Physical Chemistry 2 Quiz: Dimensional Analysis In Spectroscopy
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Dimensional Analysis In SpectroscopyQuestion 1 of 15

In photoelectron spectroscopy, the kinetic energy of ejected electrons is given by KE=hνBEKE = h\nu - BE, where BEBE is the binding energy. If X-rays with energy 1486.6 eV1486.6 \text{ eV} are used and an electron is detected with kinetic energy 1200.3 eV1200.3 \text{ eV}, what is the binding energy in both eV and wavenumbers?

286.3 eV286.3 \text{ eV} and 2.309×106 cm12.309 \times 10^6 \text{ cm}^{-1}
286.3 eV286.3 \text{ eV} and 2.309×103 cm12.309 \times 10^3 \text{ cm}^{-1}
2686.9 eV2686.9 \text{ eV} and 2.166×107 cm12.166 \times 10^7 \text{ cm}^{-1}
286.3 eV286.3 \text{ eV} and 2.309×104 cm12.309 \times 10^4 \text{ cm}^{-1}
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Physical Chemistry 2 Quiz

Physical Chemistry 2 Quiz: Dimensional Analysis In Spectroscopy

Practice Dimensional Analysis In Spectroscopy in Physical Chemistry 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Dimensional Analysis In Spectroscopy, giving you a quick way to practice the rules, question types, and explanations that matter most for Physical Chemistry 2.

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Question 1

In photoelectron spectroscopy, the kinetic energy of ejected electrons is given by KE=hνBEKE = h\nu - BE, where BEBE is the binding energy. If X-rays with energy 1486.6 eV1486.6 \text{ eV} are used and an electron is detected with kinetic energy 1200.3 eV1200.3 \text{ eV}, what is the binding energy in both eV and wavenumbers?

  1. 286.3 eV286.3 \text{ eV} and 2.309×106 cm12.309 \times 10^6 \text{ cm}^{-1} (correct answer)
  2. 286.3 eV286.3 \text{ eV} and 2.309×103 cm12.309 \times 10^3 \text{ cm}^{-1}
  3. 2686.9 eV2686.9 \text{ eV} and 2.166×107 cm12.166 \times 10^7 \text{ cm}^{-1}
  4. 286.3 eV286.3 \text{ eV} and 2.309×104 cm12.309 \times 10^4 \text{ cm}^{-1}
Explanation: The binding energy is BE=1486.61200.3=286.3 eVBE = 1486.6 - 1200.3 = 286.3 \text{ eV}. To convert to wavenumbers: ν~=BE/(hc)=286.3 eV/(1.24×104 eVcm)=2.309×106 cm1\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.

Question 2

A researcher measures the rotational constant BeB_e of a diatomic molecule as 1.93 cm11.93 \text{ cm}^{-1}. Using the relationship Be=h8π2IcB_e = \frac{h}{8\pi^2 I c}, what is the moment of inertia in both kgm2\text{kg}\cdot\text{m}^2 and amuA˚2\text{amu}\cdot\text{Å}^2?

  1. 2.74×1046 kgm22.74 \times 10^{-46} \text{ kg}\cdot\text{m}^2 and 1.65 amuA˚21.65 \text{ amu}\cdot\text{Å}^2
  2. 2.74×1047 kgm22.74 \times 10^{-47} \text{ kg}\cdot\text{m}^2 and 16.5 amuA˚216.5 \text{ amu}\cdot\text{Å}^2
  3. 4.32×1048 kgm24.32 \times 10^{-48} \text{ kg}\cdot\text{m}^2 and 2.60 amuA˚22.60 \text{ amu}\cdot\text{Å}^2
  4. 2.74×1047 kgm22.74 \times 10^{-47} \text{ kg}\cdot\text{m}^2 and 1.65 amuA˚21.65 \text{ amu}\cdot\text{Å}^2 (correct answer)
Explanation: 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 BeB_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=h8π2BecI = \frac{h}{8\pi^2 B_e c}. Substituting the constants (h=6.626×1034 Jsh = 6.626 \times 10^{-34} \text{ J}\cdot\text{s}, c=2.998×1010 cm/sc = 2.998 \times 10^{10} \text{ cm/s}) and Be=1.93 cm1B_e = 1.93 \text{ cm}^{-1}: I=6.626×10348π2×1.93×2.998×1010=2.74×1047 kgm2I = \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 amuA˚2\text{amu}\cdot\text{Å}^2 using the conversion factor 1 kgm2=6.022×1023 amuA˚21 \text{ kg}\cdot\text{m}^2 = 6.022 \times 10^{23} \text{ amu}\cdot\text{Å}^2: I=2.74×1047×6.022×1023=1.65 amuA˚2I = 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 (104610^{-46} instead of 104710^{-47}), suggesting an arithmetic error. Answer B correctly calculates the SI units but has the wrong conversion to amuA˚2\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.

