Dimensional Analysis in Spectroscopy - Physical Chemistry 2

A researcher measures the rotational constant $B_e$ of a diatomic molecule as $1.93 \text{ cm}^{-1}$ . Using the relationship $B_e = \frac{h}{8\pi^2 I c}$ , what is the moment of inertia in both $\text{kg}\cdot\text{m}^2$ and $\text{amu}\cdot\text{Å}^2$ ?
Answer: $2.74 \times 10^{-47} \text{ kg}\cdot\text{m}^2$ and $1.65 \text{ amu}\cdot\text{Å}^2$ When you encounter rotational spectroscopy problems, you're dealing with the relationship between a molecule's rotational constant and its moment of inertia. The rotational constant $B_e$ tells us how much energy spacing exists between rotational levels, which directly relates to how the molecule's mass is distributed around its rotation axis. To find the moment of inertia, rearrange the given equation: $I = \frac{h}{8\pi^2 B_e c}$ . Substituting the constants ( $h = 6.626 \times 10^{-34} \text{ J}\cdot\text{s}$ , $c = 2.998 \times 10^{10} \text{ cm/s}$ ) and $B_e = 1.93 \text{ cm}^{-1}$ : $I = \frac{6.626 \times 10^{-34}}{8\pi^2 \times 1.93 \times 2.998 \times 10^{10}} = 2.74 \times 10^{-47} \text{ kg}\cdot\text{m}^2$ Converting to $\text{amu}\cdot\text{Å}^2$ using the conversion factor $1 \text{ kg}\cdot\text{m}^2 = 6.022 \times 10^{23} \text{ amu}\cdot\text{Å}^2$ : $I = 2.74 \times 10^{-47} \times 6.022 \times 10^{23} = 1.65 \text{ amu}\cdot\text{Å}^2$ Answer A has the wrong power of 10 in the SI units ( $10^{-46}$ instead of $10^{-47}$ ), suggesting an arithmetic error. Answer B correctly calculates the SI units but has the wrong conversion to $\text{amu}\cdot\text{Å}^2$ (off by a factor of 10). Answer C contains errors in both calculations, with incorrect powers of 10 throughout. Always double-check your unit conversions in spectroscopy problems—the large conversion factors between SI and spectroscopic units make it easy to drop or add powers of 10. Keep a reliable reference for these conversion factors during calculations.
In terahertz time-domain spectroscopy, the complex refractive index $\tilde{n}(\omega) = n(\omega) + i\kappa(\omega)$ is extracted from the transmission $\tilde{T}(\omega) = \frac{4\tilde{n}}{(\tilde{n}+1)^2} e^{i\omega \tilde{n} d/c}$ where $d$ is sample thickness and $c$ is the speed of light. If the extinction coefficient $\kappa$ is related to the absorption coefficient by $\alpha = \frac{4\pi\kappa\nu}{c}$ , what are the units of the expression $\frac{\alpha \cdot c \cdot d}{\nu \cdot \kappa}$ ?
Answer: $\text{dimensionless}$ (no units) When you encounter dimensional analysis problems in spectroscopy, the key is systematically tracking the units of each variable and applying the fundamental principle that mathematical relationships must be dimensionally consistent. Let's work through this step by step. From the given relationship $\alpha = \frac{4\pi\kappa\nu}{c}$ , we can identify the units: $\alpha$ (absorption coefficient) has units of m⁻¹, $\kappa$ is dimensionless (extinction coefficient), $\nu$ has units of Hz (frequency), and $c$ has units of ms⁻¹. The sample thickness $d$ has units of m. Now let's analyze $\frac{\alpha \cdot c \cdot d}{\nu \cdot \kappa}$ : - Numerator: $\alpha \cdot c \cdot d = \text{m}^{-1} \cdot \text{ms}^{-1} \cdot \text{m} = \text{s}^{-1}$ - Denominator: $\nu \cdot \kappa = \text{Hz} \cdot \text{dimensionless} = \text{s}^{-1}$ - Final result: $\frac{\text{s}^{-1}}{\text{s}^{-1}} = \text{dimensionless}$ The answer is A - the expression is dimensionless. Answer B (m) would suggest the expression represents a length, but our calculation shows all length units cancel out. Answer C (Hz) would mean the expression has frequency units, but the frequency terms cancel with other s⁻¹ terms. Answer D (ms⁻¹) would indicate velocity units, but again, all such units cancel in our calculation. Study tip: In spectroscopy problems involving multiple physical constants, always write out the units explicitly for each term before combining them. This systematic approach prevents errors and helps you recognize when expressions should be dimensionless - a common feature in fundamental relationships.
