Department of Chemistry The Intersection Potential Energy ...rossi.chemistry.uconn.edu/chem5326/files/pes-ts-mep...2017/02/04 · B = 1:38 10 38J=K is the Boltzmann constant, hPlanck’s

Department of Chemistry

The Intersectionof

Computational Chemistryand

Experiment

Potential Energy Surfaces and Reaction Pathways

Angelo R. RossiDepartment of Chemistry

The University of [email protected]

Spring 2017

Last Updated: February 4, 2017 at 6:32am

Potential Energy Surfaces (PES)

For the reactionA −→ B

a PES might look as shown in Figure 1

Figure 1: Transition State, Saddle Point, Reaction Pathway

• The minima are represented by A and B, which could be reactant and product, or two conformers.

• The reaction path is defined as the pathway between the two minima.

• The reactant passes through a maximum in energy the (saddle point) along the reaction pathway,connecting the two minima.

• There are minima in all other directions perpendicular to the saddle point on the reaction pathway.

• A transition state is defined as the geometry at the peak of the free energy (∆G) profile, while atransition structure defines the geometry at the peak of the potential energy (∆E) profile.

A transition state is a first order saddle point on a potential energy surface (PES).

Last Updated: February 4, 2017 at 6:32am 2

The Minima on a Potential Energy Surface

Following minimization, a molecule which is at a minimum on the PES is characterized by frequencieswhich are all positive, implying positive force constants. This means that the energy is a minimum inall directions. In order to verify if a stationary point is a minimum, a vibrational frequency calculationmust be performed at the same computational level as for the geometry optimization to ensure that allvibrational frequencies are positive.

The Nature of Transition States

The vibrational spectrum of a transition state is characterized by one imaginary frequency, implying anegative force constant. This means that in one direction in nuclear-configuration space, the energy hasa maximum, while in all other (orthogonal) directions, the energy is a minimum. In order to verify ifa stationary point is a transition state, which connects reactants and products, a vibrational frequencycalculation must be performed at the same computational level as for the geometry optimization. Thenormal mode corresponding to the imaginary frequency in the transition state usually reflects the changein geometry in going from reactants to products.

Locating transition states and estimating reaction barriers can be a rather difficult task. The first step isto obtain a reasonable guess for the structure of the transition state. In general, our intuition is not aswell developed for transition states as it is for ground states. In addition, transition-state energies can beclose to excited states, and this often implies that the usual MO-based computational methods may notbe as accurate for transition states than they are for minima on a PES. A fairly high level of theory isoften required to get a good quantitative estimate of a reaction barrier.

But, these exercises will be mainly focused on locating the transition state at lower levels of theory. Oncea transition state is found at a lower level of theory, the calculation can be refined by re-optimizing thestructure at a higher level of theory.

Intrinsic Reaction Coordinate

Another feature that can be very useful when examining transition states is the Intrinsic Reaction Coor-dinate (IRC) which is related to the minimum energy path (MEP) and is defined as the steepest descentpath starting from the transition state, initially going in the direction of negative curvature in the Hessian.The path can be followed backwards (to reactants) and forward (to products). Following this path canindicate, if the transition state that connects the two minima of interest, has really been found.

The IRC is defined in terms of mass-weighted Cartesian coordinates in the Gaussian program, and it hasa well-defined step length. Computationally, once a transition state is found, it is possible to restart fromthe checkpoint file and perform and IRC.


Statistical Thermodynamics

Briefly, the total energy of a molecule consists, to a good approximation, of additive contributions from itselectronic energy, nuclear repulsion, the translational energy, the rotational energy, and vibrational energy.The vibrational energy contribution may be divided into a component corresponding to the zero-pointenergy, Ezpe, which is the sum of the energies of all ν = 0 vibrational levels and a thermal correction (Evib∗

), derived from a Boltzmann-population for the higher vibrational levels of the system.

Etot = Eelec + Etrans + Erot + Evib∗ + Ezpe (1)

Contributions from translational, rotational and excited vibrational states are frequently negligible incomparison to the energies of different isomers, but the zero-point correction, where νk is the kth vibrationalfrequency of the molecule,

Ezpe =3N−6∑k=1

hνk (2)

may be important.