Question 3

A researcher calculates the oscillator strength ff using the formula f=4.32×109nε(ν~)dν~f = \frac{4.32 \times 10^{-9}}{n} \int \varepsilon(\tilde{\nu}) \, d\tilde{\nu}, where nn is the refractive index, ε\varepsilon is the molar absorptivity, and ν~\tilde{\nu} is the wavenumber. If ε\varepsilon has units M1cm1\text{M}^{-1}\text{cm}^{-1} and the integral evaluates to 2.5×106 M1cm22.5 \times 10^6 \text{ M}^{-1}\text{cm}^{-2}, what are the units of the numerical constant 4.32×1094.32 \times 10^{-9}?

  1. Mcm2\text{M}\text{cm}^2 (concentration times area) (correct answer)
  2. dimensionless\text{dimensionless} (no units)
  3. M1cm2\text{M}^{-1}\text{cm}^{-2} (inverse concentration per area)
  4. Mcm1\text{M}\text{cm}^{-1} (concentration per length)
  5. cm2\text{cm}^2 (area units only)
Explanation: 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 ff must be dimensionless—it's a pure number representing the relative strength of an electronic transition. To find the units of the constant 4.32×1094.32 \times 10^{-9}, you need to work backwards from the requirement that ff be dimensionless. Let's analyze the units in the formula f=4.32×109nε(ν~)dν~f = \frac{4.32 \times 10^{-9}}{n} \int \varepsilon(\tilde{\nu}) \, d\tilde{\nu}. The refractive index nn is dimensionless. The integral ε(ν~)dν~\int \varepsilon(\tilde{\nu}) \, d\tilde{\nu} has units of M1cm1×cm1=M1cm2\text{M}^{-1}\text{cm}^{-1} \times \text{cm}^{-1} = \text{M}^{-1}\text{cm}^{-2} (since dν~d\tilde{\nu} has wavenumber units of cm1\text{cm}^{-1}). For ff to be dimensionless, the constant must cancel out these units: M1cm2×constant units=dimensionless\text{M}^{-1}\text{cm}^{-2} \times \text{constant units} = \text{dimensionless}. Therefore, the constant needs units of Mcm2\text{M}\text{cm}^2. Choice A is correct—the constant has units of Mcm2\text{M}\text{cm}^2. Choice B incorrectly assumes the constant is dimensionless, which would leave ff 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.

Question 4

In surface-enhanced Raman spectroscopy (SERS), the enhancement factor GG is defined as G=ISERS/NSERSInormal/NnormalG = \frac{I_{\text{SERS}}/N_{\text{SERS}}}{I_{\text{normal}}/N_{\text{normal}}} where II represents intensity and NN represents the number of molecules. If both intensities have units countss1\text{counts}\text{s}^{-1} and the molecular surface density is ρs=1012 moleculescm2\rho_s = 10^{12} \text{ molecules}\text{cm}^{-2}, what must be the units of the effective area AeffA_{\text{eff}} used in calculating NSERS=ρsAeffN_{\text{SERS}} = \rho_s \cdot A_{\text{eff}}?