In infrared spectroscopy, the wavenumber of a vibrational transition is given by $\tilde{\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}$ , where $k$ is the force constant and $\mu$ is the reduced mass. For a C-H stretch with $\tilde{\nu} = 2900 \text{ cm}^{-1}$ , what is the force constant in both N/m and mdyn/Å?
Answer: $510 \text{ N/m}$ and $5.10 \text{ mdyn/Å}$ In vibrational spectroscopy, the wavenumber relates directly to molecular vibrations through the harmonic oscillator model. When you see this equation, recognize that you're converting between spectroscopic data (wavenumber) and molecular properties (force constant). To find the force constant, rearrange the given equation: $k = (2\pi c \tilde{\nu})^2 \mu$ . For a C-H bond, the reduced mass is $\mu = \frac{m_C \times m_H}{m_C + m_H} = \frac{12 \times 1}{12 + 1} = 0.923$ amu. Convert this to kg: $\mu = 0.923 \times 1.66 \times 10^{-27} = 1.53 \times 10^{-27}$ kg. With $\tilde{\nu} = 2900 \text{ cm}^{-1} = 2.9 \times 10^5 \text{ m}^{-1}$ and $c = 3 \times 10^8 \text{ m/s}$ : $k = (2\pi \times 3 \times 10^8 \times 2.9 \times 10^5)^2 \times 1.53 \times 10^{-27} = 510 \text{ N/m}$ Converting to mdyn/Å: $1 \text{ N/m} = 10^{-2} \text{ mdyn/Å}$ , so $510 \text{ N/m} = 5.10 \text{ mdyn/Å}$ . Option A incorrectly converts units by a factor of 10 (uses $10^{-1}$ instead of $10^{-2}$ ). Option C underestimates the force constant by a factor of 10, likely from an error in the reduced mass calculation. Option D has the correct N/m value but makes the same unit conversion error as option A. Remember: C-H bonds are relatively stiff (high force constants around 500 N/m), and always double-check unit conversions between N/m and mdyn/Å using the factor $10^{-2}$ .
A researcher calculates the oscillator strength $f$ using the formula $f = \frac{4.32 \times 10^{-9}}{n} \int \varepsilon(\tilde{\nu}) \, d\tilde{\nu}$ , where $n$ is the refractive index, $\varepsilon$ is the molar absorptivity, and $\tilde{\nu}$ is the wavenumber. If $\varepsilon$ has units $\text{M}^{-1}\text{cm}^{-1}$ and the integral evaluates to $2.5 \times 10^6 \text{ M}^{-1}\text{cm}^{-2}$ , what are the units of the numerical constant $4.32 \times 10^{-9}$ ?
Answer: $\text{M}\text{cm}^2$ (concentration times area) When you encounter problems involving oscillator strength calculations, you're dealing with a fundamental relationship in spectroscopy that connects quantum mechanical properties to measurable optical parameters. The key insight is that oscillator strength $f$ must be dimensionless—it's a pure number representing the relative strength of an electronic transition. To find the units of the constant $4.32 \times 10^{-9}$ , you need to work backwards from the requirement that $f$ be dimensionless. Let's analyze the units in the formula $f = \frac{4.32 \times 10^{-9}}{n} \int \varepsilon(\tilde{\nu}) \, d\tilde{\nu}$ . The refractive index $n$ is dimensionless. The integral $\int \varepsilon(\tilde{\nu}) \, d\tilde{\nu}$ has units of $\text{M}^{-1}\text{cm}^{-1} \times \text{cm}^{-1} = \text{M}^{-1}\text{cm}^{-2}$ (since $d\tilde{\nu}$ has wavenumber units of $\text{cm}^{-1}$ ). For $f$ to be dimensionless, the constant must cancel out these units: $\text{M}^{-1}\text{cm}^{-2} \times \text{constant units} = \text{dimensionless}$ . Therefore, the constant needs units of $\text{M}\text{cm}^2$ . Choice A is correct—the constant has units of $\text{M}\text{cm}^2$ . Choice B incorrectly assumes the constant is dimensionless, which would leave $f$ with units. Choice C gives the same units as the integral, which would square those units instead of canceling them. Choice D provides units that don't properly cancel the integral's dimensions. Remember: in spectroscopy equations, always check that your final physical quantities have the correct dimensions. Oscillator strengths, quantum yields, and similar parameters should be dimensionless numbers.