After determining several stationary points and their character on the potential energy surface whichcorrespond to different conformers or electronic states, it may be of interest to determine the populationof these states at a specified temperature. In this situation, Boltzmann statistics can be employed. InBoltzmann statistics the fractional population of the ith state is given by:

Ni

N=

e− εi

kBT∑j

e−

εjkBT

(3)

where Ni is the number of molecules in the ith energy level, εi, N , the total number molecules, kB is theBoltzmann constant, and T the temperature in Kelvin where the sum includes all energy states.

A word of caution: to find a transition state with quantum chemical programs is not an easy task. Sincea calculation has no way to guess a transition state from a stable structure, it is important to guide thecalculation to the desired transition state by providing structures that are close to the final TS. However,this means necessarily, that the outcome of the calculations depends on skill to guess a reasonable TS.There is no guarantee that the calculation, based on the initial guess, will lead to a TS with the lowestenergy. If the calculation does not find a desired TS, it may be necessary to start over and provide a betterstarting structure. In calculating chemical reactions, chemical intuition, experience and luck are crucialfor success.


Transition State Theory

From a microscopic point of view, one is interested in understanding the value of the kinetic reaction rateconstant, k, from first principles. Phenomenologically, one already knows that k is temberature dependentand obeys the Arrhenius equation

k = Ae−EaRT (4)

where A is a pre-exponential factor, R the gas constant, T the temperature, and Ea is a key quantity,called the activation energy. The larger the value of Ea, the slower the reaction.

More detailed insight into reaction rates can be obtained from the transition state theory of Eyring. In thiscase, one assumes that two reactants, A and B, pass through a special geometrical arrangement, (AB)‡-thetransition state, before decaying to the products C and D according to the equation

A + Bk1

k−1

(AB)‡ → C + D (5)

The assumption of a pseudo-steady state concentration for (AB)‡, leads to the rate constant for bimoleculardecay is given by:

k = K∗k∗ (6)

where K∗ (equilbrium constant for transition state) = k1

k−1.

From Quantum Mechanics,

k∗ = κkBT

h(7)

where kB = 1.38× 10−38J/K is the Boltzmann constant, h Planck’s constant, and κ is a transmissioncoefficient, which is usually close to unity. The equilibrium constant, K∗, is related to the free energy ofthe transition state relative to the reactant energies is by:

K∗ = e−∆G‡RT (8)

and

k =kBT

he−

∆G‡RT =

kBT

he−

∆H‡RT e

∆S‡R (9)

Where ∆H‡ and ∆S‡ are the enthalpy and entropy of the transition state. Thus, one identifies theparameters of the Arrhenius equation with:

A =kBT

he

∆S‡R , Ea = ∆H‡ (10)

∆H‡ can be calculated, but ∆S‡ is slightly more difficult to obtain.


Exercises

For all exercises, a detailed reaction energy diagram similar to the one shown in Figure 2 should be providedwhen appropriate. Please be sure to label it properly with reactants, products, activation energy and otheradditional information.

Transition State

Reactants

Products

Reaction Pathway

Energ

y

Figure 2: A Typical Energy vs Reaction Diagram

Additional Information

1. Make sure that there really is a transition state and not a reactive intermediate.

If the transition state is a minimum, find the transition-state structure between reactant and product,and the transition state structure for dissociation.

2. The Intrinsic Reaction Coordinate (IRC) pathway from reactants to products should be calculatedwhen appropriate.

The IRC is related to the minimum energy path (MEP) which is defined as the steepest descentpath starting from the TS and going in either the backward direction (i.e. to reactants) or forwarddirection (i.e. to products).

Following this path can indicate if a TS was found that connects the two minima of interest.


Inversion of Ammonia, NH3

A simple example of tracing a reaction path involves the inversion of ammonia, which proceeds from aninitial pyramidal C3v equilibrium geometry through a planar D3h transition state, and back to an invertedpyramidal C3v equilibrium geometry as shown in Figure 3:

Figure 3: Inversion of Ammonia, NH3

1. Optimize both the ground state and the transition state by imposing the proper symmetries.

2. Calculate vibrational frequencies for both the ground state and transition state.

Label the frequencies according to the appropriate symmetry representations.

3. Locate the single imaginary frequency, and examine the corresponding normal mode. Verify that itcorresponds to the inversion mode.

What is the symmetry respresentation of the imaginary frequency. Does it make sense? Explain.