  1. cm2\text{cm}^2 (area units) (correct answer)
  2. moleculescm2\text{molecules}\text{cm}^{-2} (surface density units)
  3. dimensionless\text{dimensionless} (no units)
  4. molecules1cm2\text{molecules}^{-1}\text{cm}^2 (area per molecule)
  5. cm2molecules1\text{cm}^2\text{molecules}^{-1} (area per molecule)
Explanation: 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 GG compares normalized intensities, so it must be dimensionless. Since both ISERSI_{\text{SERS}} and InormalI_{\text{normal}} have units of countss1\text{counts}\text{s}^{-1}, the ratio ISERSInormal\frac{I_{\text{SERS}}}{I_{\text{normal}}} is dimensionless. Therefore, the ratio NSERSNnormal\frac{N_{\text{SERS}}}{N_{\text{normal}}} must also be dimensionless, meaning both NN values must have identical units. Since NN represents the number of molecules, it should be dimensionless (a pure count). For NSERS=ρsAeffN_{\text{SERS}} = \rho_s \cdot A_{\text{eff}} to yield a dimensionless result, we need: [molecules]=[moleculescm2]×[Aeff][\text{molecules}] = [\text{molecules}\text{cm}^{-2}] \times [A_{\text{eff}}] Solving for AeffA_{\text{eff}}: [Aeff]=[molecules][moleculescm2]=cm2[A_{\text{eff}}] = \frac{[\text{molecules}]}{[\text{molecules}\text{cm}^{-2}]} = \text{cm}^2 Choice A is correct because AeffA_{\text{eff}} represents an actual area measurement. Choice B gives the wrong units (moleculescm2\text{molecules}\text{cm}^{-2}) - that's already the units of ρs\rho_s. Choice C (dimensionless) would make NSERSN_{\text{SERS}} have the same units as ρs\rho_s, which is incorrect. Choice D (molecules1cm2\text{molecules}^{-1}\text{cm}^2) would create units that don't cancel properly with ρs\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.

Question 5

In Z-scan measurements, the nonlinear absorption coefficient βeff\beta_{\text{eff}} is determined from T=11+βeffI0LeffT = \frac{1}{1 + \beta_{\text{eff}} I_0 L_{\text{eff}}} where TT is transmittance, I0I_0 is peak intensity, and Leff=1eα0Lα0L_{\text{eff}} = \frac{1-e^{-\alpha_0 L}}{\alpha_0} is the effective length with α0\alpha_0 being the linear absorption coefficient. If α0=0.1 cm1\alpha_0 = 0.1 \text{ cm}^{-1}, L=2 mmL = 2 \text{ mm}, and I0=108 Wcm2I_0 = 10^8 \text{ W}\text{cm}^{-2}, what are the units of βeff\beta_{\text{eff}}?

  1. cmW1\text{cm}\text{W}^{-1} (length per power) (correct answer)
  2. cm2W1\text{cm}^2\text{W}^{-1} (area per power)
  3. cm1W1\text{cm}^{-1}\text{W}^{-1} (inverse length per power)
  4. W1\text{W}^{-1} (inverse power)
  5. dimensionless\text{dimensionless} (no units)
Explanation: 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 βeff\beta_{\text{eff}}, rearrange the transmittance equation: βeff=1TTI0Leff\beta_{\text{eff}} = \frac{1-T}{T \cdot I_0 \cdot L_{\text{eff}}}. Since transmittance TT is dimensionless (it's a ratio), the numerator (1T)(1-T) is also dimensionless. For the equation to be dimensionally consistent, the denominator must also be dimensionless, which means βeffI0Leff\beta_{\text{eff}} \cdot I_0 \cdot L_{\text{eff}} must be dimensionless. Given that I0I_0 has units of Wcm2\text{W}\text{cm}^{-2} and LeffL_{\text{eff}} has units of length (cm), we have: [βeff]Wcm2cm=dimensionless[\beta_{\text{eff}}] \cdot \text{W}\text{cm}^{-2} \cdot \text{cm} = \text{dimensionless}. This simplifies to [βeff]Wcm1=dimensionless[\beta_{\text{eff}}] \cdot \text{W}\text{cm}^{-1} = \text{dimensionless}, so [βeff]=cmW1[\beta_{\text{eff}}] = \text{cm}\text{W}^{-1}. Answer A (cmW1\text{cm}\text{W}^{-1}) is correct. Answer B (cm2W1\text{cm}^2\text{W}^{-1}) would arise if you mistakenly thought intensity had units of Wcm1\text{W}\text{cm}^{-1} instead of Wcm2\text{W}\text{cm}^{-2}. Answer C (cm1W1\text{cm}^{-1}\text{W}^{-1}) incorrectly treats length as having inverse units. Answer D (W1\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.