In electron spin resonance (ESR), the resonance condition is $h\nu = g\mu_B B$ . If the g-factor is measured as $2.003$ and the magnetic field is $0.335 \text{ T}$ , what microwave frequency should be used? Express the answer in both GHz and wavenumbers.
Answer: $9.39 \text{ GHz}$ and $0.313 \text{ cm}^{-1}$ When you encounter ESR problems, you're dealing with the interaction between an unpaired electron's magnetic moment and an external magnetic field. The resonance condition $h\nu = g\mu_B B$ tells you that the photon energy must match the energy gap between electron spin states. To find the microwave frequency, rearrange the equation: $\nu = \frac{g\mu_B B}{h}$ . Using the given values ( $g = 2.003$ , $B = 0.335$ T) and constants ( $\mu_B = 9.274 \times 10^{-24}$ J/T, $h = 6.626 \times 10^{-34}$ J·s): $\nu = \frac{(2.003)(9.274 \times 10^{-24})(0.335)}{6.626 \times 10^{-34}} = 9.39 \times 10^9 \text{ Hz} = 9.39 \text{ GHz}$ For wavenumbers, use $\tilde{\nu} = \frac{\nu}{c}$ where $c = 3.00 \times 10^{10}$ cm/s: $\tilde{\nu} = \frac{9.39 \times 10^9}{3.00 \times 10^{10}} = 0.313 \text{ cm}^{-1}$ Option A gives double the correct frequency (18.8 GHz), suggesting an error like using $2g$ instead of $g$ . Option B has the right frequency but wrong wavenumber (3.13 instead of 0.313), likely from using the wrong speed of light units or decimal placement. Option D shows the same frequency error as B but with an order-of-magnitude error in wavenumbers (31.3). Remember that ESR frequencies are typically in the microwave range (X-band around 9-10 GHz), and always check your unit conversions carefully—wavenumbers require the speed of light in cm/s, not m/s.
In Z-scan measurements, the nonlinear absorption coefficient $\beta_{\text{eff}}$ is determined from $T = \frac{1}{1 + \beta_{\text{eff}} I_0 L_{\text{eff}}}$ where $T$ is transmittance, $I_0$ is peak intensity, and $L_{\text{eff}} = \frac{1-e^{-\alpha_0 L}}{\alpha_0}$ is the effective length with $\alpha_0$ being the linear absorption coefficient. If $\alpha_0 = 0.1 \text{ cm}^{-1}$ , $L = 2 \text{ mm}$ , and $I_0 = 10^8 \text{ W}\text{cm}^{-2}$ , what are the units of $\beta_{\text{eff}}$ ?
Answer: $\text{cm}\text{W}^{-1}$ (length per power) Z-scan measurements are a key technique in nonlinear optics for determining how materials absorb light at high intensities. When analyzing these measurements, dimensional analysis becomes crucial for understanding the physical meaning of the parameters. To find the units of $\beta_{\text{eff}}$ , rearrange the transmittance equation: $\beta_{\text{eff}} = \frac{1-T}{T \cdot I_0 \cdot L_{\text{eff}}}$ . Since transmittance $T$ is dimensionless (it's a ratio), the numerator $(1-T)$ is also dimensionless. For the equation to be dimensionally consistent, the denominator must also be dimensionless, which means $\beta_{\text{eff}} \cdot I_0 \cdot L_{\text{eff}}$ must be dimensionless. Given that $I_0$ has units of $\text{W}\text{cm}^{-2}$ and $L_{\text{eff}}$ has units of length (cm), we have: $[\beta_{\text{eff}}] \cdot \text{W}\text{cm}^{-2} \cdot \text{cm} = \text{dimensionless}$ . This simplifies to $[\beta_{\text{eff}}] \cdot \text{W}\text{cm}^{-1} = \text{dimensionless}$ , so $[\beta_{\text{eff}}] = \text{cm}\text{W}^{-1}$ . Answer A ( $\text{cm}\text{W}^{-1}$ ) is correct. Answer B ( $\text{cm}^2\text{W}^{-1}$ ) would arise if you mistakenly thought intensity had units of $\text{W}\text{cm}^{-1}$ instead of $\text{W}\text{cm}^{-2}$ . Answer C ( $\text{cm}^{-1}\text{W}^{-1}$ ) incorrectly treats length as having inverse units. Answer D ( $\text{W}^{-1}$ ) ignores the length dimension entirely. Remember: always use dimensional analysis to check your nonlinear optics calculations. The physical units must balance on both sides of any equation.