4. Perform an IRC calculation which shows the path connecting the two pyramidal structures.

Use an appropriate step size, and sufficient steps to unambiguously establish the reaction path.

5. Locate the energies of reactants, products and transition state in their respective .log files.

Calculate the activation energy, as well as the ∆G between reactant and product. The output fromfrequency calculations should provide information about ∆G energies.

Using these results and those of the IRC calculation construct a reaction energy diagram, similar tothe one in Figure 2, showing the reactant, transition state, and product for the inversion of ammonia,and be sure to include energies and label everything appropriately.


Isomerization of H− C ≡ N to C ≡ N− H

As an example of a transition state search, consider the process of hydrogen migration shown in Figure 4:

Figure 4: Isomerization of H− C ≡ N to C ≡ N− H

1. In this case, the minima of H−C−−−N and C−−−N−H can be obtained in the usual fashion, and verifiedas minima by calculating vibrational frequencies.

2. Next, construct a reasonable estimate of the transition state geometry by changing the < H−C−Nangle from 180◦ to about 60◦, but the H−C and N−H bond distances should be elongated as well.

Then, perform a transition-state optimization using this initial geometry.

3. Optimizing to a transition state always requires the additional calculation of vibrational frequencies,because verification that there is precisely one imaginary frequency must be ascertained.

After the calculation finishes in Step #2 finishes successfully, perform a frequency calculation, andview the vibrational frequencies using visualization software.

One frequency should be imaginary, and the corresponding normal mode represents the hydrogenmigration. Also, examine the other normal modes in the transition state.

4. Perform IRC calculations, in both the forwared and reverse directions, which shows the path con-necting the H-CN and CN-H structures.

Use an appropriate step size, and sufficient steps to unambiguously establish the reaction path.

5. Locate the energies of reactants, products and transition state in their respective .log files.

Calculate the activation energy, as well as the ∆G between reactant and product. The output fromfrequency calculations should provide information about ∆G energies.

Using these results and those of the IRC calculation construct a reaction energy diagram, similarto the one in Figure 2, showing the reactant, transition state, and product for HCN isomerizationreaction, and be sure to include energies and label everything appropriately.

Isomerization of CH3 − C ≡ N to C ≡ N− CH3

1. Repeat the above steps for the isomerization of methyl isocyanide (CH3 − C ≡ N) to C ≡ N− CH3.

What energetic changes would you expect when H− is replaced with H3C−?

Explain those differences on the basis of changes in electronic structure of the substituents.


Conformational Analysis

Organic molecules can assume different spatial arrangements (conformations) which are generated byrotation about single bonds. A detailed analysis of the various conformations adopted by individualmolecules is termed conformational analysis. Conformational analysis is an important tool for chemiststrying to unravel the complex structure of both organic and bio-organic molecules in an effort to obtain aclearer understanding of the reactivity and interaction with other molecules.

Rotation About the Central C-C Bond in Butadiene, Acrolein, and Glyoxal

The goal of this exercise is to gain experience in locating different minima on a potential energy surface,to characterize their nature using frequency calculations, and to understand the chemical implications ofthe different minima. Thermal corrections to the total energy resulting from the calculation of vibrationalfrequenciesat the stationary points on a potential energy surface will also be performed.

Glyoxal Butadiene Acroleintrans 0 0 0cis 16a 17b ∼6c

Table 1: Relative Energies of Isomers in kJ/mol for Glyoxal, Butadiene, and Acrolein; (a) Butz KW;Krajnovich DJ; Parmenter CS, Journal of Chemical Physics, 93, 1557-1567 (1990); (b) Engeln R; ConsalvoD; Reuss J, Chemical Physics, 160 , 427-433 (1992); (c) Scuseria, ES; Schaefer III, HF Journal of theAmerican Chemical Society, 111, 7761-7765 (1989)

Rotation Potential Energy Curves, Optimization, and Frequency Calculations

• Calculate rigid and relaxed potential energy curves (PES) for rotation about the central C-C bondfor butadiene, acrolein, and glyoxal.

• Determine whether the calculated stationary points, i.e. the cis conformers , the trans conform-ers , and the maximum energy conformers, are local minima or transition states. To this end,frequency calculations should be performed.

Use chemical intuition in order to guess a starting geometry that leads to convergence to a transitionstate structure, and confirm the results by performing frequency analysis.