Question 6

In terahertz time-domain spectroscopy, the complex refractive index n~(ω)=n(ω)+iκ(ω)\tilde{n}(\omega) = n(\omega) + i\kappa(\omega) is extracted from the transmission T~(ω)=4n~(n~+1)2eiωn~d/c\tilde{T}(\omega) = \frac{4\tilde{n}}{(\tilde{n}+1)^2} e^{i\omega \tilde{n} d/c} where dd is sample thickness and cc is the speed of light. If the extinction coefficient κ\kappa is related to the absorption coefficient by α=4πκνc\alpha = \frac{4\pi\kappa\nu}{c}, what are the units of the expression αcdνκ\frac{\alpha \cdot c \cdot d}{\nu \cdot \kappa}?

  1. dimensionless\text{dimensionless} (no units) (correct answer)
  2. m\text{m} (length units)
  3. Hz\text{Hz} (frequency units)
  4. ms1\text{m}\text{s}^{-1} (velocity units)
  5. s\text{s} (time units)
Explanation: 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 α=4πκνc\alpha = \frac{4\pi\kappa\nu}{c}, we can identify the units: α\alpha (absorption coefficient) has units of m⁻¹, κ\kappa is dimensionless (extinction coefficient), ν\nu has units of Hz (frequency), and cc has units of ms⁻¹. The sample thickness dd has units of m. Now let's analyze αcdνκ\frac{\alpha \cdot c \cdot d}{\nu \cdot \kappa}:
  • Numerator: αcd=m1ms1m=s1\alpha \cdot c \cdot d = \text{m}^{-1} \cdot \text{ms}^{-1} \cdot \text{m} = \text{s}^{-1}
  • Denominator: νκ=Hzdimensionless=s1\nu \cdot \kappa = \text{Hz} \cdot \text{dimensionless} = \text{s}^{-1}
  • Final result: s1s1=dimensionless\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.

Question 7

In electroabsorption spectroscopy, the change in absorption coefficient under an applied electric field FF is given by Δα=α0+α1F+α2F2\Delta \alpha = \alpha_0 + \alpha_1 F + \alpha_2 F^2 where α0\alpha_0, α1\alpha_1, and α2\alpha_2 are field-independent coefficients. If Δα\Delta \alpha and the baseline absorption α0\alpha_0 both have units cm1\text{cm}^{-1} and the electric field strength F=105 Vcm1F = 10^5 \text{ V}\text{cm}^{-1}, what are the units of the quadratic electroabsorption coefficient α2\alpha_2?

  1. cmV2\text{cm}\text{V}^{-2} (length per voltage squared) (correct answer)
  2. cm1V2\text{cm}^{-1}\text{V}^{-2} (inverse length per voltage squared)
  3. cm3V2\text{cm}^{-3}\text{V}^{-2} (inverse volume per voltage squared)
  4. cm3V2\text{cm}^3\text{V}^{-2} (volume per voltage squared)
  5. dimensionless\text{dimensionless} (no units)
Explanation: 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 Δα=α0+α1F+α2F2\Delta \alpha = \alpha_0 + \alpha_1 F + \alpha_2 F^2, every term on the right side must have the same units as Δα\Delta \alpha, which is cm1\text{cm}^{-1}. This is a fundamental requirement in physics equations. For the quadratic term α2F2\alpha_2 F^2 to equal cm1\text{cm}^{-1}, we need: [α2]×[F2]=cm1[\alpha_2] \times [F^2] = \text{cm}^{-1} Since F=105 Vcm1F = 10^5 \text{ V}\text{cm}^{-1}, we have [F2]=V2cm2[F^2] = \text{V}^2\text{cm}^{-2} Therefore: [α2]=cm1V2cm2=cm1×V2×cm2=cmV2[\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 (cm1V2\text{cm}^{-1}\text{V}^{-2}) would give α2F2\alpha_2 F^2 units of cm3\text{cm}^{-3}, not cm1\text{cm}^{-1}. C (cm3V2\text{cm}^{-3}\text{V}^{-2}) would result in α2F2\alpha_2 F^2 having units of cm5\text{cm}^{-5}, completely wrong. D (cm3V2\text{cm}^3\text{V}^{-2}) would make α2F2\alpha_2 F^2 have units of cm\text{cm}, not the required cm1\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.

Question 8

In single-molecule fluorescence correlation spectroscopy, the autocorrelation function G(τ)=1+1N(1+ττD)1(1+τω2τD)1/2G(\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 DD is related to the diffusion time by τD=w024D\tau_D = \frac{w_0^2}{4D} where w0w_0 is the beam waist radius, and D=2.3×1010 m2s1D = 2.3 \times 10^{-10} \text{ m}^2\text{s}^{-1}, what are the units of the ratio τDDw02\frac{\tau_D \cdot D}{w_0^2}?