In single-molecule fluorescence correlation spectroscopy, the autocorrelation function $G(\tau) = 1 + \frac{1}{N} \left(1 + \frac{\tau}{\tau_D}\right)^{-1} \left(1 + \frac{\tau}{\omega^2 \tau_D}\right)^{-1/2}$ describes molecular diffusion. If the diffusion coefficient $D$ is related to the diffusion time by $\tau_D = \frac{w_0^2}{4D}$ where $w_0$ is the beam waist radius, and $D = 2.3 \times 10^{-10} \text{ m}^2\text{s}^{-1}$ , what are the units of the ratio $\frac{\tau_D \cdot D}{w_0^2}$ ?
Answer: $\text{dimensionless}$ (no units) When tackling dimensional analysis problems in physical chemistry, you need to systematically track the units of each variable and see how they combine mathematically. Let's work through this step by step. You're given that $\tau_D = \frac{w_0^2}{4D}$ , so we can identify the units of each component: - $\tau_D$ has units of time: $\text{s}$ - $D$ has units of $\text{m}^2\text{s}^{-1}$ - $w_0$ has units of length: $\text{m}$ Now let's substitute into the ratio $\frac{\tau_D \cdot D}{w_0^2}$ : $\frac{\tau_D \cdot D}{w_0^2} = \frac{(\text{s}) \cdot (\text{m}^2\text{s}^{-1})}{(\text{m})^2} = \frac{\text{s} \cdot \text{m}^2 \cdot \text{s}^{-1}}{\text{m}^2} = \frac{\text{m}^2}{\text{m}^2} = 1$ The seconds cancel out ( $\text{s} \cdot \text{s}^{-1} = 1$ ), and the square meters cancel out, leaving us with a dimensionless quantity. Choice A is correct because all units cancel completely. Choice B ( $\text{s}^{-1}$ ) incorrectly assumes only the length units cancel. Choice C ( $\text{m}^2\text{s}^{-1}$ ) mistakenly treats this as if you're just looking at the diffusion coefficient units. Choice D ( $\text{s}$ ) incorrectly assumes only some of the time units cancel. Key strategy: In dimensional analysis problems, write out every unit explicitly and work through the cancellations step-by-step. Don't rush—methodical unit tracking prevents careless errors and reveals when expressions should be dimensionless.
A mass spectrometer operates with an electric field of $2.5 \times 10^4 \text{ V/m}$ and accelerates ions through a potential difference of $3000 \text{ V}$ . An ion with $m/z = 120$ reaches the detector. What is the final velocity of this ion in m/s, and what assumption about initial conditions is required for dimensional consistency?
Answer: $2.24 \times 10^5 \text{ m/s}$ assuming the ion starts from rest When ions are accelerated through an electric potential in a mass spectrometer, you're dealing with energy conservation where electrical potential energy converts to kinetic energy. The key relationship is: $qV = \frac{1}{2}mv^2$ , where $q$ is the ion charge, $V$ is the potential difference, $m$ is the mass, and $v$ is the final velocity. For an ion with $m/z = 120$ , assuming it's singly charged ( $z = 1$ ), the mass is $120 \text{ u} = 120 \times 1.66 \times 10^{-27} \text{ kg} = 1.99 \times 10^{-25} \text{ kg}$ . The charge is $1.60 \times 10^{-19} \text{ C}$ . Solving for velocity: $v = \sqrt{\frac{2qV}{m}} = \sqrt{\frac{2 \times 1.60 \times 10^{-19} \times 3000}{1.99 \times 10^{-25}}} = 2.24 \times 10^5 \text{ m/s}$ The assumption "starts from rest" is crucial for dimensional consistency because our energy equation assumes zero initial kinetic energy. Without this assumption, you'd need additional terms accounting for initial velocity. Answer A doubles the correct value, likely from incorrectly using $2qV$ instead of $qV$ in the energy equation. Answer B gives the correct calculation but incorrectly assumes constant temperature, which is irrelevant to the energy conversion process. Answer D halves the correct value, possibly from using $\frac{1}{4}mv^2$ instead of $\frac{1}{2}mv^2$ . Remember: in mass spectrometry problems, always start with energy conservation and clearly state your initial condition assumptions. The "from rest" assumption is standard and necessary for the basic kinetic energy equation to apply directly.
