*24-9 . A typical simple infrared spectrophotometer covers a wavelength range from 3 to 15 jum Express its range (a) in wavenumbers and (b)in hertz. 24-10 Calculate the frequency in hertz and the energy in joules of an X-ray photon with a wavelength of 2.70stackrel (circ )(A) 2411. Calculate the wavelength and the energy in joules associated with a signal at 220 MHz.

Let's correct and clarify the calculations for each part of the problem:<br /><br />### 24-9<br />(a) **Wavenumbers:**<br />Wavenumber (\(\tilde{\nu}\)) is the reciprocal of wavelength (\(\lambda\)):<br /><br />\[<br />\tilde{\nu} = \frac{1}{\lambda}<br />\]<br /><br />Given the wavelength range from 3 to 15 \(\mu m\):<br /><br />\[<br />\tilde{\nu}_{min} = \frac{1}{3 \times 10^{-6}} = 3.33 \times 10^5 \, \text{cm}^{-1}<br />\]<br /><br />\[<br />\tilde{\nu}_{max} = \frac{1}{15 \times 10^{-6}} = 6.67 \times 10^4 \, \text{cm}^{-1}<br />\]<br /><br />So, the range in wavenumbers is approximately \(6.67 \times 10^4 \, \text{cm}^{-1}\) to \(3.33 \times 10^5 \, \text{cm}^{-1}\).<br /><br />(b) **Hertz:**<br />Frequency (\(f\)) is the speed of light (\(c\)) divided by the wavelength (\(\lambda\)):<br /><br />\[<br />f = \frac{c}{\lambda}<br />\]<br /><br />Given the speed of light \(c = 3 \times 10^8 \, \text{m/s}\):<br /><br />\[<br />f_{min} = \frac \times 10^8}{3 \times 10^{-6}} = 1 \times 10^{14} \, \text{Hz}<br />\]<br /><br />\[<br />f_{max} = \frac{3 \times 10^8}{15 \times 10^{-6}} = 2 \times 10^{13} \, \text{Hz}<br />\]<br /><br />So, the range in hertz is approximately \(2 \times 10^{13} \, \text{Hz}\) to \(1 \times 10^{14} \, \text{Hz}\).<br /><br />### 24-10<br />**Frequency and Energy of an X-ray Photon:**<br /><br />Given the wavelength \(\lambda = 2.70 \, \text{Å} = 2.70 \times 10^{-10} \, \text{m}\):<br /><br />(a) **Frequency:**<br /><br />\[<br />f = \frac{c}{\lambda} = \frac{3 \times 10^8}{2.70 \times 10^{-10}} = 1.11 \times 10^{18} \, \text{Hz}<br />\]<br /><br />(b) **Energy:**<br /><br />Using Planck's constant \(h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}\):<br /><br />\[<br />E = h \cdot f = 6.626 \times 10^{-34} \times 1.11 \times 10^{18} = 7.35 \times 10^{-16} \, \text{J}<br />\]<br /><br />So, the frequency is \(1.11 \times 10^{18} \, \text{Hz}\) and the energy is \(7.35 \times 10^{-16} \, \text{J}\).<br /><br />### 24-11<br />**Wavelength and Energy of a Signal at 220 MHz:**<br /><br />(a) **Wavelength:**<br /><br />\[<br />\lambda = \frac{c}{f} = \frac{3 \times 10^8}{220 \times 10^6} = 1.36 \times 10^{-3} \, \text{m} = 1.36 \, \text{mm}<br />\]<br /><br />(b) **Energy:**<br /><br />\[<br />E = h \cdot f = 6.626 \times 10^{-34} \times 220 \times 10^6 = 1.46 \times 10^{-26} \, \text{J}<br />\]<br /><br />So, the wavelength is \(1.36 \, \text{mm}\) and the energy is \(1.46 \times 10^{-26} \, \text{J}\).