For either butadiene, acrolein, or glyoxal, a reasonable starting geometry for a transition state struc-ture can be obtained by rotating a −HC−−CH2 or −C−−O group about the central C-C bond by 90◦,so that it is arranged perpendicular to the other −HC−−CH2 or −C−−O group.


butadienetrans cis

acroleintrans cis

glyoxaltrans cis

Figure 5: trans- and cis-Conformations of Butadiene(top), Acrolein(center), and Glyoxal(bottom)


Conformational Analysis (continued)

Rotation Potential Energy Curves, Optimization, and Frequency Calculations (continued)

• Provide structural details of the transition state structures, along with evidence that they are, infact, transition state structures.

• Compare the relative energies of each conformer, and discuss the significance.

• Does the inclusion of zero-point energy significantly improve the relative energies of the conformersdiscussed in the previous question?

Compare the magnitude of the thermal correction to that of the ZPE corrections. Which contributionis more significant for relative isomer energies?

Rotational Potential Energy Curves: Rotation About the Central C-C Bond

• Construct a graph of dihedral angle versus relative energy of all three molecules for both rigidrotational and relaxed rotational potential energy surfaces.

• Discuss similarities and differences, and explain the results. Which type of PES gives lower rotationalenergy barriers and why?

• Rationalize a comparison of the rotational energy barriers in terms of the lowest occupied molecularorbitals.

Thermodynamics and Kinetics

• Calculate the fractional population of each isomeric form using Boltzmann statistics ( Equation 3).

Will the calculated ratios of the different isomers be in the same proportion as the experiment?Discuss possible sources of deviations from the expected ratios.

• Using the optimized geometries of the glyoxal isomers, which have been calculated previously, de-termine how quickly glyoxal interconverts at room temperature by calculating the activation energyand the rate constant ( Equation 9 and Equation 10).


H/D Kinetic Isotope Effect

Below is an example of a potential energy curve of potential energy versus internuclear separation with thezero point vibrational energies of two isotopic molecules involving X-H and X-D, where X is a group/atomthat is much heavier than H or D.

Figure 6: Potential Energy Curve Showing Differences in Zero Point Energies for H and DIsotopes.

EoD and Eo

H correspond to the zero point energies of deuterium and hydrogen. The zero point energy (ZPE)is the lowest possible energy of a molecule, and is dependent upon the reduced mass of the molecule.

The heavier the molecule or atom, the lower the frequency of vibration and the smaller the zero pointenergy. Lighter molecules or atoms have a greater frequency of vibration and a higher zero point energy.As shown in the figure, deuterium is heavier than hydrogen and therefore has the lower zero point energy.

The lower zero point energy (ZPE) for deuterium results in a larger bond dissociation energy (X-D) fordeuterium than for hydrogen (X-H).

This difference in energy because of isotopic replacement results in differing rates of reaction, an effectwhich is measured as the Kinetic Isotope Effect (KIE).

Subsequent to determining a relevant transition state for a reaction, it is possible calculate the KineticIsotope Effect (KIE), which is the change in the rate of a chemical reaction when one of the atoms in thereactants is substituted with one of its isotopes.


Figure 7: Differences in Activation Energies (EA) Between Reactant and Transition StateResulting from Differences in Zero Point Energies for H and D Isotopes.


Calculate the H/D effect on the reaction rate of the reaction

CH4 + ·OH→ H2O + ·CH3

CH3D + ·OH→ HDO + ·CH3

(11)

1. Build all molecules in Equation 11 using Z-matrix format, then perform a geometry optimization onthem.

2. Consider the two transition-state structures:

H3C · · ·H · · ·O−H

H3C · · ·D · · ·O−H(12)

Determine the geometry and energy for the H isotopic transition state structure. In the transitionstate, one of the C-H bonds must be significantly stretched, i.e. C· · ·H, and the H· · ·O bond shouldalready be partially formed. Such a starting structure should be used.

A rule of thumb is to stretch the bond to ∼1.5 times its equilibrium value.

3. Since the zero point energy (ZPE) is isotope sensitive, the magnitude of the barrier is also isotopedependent, and a kinetic isotope effect may be observed.