In surface-enhanced Raman spectroscopy (SERS), the enhancement factor $G$ is defined as $G = \frac{I_{\text{SERS}}/N_{\text{SERS}}}{I_{\text{normal}}/N_{\text{normal}}}$ where $I$ represents intensity and $N$ represents the number of molecules. If both intensities have units $\text{counts}\text{s}^{-1}$ and the molecular surface density is $\rho_s = 10^{12} \text{ molecules}\text{cm}^{-2}$ , what must be the units of the effective area $A_{\text{eff}}$ used in calculating $N_{\text{SERS}} = \rho_s \cdot A_{\text{eff}}$ ?
Answer: $\text{cm}^2$ (area units) When working with SERS enhancement factors, you're dealing with dimensional analysis applied to spectroscopic measurements. The key insight is recognizing that each term in the enhancement factor equation must be dimensionally consistent. Let's trace through the units systematically. The enhancement factor $G$ compares normalized intensities, so it must be dimensionless. Since both $I_{\text{SERS}}$ and $I_{\text{normal}}$ have units of $\text{counts}\text{s}^{-1}$ , the ratio $\frac{I_{\text{SERS}}}{I_{\text{normal}}}$ is dimensionless. Therefore, the ratio $\frac{N_{\text{SERS}}}{N_{\text{normal}}}$ must also be dimensionless, meaning both $N$ values must have identical units. Since $N$ represents the number of molecules, it should be dimensionless (a pure count). For $N_{\text{SERS}} = \rho_s \cdot A_{\text{eff}}$ to yield a dimensionless result, we need: $[\text{molecules}] = [\text{molecules}\text{cm}^{-2}] \times [A_{\text{eff}}]$ Solving for $A_{\text{eff}}$ : $[A_{\text{eff}}] = \frac{[\text{molecules}]}{[\text{molecules}\text{cm}^{-2}]} = \text{cm}^2$ Choice A is correct because $A_{\text{eff}}$ represents an actual area measurement. Choice B gives the wrong units ( $\text{molecules}\text{cm}^{-2}$ ) - that's already the units of $\rho_s$ . Choice C (dimensionless) would make $N_{\text{SERS}}$ have the same units as $\rho_s$ , which is incorrect. Choice D ( $\text{molecules}^{-1}\text{cm}^2$ ) would create units that don't cancel properly with $\rho_s$ . Study tip: In dimensional analysis problems, always work backwards from what the final result should be, then solve for the unknown quantity's units.
A researcher measures the fluorescence lifetime of a compound and reports the value as $2.5 \times 10^{-8}$ seconds. When calculating the radiative rate constant using $k_r = \phi_f / \tau$ , where $\phi_f = 0.85$ is the quantum yield, what are the correct units for $k_r$ and its numerical value?
Answer: $3.4 \times 10^7 \text{ s}^{-1}$ with units of inverse time The radiative rate constant $k_r$ has units of $\text{s}^{-1}$ since it equals a dimensionless quantum yield divided by time. Calculating: $k_r = 0.85/(2.5 \times 10^{-8} \text{ s}) = 3.4 \times 10^7 \text{ s}^{-1}$ . Choice B has incorrect units (should be inverse time). Choice C uses the lifetime value instead of calculating the rate constant. Choice D incorrectly multiplies instead of divides the quantum yield by lifetime.
A UV-Vis absorption spectrum shows a peak with molar absorptivity $\varepsilon = 2.4 \times 10^4 \text{ L mol}^{-1} \text{cm}^{-1}$ at $\lambda = 280 \text{ nm}$ . Using the relationship $f = 4.32 \times 10^{-9} \int \varepsilon(\tilde{\nu}) d\tilde{\nu}$ , what are the units of the oscillator strength $f$ and the integral?