  1. dimensionless\text{dimensionless} (no units) (correct answer)
  2. s1\text{s}^{-1} (inverse time)
  3. m2s1\text{m}^2\text{s}^{-1} (area per time)
  4. s\text{s} (time units)
  5. m2s\text{m}^{-2}\text{s} (time per area)
Explanation: 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 τD=w024D\tau_D = \frac{w_0^2}{4D}, so we can identify the units of each component:
  • τD\tau_D has units of time: s\text{s}
  • DD has units of m2s1\text{m}^2\text{s}^{-1}
  • w0w_0 has units of length: m\text{m}
Now let's substitute into the ratio τDDw02\frac{\tau_D \cdot D}{w_0^2}: τDDw02=(s)(m2s1)(m)2=sm2s1m2=m2m2=1\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 (ss1=1\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 (s1\text{s}^{-1}) incorrectly assumes only the length units cancel. Choice C (m2s1\text{m}^2\text{s}^{-1}) mistakenly treats this as if you're just looking at the diffusion coefficient units. Choice D (s\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.

Question 9

In NMR spectroscopy, the chemical shift in ppm is related to the absolute frequency difference by δ=ννrefν0×106\delta = \frac{\nu - \nu_{\text{ref}}}{\nu_0} \times 10^6. If a proton signal appears at δ=7.25 ppm\delta = 7.25 \text{ ppm} in a 400 MHz400 \text{ MHz} spectrometer, what is the absolute frequency difference from TMS in both Hz and rad/s?

  1. 29000 Hz29000 \text{ Hz} and 1.82×105 rad/s1.82 \times 10^5 \text{ rad/s}
  2. 2900 Hz2900 \text{ Hz} and 1.82×103 rad/s1.82 \times 10^3 \text{ rad/s}
  3. 290 Hz290 \text{ Hz} and 1.82×103 rad/s1.82 \times 10^3 \text{ rad/s}
  4. 2900 Hz2900 \text{ Hz} and 1.82×104 rad/s1.82 \times 10^4 \text{ rad/s} (correct answer)
Explanation: 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 δ=ννrefν0×106\delta = \frac{\nu - \nu_{\text{ref}}}{\nu_0} \times 10^6, you can rearrange to find the absolute frequency difference: ννref=δ×ν0106\nu - \nu_{\text{ref}} = \frac{\delta \times \nu_0}{10^6}. With δ=7.25 ppm\delta = 7.25 \text{ ppm} and ν0=400 MHz=4×108 Hz\nu_0 = 400 \text{ MHz} = 4 \times 10^8 \text{ Hz}, the calculation becomes: ννref=7.25×4×108106=2900 Hz\nu - \nu_{\text{ref}} = \frac{7.25 \times 4 \times 10^8}{10^6} = 2900 \text{ Hz}. To convert to rad/s, multiply by 2π2\pi: 2900×2π=18,220 rad/s=1.82×104 rad/s2900 \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 10310^3 instead of 10410^4. 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 2π2\pi factor to avoid unit conversion errors.

Question 10

A researcher measures the fluorescence lifetime of a compound and reports the value as 2.5×1082.5 \times 10^{-8} seconds. When calculating the radiative rate constant using kr=ϕf/τk_r = \phi_f / \tau, where ϕf=0.85\phi_f = 0.85 is the quantum yield, what are the correct units for krk_r and its numerical value?

  1. 3.4×107 s13.4 \times 10^7 \text{ s}^{-1} with units of inverse time (correct answer)
  2. 3.4×107 s3.4 \times 10^7 \text{ s} with units of time
  3. 2.1×108 s12.1 \times 10^{-8} \text{ s}^{-1} with units of inverse time
  4. 2.9×109 s12.9 \times 10^{-9} \text{ s}^{-1} with units of inverse time
Explanation: The radiative rate constant krk_r has units of s1\text{s}^{-1} since it equals a dimensionless quantum yield divided by time. Calculating: kr=0.85/(2.5×108 s)=3.4×107 s1k_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.

