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Using 
and the speed of light as 3.00 × 108 m s−1 to three significant figures (as the wavelength is given to three significant figures) and working in metre units
1.2 The X-ray absorption edge of Zn is at 9.6659 keV. Calculate the X-ray wavelength equivalent.
From Table 1.1 1 eV is equivalent to 1239.81 nm. Therefore 9.6659 × 103 eV is equivalent to 1239.81× 10−9 / 9.6659 × 103 = 1.2827 × 10−10 m; equivalent to 128.27 pm or 1.2827 Å.
1.3 Convert 400 nm into wavenumber units.
The‘wavenumber’ is defined as the number of waves in unit length (1/λ) so for a wavelength of 400 × 10−9 m, 
= 1/400 × 10−9 = 2500000 m−1 = 25,000 cm−1.
1.4 In which region of the electromagnetic spectrum is an emission from a neodynium-YAG laser corresponding to a transition between electronic energy levels at 11,502 and 2,111 cm−1?
The energy difference between the two levels is 11,502 – 2,111 cm−1 = 9391 cm−1. We can convert this to a wavelength using the knowledge that wavenumber is number of waves per unit length. 9391 cm−1 = 939100 m−1 . So the wavelength is 1/939100 m = 1.065 × 10−6 m or 1065 nm. This wavelength is in the near infrared (NIR) portion of the electromagnetic spectrum.
1.5 If 2 hours of synchrotron X-ray beam time was available what size crystal could be studied assuming zero read out time from the detectors?
For a crystal of dimensions 100 ×100 ×100 μm the collection of a complete X-ray diffraction pattern on a synchrotron source was calculated as 0.4 s. 2 hours is 2 × 60 ×60 = 7200 s. Therefore there is a factor of 7200/0.4 = 18000 increase in instrument time over a 100 ×100 ×100 μm crystal and the study of a crystal smaller by this factor in volume should be possible. The volume of the original crystal was 106 (μm)3 to the volume of a crystal that could be studied in 2 h would be 106/18000 (μm)3 = 55.56 (μm)3. And, if a cube, the dimension of the cube side would be [55.56 (μm)3 ]⅓ = 3.82 μm.
1.6 Calculate the kinetic energy of a neutron moving at 1000 m s−1. What type of molecular energies (electronic, vibrational or rotational) does this correspond to?
The kinetic energy, KE, is given by the expression KE = ½mv2. Therefore KE = ½ × 1.675´10−27 kg × (1000)2 m2 s−2 = 8.375 × 10−22 kg m2 s−2 = 8.375 × 10−22 J. For a mole of compound this needs to be multiplied by Avogadro’s number, 6.022 ×1023 mol−1 , to give 8.375 × 10−22 J × 6.022 ×1023 mol−1 = 504.3 Jmol−1 . Equivalent to around 0.5 kJmol−1. Consulting Figure 1.2 shows that this energy corresponds to low energy molecular vibrations. Inelastic scattering of neutrons can be used to study these vibrations.