Answer: $f$ is dimensionless; integral has units $\text{L mol}^{-1} \text{cm}^{-2}$ The oscillator strength $f$ is fundamentally dimensionless as it represents the ratio of actual to classical oscillator strength. Since $\varepsilon$ has units $\text{L mol}^{-1} \text{cm}^{-1}$ and $\tilde{\nu}$ has units $\text{cm}^{-1}$ , the integral $\int \varepsilon(\tilde{\nu}) d\tilde{\nu}$ has units $\text{L mol}^{-1} \text{cm}^{-1} \times \text{cm}^{-1} = \text{L mol}^{-1} \text{cm}^{-2}$ . The constant $4.32 \times 10^{-9}$ must have units $\text{mol L}^{-1} \text{cm}^{2}$ to make $f$ dimensionless. Choices B and D have incorrect units for the integral, while choice C incorrectly assigns units to $f$ .
In NMR spectroscopy, the chemical shift in ppm is related to the absolute frequency difference by $\delta = \frac{\nu - \nu_{\text{ref}}}{\nu_0} \times 10^6$ . If a proton signal appears at $\delta = 7.25 \text{ ppm}$ in a $400 \text{ MHz}$ spectrometer, what is the absolute frequency difference from TMS in both Hz and rad/s?
Answer: $2900 \text{ Hz}$ and $1.82 \times 10^4 \text{ rad/s}$ When you encounter NMR chemical shift problems, you're working with the relationship between the dimensionless ppm scale and actual frequency differences. The key is understanding that chemical shift in ppm represents parts per million of the spectrometer's operating frequency. Starting with the given equation $\delta = \frac{\nu - \nu_{\text{ref}}}{\nu_0} \times 10^6$ , you can rearrange to find the absolute frequency difference: $\nu - \nu_{\text{ref}} = \frac{\delta \times \nu_0}{10^6}$ . With $\delta = 7.25 \text{ ppm}$ and $\nu_0 = 400 \text{ MHz} = 4 \times 10^8 \text{ Hz}$ , the calculation becomes: $\nu - \nu_{\text{ref}} = \frac{7.25 \times 4 \times 10^8}{10^6} = 2900 \text{ Hz}$ . To convert to rad/s, multiply by $2\pi$ : $2900 \times 2\pi = 18,220 \text{ rad/s} = 1.82 \times 10^4 \text{ rad/s}$ . This confirms answer D is correct. Answer A miscalculates the frequency difference by a factor of 10, giving 29,000 Hz instead of 2,900 Hz, but coincidentally gets the right order of magnitude for rad/s. Answer B correctly calculates 2,900 Hz but makes an error converting to rad/s, getting $10^3$ instead of $10^4$ . Answer C contains both errors: wrong frequency (290 Hz, off by a factor of 10) and wrong rad/s conversion. Remember that chemical shifts scale directly with spectrometer frequency - a 7.25 ppm shift always means 7.25 parts per million of whatever the operating frequency is. Practice converting between Hz and rad/s using the $2\pi$ factor to avoid unit conversion errors.
In electroabsorption spectroscopy, the change in absorption coefficient under an applied electric field $F$ is given by $\Delta \alpha = \alpha_0 + \alpha_1 F + \alpha_2 F^2$ where $\alpha_0$ , $\alpha_1$ , and $\alpha_2$ are field-independent coefficients. If $\Delta \alpha$ and the baseline absorption $\alpha_0$ both have units $\text{cm}^{-1}$ and the electric field strength $F = 10^5 \text{ V}\text{cm}^{-1}$ , what are the units of the quadratic electroabsorption coefficient $\alpha_2$ ?