Rates of reaction are determined, in part, by the energy barrier for the reaction (∆G‡) as shownin Figure 7, which involve changes in the the zero point energies (ZPEs), but ∆G‡ can be approxi-mated as:

∆G‡ ≈ (PE‡ − PEReactants)H − (PE‡ − PEReactants)D) + (∆ ZPE‡H −∆ ZPE‡D)Reactants (13)

A transition-state stucture is bound together very weakly along a specific reaction path, but zeropoint energies for transition-state structures may not be the same for the both isotopes.

(∆ ZPE‡H = ZPE‡H − ZPEH)Reactants 6= (∆ ZPE‡D = ZPE‡D − ZPED)Reactants (14)

If one assumes that the potential energy surface (PES) is approximately the same for all isotopicderivatives, then

(PE‡ − PEReactants)H ≈ (PE‡ − PEReactant)D (15)

Then,

∆G‡ ≈ (∆ ZPE‡H −∆ ZPE‡D)Reactants (16)

Using Equation 9, the kinetic isotope effect for a reaction involving H-migration, followed by D-substitution, can then be approximated as

kHkD

= e−(∆ZPE

‡H−∆ZPE

‡D

)ReactantsRT (17)

Calculate the ratio of reaction rates, and predict the resulting isotope effect, kH

kD, using Equation 17.


4. Thermal effects on the Kinetic Isotope Effect can also be explored to predict a unimolecular rateconstant, k by using Equation 9, where kB is the Boltzmann constant, and ∆G‡ is the free energy ofactivation.

Calculated values of ∆H‡ and ∆S‡ can be obtained, following a frequency analysis. Calculate theactivation enthalpy and entropy of the optimized transition state structure.

Then, calculate the ratio of reaction rates using Equation 9, and predict the resulting isotope effect,kH

kD:

kHkD

= e−(∆G‡H−∆G‡D)

RT (18)

Calculate the kinetic isotope effect, kH

kDat different temperatures, e.g. 0 K, 320 K, and 340 K, and

then compare these isotopic ratios to the calculated value at 300 K. Explain what this means for theabove reaction.

Further Studies

1. Think of another possible transition state structure for Equation 11, and try to guide the optimizationto this transition state. Recalculate ratio of reaction rates using Equation 17 and Equation 18.

2. Another reaction which can be influenced by deuterium substitution involves revisiting the isomer-ization

H−C ≡ N→ C ≡ N−H (19)

shown in Figure 3.

Explore the energetic effects of this reaction when H is replaced by D:

D−C ≡ N→ C ≡ N−D (20)

3. While it is always possible that deviations between theory and experiment represent inaccuracy in themodelling, what physical effects might be responsible for differences between theory and experiment?


Appendix

The Uncertainty Principle and Zero Point Energy

The Classical Harmonic Oscillator

A classical harmonic oscillator is described by kinetic energy is T = 12mx2 = p2

2mand potential energy

V = 12kx2:

E =p2

2m+

1

2kx2 (21)

If the system has a finite energy, E, the motion is bounded by two values, ±x0, such that V (x0) = E.

The energy is constant, since it is a conservative system, with no dissipation. Most of the time the particleis in the position x0, since there the velocity is zero, while at x = 0, the velocity is maximum.

Quantum Harmonic Oscillator: Schrodinger Equation

In quantum mechanics, the momentum is an operator which operates on the wave function, ψ. Themomentum operator is a differential operator, and for one particle in one dimension, is given as

p = −i~ ∂∂x

(22)

The Schriodinger equation for a harmonic oscillator may be obtained by using the classical spring potential,

V =1

2kx2 =

1

2mω2x2 because ω =

√k

m(angular frequency) (23)

The Schrodinger form of the harmonic oscillator equation becomes

− ~2

2m

∂2ψ(x)

∂x2+

1

2mω2x2ψ(x) = Eψ(x) (24)

The Uncertainty Principle

The position and momentum of a particle cannot be simultaneously measured with arbitrarily high preci-sion. There is a minimum value for the product of the uncertainties of these two measurements,

∆x∆p >~2

(25)

as well as a minimum value for the product of the uncertainties of the energy and time

∆E∆t >~2

(26)


Quantum Harmonic Oscillator: Energy Minimum from Uncertainty Principle

The ground state energy for the quantum harmonic oscillator can be shown to be the minimum energyallowed by the uncertainty principle.