Question 11

In infrared spectroscopy, the wavenumber of a vibrational transition is given by ν~=12πckμ\tilde{\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}, where kk is the force constant and μ\mu is the reduced mass. For a C-H stretch with ν~=2900 cm1\tilde{\nu} = 2900 \text{ cm}^{-1}, what is the force constant in both N/m and mdyn/Å?

  1. 510 N/m510 \text{ N/m} and 51.0 mdyn/A˚51.0 \text{ mdyn/Å}
  2. 510 N/m510 \text{ N/m} and 5.10 mdyn/A˚5.10 \text{ mdyn/Å} (correct answer)
  3. 51.0 N/m51.0 \text{ N/m} and 0.51 mdyn/A˚0.51 \text{ mdyn/Å}
  4. 5100 N/m5100 \text{ N/m} and 51.0 mdyn/A˚51.0 \text{ mdyn/Å}
Explanation: 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=(2πcν~)2μk = (2\pi c \tilde{\nu})^2 \mu. For a C-H bond, the reduced mass is μ=mC×mHmC+mH=12×112+1=0.923\mu = \frac{m_C \times m_H}{m_C + m_H} = \frac{12 \times 1}{12 + 1} = 0.923 amu. Convert this to kg: μ=0.923×1.66×1027=1.53×1027\mu = 0.923 \times 1.66 \times 10^{-27} = 1.53 \times 10^{-27} kg. With ν~=2900 cm1=2.9×105 m1\tilde{\nu} = 2900 \text{ cm}^{-1} = 2.9 \times 10^5 \text{ m}^{-1} and c=3×108 m/sc = 3 \times 10^8 \text{ m/s}: k=(2π×3×108×2.9×105)2×1.53×1027=510 N/mk = (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 N/m=102 mdyn/A˚1 \text{ N/m} = 10^{-2} \text{ mdyn/Å}, so 510 N/m=5.10 mdyn/A˚510 \text{ N/m} = 5.10 \text{ mdyn/Å}. Option A incorrectly converts units by a factor of 10 (uses 10110^{-1} instead of 10210^{-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 10210^{-2}.

Question 12

A mass spectrometer operates with an electric field of 2.5×104 V/m2.5 \times 10^4 \text{ V/m} and accelerates ions through a potential difference of 3000 V3000 \text{ V}. An ion with m/z=120m/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?

  1. 4.48×105 m/s4.48 \times 10^5 \text{ m/s} assuming the ion starts from rest
  2. 2.24×105 m/s2.24 \times 10^5 \text{ m/s} assuming constant ion temperature throughout acceleration
  3. 2.24×105 m/s2.24 \times 10^5 \text{ m/s} assuming the ion starts from rest (correct answer)
  4. 1.12×105 m/s1.12 \times 10^5 \text{ m/s} assuming negligible initial kinetic energy compared to final
Explanation: 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=12mv2qV = \frac{1}{2}mv^2, where qq is the ion charge, VV is the potential difference, mm is the mass, and vv is the final velocity. For an ion with m/z=120m/z = 120, assuming it's singly charged (z=1z = 1), the mass is 120 u=120×1.66×1027 kg=1.99×1025 kg120 \text{ u} = 120 \times 1.66 \times 10^{-27} \text{ kg} = 1.99 \times 10^{-25} \text{ kg}. The charge is 1.60×1019 C1.60 \times 10^{-19} \text{ C}. Solving for velocity: v=2qVm=2×1.60×1019×30001.99×1025=2.24×105 m/sv = \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 2qV2qV instead of qVqV 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 14mv2\frac{1}{4}mv^2 instead of 12mv2\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.

Question 13

In electron spin resonance (ESR), the resonance condition is hν=gμBBh\nu = g\mu_B B. If the g-factor is measured as 2.0032.003 and the magnetic field is 0.335 T0.335 \text{ T}, what microwave frequency should be used? Express the answer in both GHz and wavenumbers.