Answer: $\text{cm}\text{V}^{-2}$ (length per voltage squared) When you encounter electroabsorption spectroscopy problems, you're dealing with how electric fields modify optical properties. The key insight is using dimensional analysis to ensure each term in the equation has consistent units. Looking at the given equation $\Delta \alpha = \alpha_0 + \alpha_1 F + \alpha_2 F^2$ , every term on the right side must have the same units as $\Delta \alpha$ , which is $\text{cm}^{-1}$ . This is a fundamental requirement in physics equations. For the quadratic term $\alpha_2 F^2$ to equal $\text{cm}^{-1}$ , we need: $[\alpha_2] \times [F^2] = \text{cm}^{-1}$ Since $F = 10^5 \text{ V}\text{cm}^{-1}$ , we have $[F^2] = \text{V}^2\text{cm}^{-2}$ Therefore: $[\alpha_2] = \frac{\text{cm}^{-1}}{\text{V}^2\text{cm}^{-2}} = \text{cm}^{-1} \times \text{V}^{-2} \times \text{cm}^{2} = \text{cm}\text{V}^{-2}$ This confirms answer A is correct. B ( $\text{cm}^{-1}\text{V}^{-2}$ ) would give $\alpha_2 F^2$ units of $\text{cm}^{-3}$ , not $\text{cm}^{-1}$ . C ( $\text{cm}^{-3}\text{V}^{-2}$ ) would result in $\alpha_2 F^2$ having units of $\text{cm}^{-5}$ , completely wrong. D ( $\text{cm}^3\text{V}^{-2}$ ) would make $\alpha_2 F^2$ have units of $\text{cm}$ , not the required $\text{cm}^{-1}$ . Study tip: In spectroscopy problems involving field effects, always check dimensional consistency. Each term in a sum must have identical units, making dimensional analysis your most reliable tool for coefficient problems.
In photoelectron spectroscopy, the kinetic energy of ejected electrons is given by $KE = h\nu - BE$ , where $BE$ is the binding energy. If X-rays with energy $1486.6 \text{ eV}$ are used and an electron is detected with kinetic energy $1200.3 \text{ eV}$ , what is the binding energy in both eV and wavenumbers?
Answer: $286.3 \text{ eV}$ and $2.309 \times 10^6 \text{ cm}^{-1}$ The binding energy is $BE = 1486.6 - 1200.3 = 286.3 \text{ eV}$ . To convert to wavenumbers: $\tilde{\nu} = BE/(hc) = 286.3 \text{ eV}/(1.24 \times 10^{-4} \text{ eV}\cdot\text{cm}) = 2.309 \times 10^6 \text{ cm}^{-1}$ . Choice B uses an incorrect conversion factor (off by 1000). Choice C incorrectly adds the energies instead of subtracting. Choice D has the wrong power of 10 in the wavenumber conversion.
A researcher uses time-resolved fluorescence to measure excited-state dynamics. The fluorescence intensity follows $I(t) = I_0 e^{-t/\tau}$ with $\tau = 4.2 \text{ ns}$ . If the detection system has a time resolution of $100 \text{ ps}$ , what fraction of the total fluorescence signal is collected within the first $500 \text{ ps}$ , and what are the appropriate units for the rate constant $k = 1/\tau$ ?
Answer: $11.3\%$ of signal collected; $k$ has units $\text{s}^{-1}$ Time-resolved fluorescence spectroscopy measures how excited molecules return to their ground state, following first-order exponential decay kinetics. When you see the equation $I(t) = I_0 e^{-t/\tau}$ , you're dealing with a classic exponential decay problem where $\tau$ is the fluorescence lifetime. To find the fraction of signal collected within 500 ps, you need to calculate what percentage of the total signal has been emitted by that time. The fraction collected is: $\frac{\int_0^{500\text{ ps}} I_0 e^{-t/\tau} dt}{\int_0^{\infty} I_0 e^{-t/\tau} dt}$ This simplifies to $1 - e^{-t/\tau} = 1 - e^{-500\text{ ps}/4200\text{ ps}} = 1 - e^{-0.119} = 1 - 0.887 = 0.113$ or 11.3%. For the rate constant units, since $k = 1/\tau$ and $\tau$ has units of time, $k$ must have units of inverse time. The standard SI unit for rate constants is $\text{s}^{-1}$ , not $\text{ns}^{-1}$ . Answer A gives the correct percentage but wrong units for the rate constant. Answer C incorrectly calculates 88.7% - this represents the fraction remaining (not collected) at 500 ps, confusing $e^{-t/\tau}$ with $1-e^{-t/\tau}$ . Answer D shows 5.6%, which might result from a calculation error or unit conversion mistake. Study tip: In exponential decay problems, always distinguish between "fraction remaining" ( $e^{-t/\tau}$ ) and "fraction that has decayed" ( $1-e^{-t/\tau}$ ). Also remember that rate constants in kinetics conventionally use SI units of $\text{s}^{-1}$ .

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