E =(∆p)2

2m+

1

2mω2(∆x)2 (27)

where ∆x is the position uncertainty, and ∆p is the momentum uncertainty. The energy can be expressedin terms of the position uncertainty:

E =~2

8m(∆x)2+

1

2mω2(∆x)2 (28)

Minimizing the above energy by taking the derivative with respect to the position uncertainty, ∆x, followedby setting it equal to zero gives”

− ~2

4m(∆x)3+mω2∆x = 0 (29)

Solving for the position uncertainly gives

∆x =

√~

2mω(30)

Substituting this value into Equation 28, yields

E0 =~2

8m(∆x)2+

1

2mω2(∆x)2 =

~ω4

+~ω4

=~ω2

(31)

This significant result tells us that the energy of a system described by a harmonic oscillator potentialcannot have zero energy. Physical systems such as polyatomic molecules cannot have zero energy even atabsolute zero temperature. The energy of the ground vibrational state is often referred to as zero pointvibrational energy (ZPE).


Simple Derivation of Isotope Effect

Contrast the ratio of frequencies for HCl and DCl.

∆E = hν =h

2π

√k

µ(32)

where k is the H-Cl or D-Cl force constant and µ is the reduced mass.

∆EHCl = hνHCl =h

2π

√k

µHCl

; µHCl =mH ·mCl

mH +mCl

∼ mH ·mCl

mCl

∼ mH (33)

∆EDCl = hνDCl =h

2π

√k

µDCl

; µDCl =mD ·mCl

mD +mCl

∼ mD ·mCl

mCl

∼ mD (34)

∆EHCl

∆EDCl

=νHCl

νDCl

=

√mD

mH

∼ 1.4 (35)


Important Reminders for Gaussian 09

• If geometric or basis set data is entered into a G09 input file, as opposed to reading it from thecheckpoint file, end the input file with a blank line.

• In general, to find transition states, use opt=(ts,calcfc) keywords.

If a transition state has a different symmetry from other points along the reaction path, then optimizethe transition state in this unique symmetry, then do not use opt=(ts,calcfc).

If symmetry is preserved or there is no symmetry along the reaction path, then use opt=(ts,calcfc).

It may be necessary to eliminate a symmetry constraint entirely, and include the nosymm keywordwhen searching for a transition state.

When looking for a TS, it is often helpful to use opt=(ts,calcfc,noeigentest). The other keywordsbesides ts equest calculation of analytic force constants on the first step (calcfc), and that the jobnot stop, if other than exactly one negative force constant is found at some step (noeigentest).

• A great deal of CPU time can be saved by re-using information from previous calculations stored inthe checkpoint file. Plan your calculations to try to save time. CPU time can also be saved by usingsymmetry when appropriate.

• The keywords guess=read, geom=checkpoint, and guess=allcheck read a previously calculatedwave function, a previously calculated geometry, or both previously calculated geometry and wavefunction, respectively, from the last completed calculation.

If an optimization calculation has been just performed, then a frequency calculation should certainlyuse these keywords.

The old checkpoint file (%OldCheck) should be specified, when reading data from a previouslycompleted calculation. For a combined optimization and frequency calculation, it is also possible toinclude both the freq and opt keywords in the same job.

Frequencies must be calculated using the same level of theory and basis sets with which the geometrywas optimized, in order to produce a consistent set of vibrational frequencies.

• An optimized geometry calculation at one level of theory, can be repeated at a different level byincluding the keyword opt=(readfc), or opt=(ts,readfc) for a second pass at a transition-stateoptimization. This causes G09 to start with the force constants from the previous calculation and,may be efficient.

It is usually a good idea to work your way up from a smaller basis set to a larger, more accuratebasis set when performing a a geometry optimization so that computer time is not wasted becauseof a bad initial guess.

• An example of the G09 keyword for IRC is given by irc=(forward,maxpoints=20,stepsize=10)which involves taking 20 steps in the product (forward) direction from the TS.

• The freqchk utility of G09 may be used to rerun the thermochemical analysis, for a different tem-peratures, from the frequency data stored in a Gaussian checkpoint file. An additional frequencycalculation is not needed for this.


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Department of Chemistry The Intersection Potential Energy ...rossi.chemistry.uconn.edu/chem5326/files/pes-ts-mep...2017/02/04 · B = 1:38 10 38J=K is the Boltzmann constant, hPlanck’s