  1. 18.8 GHz18.8 \text{ GHz} and 0.627 cm10.627 \text{ cm}^{-1}
  2. 9.39 GHz9.39 \text{ GHz} and 3.13 cm13.13 \text{ cm}^{-1}
  3. 9.39 GHz9.39 \text{ GHz} and 0.313 cm10.313 \text{ cm}^{-1} (correct answer)
  4. 9.39 GHz9.39 \text{ GHz} and 31.3 cm131.3 \text{ cm}^{-1}
Explanation: 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 hν=gμBBh\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: ν=gμBBh\nu = \frac{g\mu_B B}{h}. Using the given values (g=2.003g = 2.003, B=0.335B = 0.335 T) and constants (μB=9.274×1024\mu_B = 9.274 \times 10^{-24} J/T, h=6.626×1034h = 6.626 \times 10^{-34} J·s): ν=(2.003)(9.274×1024)(0.335)6.626×1034=9.39×109 Hz=9.39 GHz\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 ν~=νc\tilde{\nu} = \frac{\nu}{c} where c=3.00×1010c = 3.00 \times 10^{10} cm/s: ν~=9.39×1093.00×1010=0.313 cm1\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 2g2g instead of gg. 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.

Question 14

A researcher uses time-resolved fluorescence to measure excited-state dynamics. The fluorescence intensity follows I(t)=I0et/τI(t) = I_0 e^{-t/\tau} with τ=4.2 ns\tau = 4.2 \text{ ns}. If the detection system has a time resolution of 100 ps100 \text{ ps}, what fraction of the total fluorescence signal is collected within the first 500 ps500 \text{ ps}, and what are the appropriate units for the rate constant k=1/τk = 1/\tau?

  1. 11.3%11.3\% of signal collected; kk has units ns1\text{ns}^{-1}
  2. 11.3%11.3\% of signal collected; kk has units s1\text{s}^{-1} (correct answer)
  3. 88.7%88.7\% of signal collected; kk has units s1\text{s}^{-1}
  4. 5.6%5.6\% of signal collected; kk has units s1\text{s}^{-1}
Explanation: 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)=I0et/τI(t) = I_0 e^{-t/\tau}, you're dealing with a classic exponential decay problem where τ\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: 0500 psI0et/τdt0I0et/τdt\frac{\int_0^{500\text{ ps}} I_0 e^{-t/\tau} dt}{\int_0^{\infty} I_0 e^{-t/\tau} dt} This simplifies to 1et/τ=1e500 ps/4200 ps=1e0.119=10.887=0.1131 - 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/τk = 1/\tau and τ\tau has units of time, kk must have units of inverse time. The standard SI unit for rate constants is s1\text{s}^{-1}, not ns1\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 et/τe^{-t/\tau} with 1et/τ1-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" (et/τe^{-t/\tau}) and "fraction that has decayed" (1et/τ1-e^{-t/\tau}). Also remember that rate constants in kinetics conventionally use SI units of s1\text{s}^{-1}.

Question 15

A UV-Vis absorption spectrum shows a peak with molar absorptivity ε=2.4×104 L mol1cm1\varepsilon = 2.4 \times 10^4 \text{ L mol}^{-1} \text{cm}^{-1} at λ=280 nm\lambda = 280 \text{ nm}. Using the relationship f=4.32×109ε(ν~)dν~f = 4.32 \times 10^{-9} \int \varepsilon(\tilde{\nu}) d\tilde{\nu}, what are the units of the oscillator strength ff and the integral?

  1. ff is dimensionless; integral has units L mol1cm2\text{L mol}^{-1} \text{cm}^{-2} (correct answer)
  2. ff is dimensionless; integral has units L mol1cm0\text{L mol}^{-1} \text{cm}^{0}
  3. ff has units mol L1\text{mol L}^{-1}; integral has units L mol1cm2\text{L mol}^{-1} \text{cm}^{-2}
  4. ff is dimensionless; integral has units L mol1cm1\text{L mol}^{-1} \text{cm}^{-1}
Explanation: The oscillator strength ff is fundamentally dimensionless as it represents the ratio of actual to classical oscillator strength. Since ε\varepsilon has units L mol1cm1\text{L mol}^{-1} \text{cm}^{-1} and ν~\tilde{\nu} has units cm1\text{cm}^{-1}, the integral ε(ν~)dν~\int \varepsilon(\tilde{\nu}) d\tilde{\nu} has units L mol1cm1×cm1=L mol1cm2\text{L mol}^{-1} \text{cm}^{-1} \times \text{cm}^{-1} = \text{L mol}^{-1} \text{cm}^{-2}. The constant 4.32×1094.32 \times 10^{-9} must have units mol L1cm2\text{mol L}^{-1} \text{cm}^{2} to make ff dimensionless. Choices B and D have incorrect units for the integral, while choice C incorrectly assigns units to ff.