5.6. Solutions to Problems 195

case that the optical constants for the absorbing centers are not known at least one

calibration experiment has to be carried out prior to using the instrument for routine

measurements. In this calibration experiment the concentration of the absorbing

particles should be known and be identical for both channels. This enables one to

measure α

M

and then to ﬁnd from Eq. (5.49) the value of α which can be used in

further measurements.

Problems

5.32. A two-channel spectrophotometer, like that of Fig. 5.27, was used for

measurement of the concentration of Ag particles homogeneously dispersed in

a partially transparent solution. Two cuvettes ﬁlled with the solution were intro-

duced in the device. The thickness of the liquid was t

1

= 1. 0 mm in the ﬁrst

vessel and t

2

= 3. 5 mm in the second. The ratio of the detector signals measured

in both channels for a wavelength of 0.59 μm was i

Det2

/i

Det1

= 0. 05. Calculate

the concentration of Ag in the solution.

[Note: Refractive index ofAg particles in the spectral interval of the measurements

is n = 0. 18 − 3. 64j.]

5.6. Solutions to Problems

5.1. According to the deﬁnition of wavenumber, we obtain for the given spectral

line N = 10, 000/0. 546075 = 18, 312. 50 cm

−1

and the optical frequency

ν =

c

λ

=

3 × 10

10

cm/s

0. 546075 × 10

−4

cm

= 5. 493751 × 10

14

Hz.

Taking into account Plank’s constant h = 6. 625 × 10

−34

J s and the conversion

factor between J and eV (1 eV = 1. 6022 × 10

−19

J) we obtain the energy of the

transition as E = hν = 2. 2716 eV.

5.2. The natural width of the spectral line is 1. 27 × 10

−4

Å (see Section 5.1).

From the deﬁnition of wavenumber one obtains N /N = λ/ λ and therefore

N =

10, 000

λ

2

λ =

1. 27 × 10

−4

0. 36

= 3. 5278 × 10

−4

cm

−1

.

Since ν = c/λ we also ﬁnd

ν =

c

λ

2

λ =

3 × 10

10

cm/s

0. 36 × 10

−8

cm

2

× 1. 27 × 10

−12

cm = 10. 58 MHz.

196 5 ♦ Optical Systems for Spectral Measurements

5.3. Expression (5.3) yields for iron (atomic weight 56) at temperature 10,000 K:

δλ

D

= 7. 18 × 10

−7

× 3, 100 ×

10, 000

56

= 0. 0297 Å.

This value is signiﬁcantly smaller (more than twice) than the wavelength differ-

ences of the triplet: (λ)

12

= 0. 36 Å; ( λ)

23

= 0. 34 Å. Therefore, the triplet is

still resolvable (if an appropriate spectral instrument is exploited).

5.4. Calculation of the Doppler broadening due to scattering on the electrons in

the corona is done according to the relation presented in the problem (electron

mass m

e

= 9. 11 × 10

−31

kg; k = 1. 3806 × 10

−23

JK):

ν

D

/ν =

1

3 × 10

8

2 × 1. 3806 × 10

−23

9. 11 × 10

−31

600, 000 = 0. 014175.

Since λ

D

/λ = ν

D

/ν, we obtain λ

D

= 55. 76 Å which is about ﬁve orders of

magnitude greater than the normal (“natural”) width of the spectral line. Therefore

absorption of photons occurs in wide spectral interval and this fact deﬁnitely can

explain why Fraunhoffer absorption lines in the corona are so weak that they are

hardly detectable.

5.5 Using Eq. (5.3) for Ne atoms (atomic weight 20) and remembering that the

main line of a He–Ne laser is 6,328 Å, we get

λ

D

= 7. 18 × 10

−7

× 6328

350

20

= 0. 019 Å

which is about 200 times greater than the natural width of the spectral line.

5.6. The reﬂectance of each surface can be found from Eq. (5.9a) which yields the

following: for Au, R = 84. 9%; for Ag, R = 94. 5%; for Cu, R = 73. 2%; and for

Ni, R = 61. 9%.

5.7. The shortest wavelength corresponds to the greatest wavenumber, hence,

using the deﬁnition of wavenumber, we ﬁnd the reference wavelength as λ

1

=

10, 000/3, 067 = 3. 2605 μm. The other wavelengths are λ

2

= 3. 2744 μm;

λ

3

= 3. 2982 μm; λ

4

= 3. 3546 μm; λ

5

= 3. 4247 μm; λ

6

= 3. 4843 μm.

Denoting the coordinate of each wavelength λ

i

in the output plane as x

i

, one can

calculate them with regard to the shortest wavelength as follows: x

i

= x

i

−x

1

=

(λ

i

− λ

1

)/(dλ/dl), where dλ/dl = 50 nm/mm. This gives x

2

= 0. 278 mm;

x

3

= 0. 754 mm; x

4

= 1. 882 mm; x

5

= 3. 284 mm; x

6

= 4. 476 mm.

5.8. Assuming that the wavelength difference between the two lines of the vio-

let doublet represents the minimum resolvable spectral interval of the system,

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