmonic oscillator model E p(r) = 1

2

kh (r− re)2, we see that kh = 2Deα

2.

From part (a), we have

kh = 2Deα2 = 52.348 N m−1.

(c) From the anharmonicity constant given (ν 0xe = 14.15 cm−1), we get

xe=ν 0xeν 0

= 14.15

938.8

=1.5072 × 10−2.

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From comparing the given expression for ω with that for ν 0 for the harmonic oscillator [see part (a)], we

see that ω = 2πν 0. Now the Morse parameter α can be evaluated from the relationship xe = α2h̄

2µω, or

α=

q 2µ(2πν 0c)xe/h̄ =

q 8π2µν 0xe/h

=s 8π2 × 1.6739 × 10−27 kg × 938.8 cm−1 × 2.998 × 1010 cm s−1 × 1.5072 × 10−26.626× 10−34 J s

=9.1986 × 109 m−1.

Converting to more commonly used units for molecules, α = 9.1986 × 10−3 pm−1.

(d) Finally, since kh = 2Deα2, we get

De= kh2α2

= 52.348 kg s−2

2 × (9.1986 × 109 m−1)2

=3.0933 × 10−19 J

= 3.0933 × 10−19 J

6.626 × 10−34 J s × 2.998 × 1010 cm s−1=15572 cm−1.

(e) We plot the function E p(r) (cm−1)= 15572 ×h

1− e−9.1986×10−3(r−236.7)

i2from 50 to 800 pm.

x 800700600500400300200100

30000

25000

20000

15000

10000

5000

0

4

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