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Chemistry LibreTexts

Solutions 21

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  • Q21.1

    What is the wavenumber and frequency of a 450-nm photon?  How does the energy of 450-nm light compare to the energy used to overcome activation barriers in chemical reactions (i.e., thermal energy \(k_bT\)) at room temperature?


    a) \[λ = \dfrac{c}{v} = \dfrac{1}{ \tilde{\nu}}\]

    \[\tilde{\nu} = \dfrac{1}{\lambda} = \dfrac{1}{(450\; \cancel{nm}) \left( \dfrac{1\, \cancel{nm}}{ 1 \times 10^{-7} \, cm} \right) } = 2.2 \times 10^4 \; cm^{-1}\] 

    \[ v = \dfrac{ c}{ λ} = \dfrac{3.00 \times 10^8 \;\cancel{m} s^{-1}} {450\; \cancel{nm} \left( \dfrac{1 \; \cancel{nm}}{1 \times 10^{-9}\; m} \right)} = 6.7 \times 10^{14} s^{-1}\]

    b) \[E = hv = (6.6x10^{-34}Js)(6.6x10^{14}s^{-1}) = 43.56x10^{-21}J\]

    \[k_BT = 4.4x10^{-23}J = 4.14 x10^{-21}J\]

    E is about 10 times bigger than room temperature thermal energy. 


    Convert the following absorbance of a sample to percent transmittance: (a) 0.0, (b) 0.12, and (c) 0%.


    \[T =\dfrac{I}{I_0} = 10^{\log I/I_o} = 10^{-A}\]

    1. \(A = 0.0\) and \(T = 10^{-0} = 1 = 100\%\)
    2. \(A = 0.12\) and \(T = 10^{-0.12} = 0.76 = 76\%\).
    3. \(A = 4.6\) and \(T = 10^{-4.6} = 2.5 \times 10^{-5} = 0.0025\%\)


    An excited molecule has an average lifetime of \(2.0 \times 10^{-8} \ seconds\), and the radiation emission is 600 nm. Find the frequency and wavelength uncertainties.


    $$\Delta \nu = \dfrac{1}{4 \pi \Delta t}$$

    $$\Delta \nu = \dfrac{1}{4 \pi \times 2 \times 10^{-8} \ s}$$

    $$\Delta \nu = 4.0 \times 10^6 \ seconds$$


    $$\Delta \lambda = \dfrac{\lambda \ \Delta \nu}{\nu}$$

    $$\Delta \lambda = \dfrac{600 \ nm \times 4.0 \times 10^6 \ s^{-1} \times 600 \times 10^9 \ m}{3.00 \times 10^8 \ \dfrac{m}{s}}$$

    $$\Delta \lambda = 4.8 \times 10^{-6} \ nm$$


    What is the concentration of a lysozyme solution if a 1.0-cm wide cuvette containing the solution transmits 25% of λ = 280 nm (UV) light? The molar extinction coefficient for lysozyme at 280 nm is 26.4 L mol-1cm-1.


    This is an application of Beer-Lambert Law

    \[A= \epsilon bc\]

    and the relationship between transmittance and absorbance,

    \[ -\log\ T = A\]

    We can solve for c:

    \[ c =\dfrac {-\log\ T}{\varepsilon b}\]

    \[ c =\dfrac {-\log\ 0.25}{(26.4\dfrac{L}{mol*cm})(1.0\ cm)}\]

    \[ c =0.0228\ \dfrac{mol}{L}\]

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    The Doppler effect results in a shift in the wavelength of a star in astronomic meaurements, but only to a broadening of a transition in a gas; why?


    The Doppler effect in a star reflects that the average velocity of the star is non-zero with respect to the observed (us). However, in a gas sample in the laboratory, the average velocity is zero (or small compared to the velocity of the star) and the broadening reflects the wide range of velocities, both toward and away from the observer, of atoms that emit photons.


    Which of these transitions are possible for the electronic absorption of an atom or molecule?
    A 1s electron in hydrogen is promoted to the 2p orbital
    A 1s electron in hydrogen is promoted to the 3d orbital
    A 2s electron in hydrogen is promoted to the 3d orbital
    An electron with the quantum numbers (\(n=2\), \(l=1\), \(m_l=1\) and \(m_s=+1/2\) in hydrogen is demoted (e.g., via the emission of a photon) to a wavefunction with the quantum numbers (\(n=1\), \(l=0\), \(m_l=0\) and \(m_s=+1/2\).


    Assuming a one photon transition

    a) Yes

    b) no

    c) no

    d) Yes