New answers tagged

3

To answer the question, over and above the errors noted in the other answer the integral is just not well defined - it is infinite. There are two ways to see this The "common sense" method. Consider $$ \int^\infty_{-\infty} \left[ A\sin(kx) + B \cos(kx)\right]^2dx$$ The quantity under the integral is positive semi-definite - it is always greater ...


1

Your attempt has multiple issues, most of which are very basic. The first issue that I noticed was how you expanded and simplified your integral. \begin{align} \int^\infty_{-\infty}[A^2\sin^2(kx) &+ AB\sin(kx)\cos(kx) \\ &+ AB\cos(kx)\sin(kx) +B^2\cos^2(kx)]\mathrm dx = 1 \tag{1} \end{align} Since, $\sin^2 x + \cos^2 x = 1$, $$\implies \int^\...


2

The curves labelled Harmonic/Morse show how the potential energy changes with internuclear separation, and the horizontal lines are the allowed total energies, potential plus kinetic. The potential energy is a maximum at the turning points (end of the horizontal lines in your picture) and zero at the $r_e$ value but because the total energy is constant the ...


6

The kinetic energy is not ignored. Each quantized state (labeled by the vibrational quantum number) has a total energy given by the expectation value of the Hamiltonian operator for the wavefunction. The contribution of the kinetic energy can be computed as the expectation value of the QM kinetic energy operator, or as the difference between the expectation ...


1

If I well understand your question you have a set of experimental points $(E_1,X_1), (E_2,X_2), ... , (E_n,X_n)$. You want to fit to this set of points the function : $$ E(X) = E_\infty + Ae^{-\alpha X} $$ So, you want to approximately determine the parameters $E_\infty$ , $A$ and $\alpha$. This requires a non-linear regression calculus. For example see ...


13

These orbitals represent the angular part of the wavefunction. The solution obtained directly from solving the Schrödinger equation produces equations containing complex numbers so cannot be drawn on normal $xyz$ axes and are hard to visualise. The angular part of the wavefunction is given by functions called Spherical Harmonics these usually are given the ...


1

The OP can understand, mathematically, why the energy of the electron in the Bohr atom is zero at infinity, so let's avoid the mathematics in order to assist the imagination. In the (small?) universe which consists of just a single hydrogen atom (plus all necessary equipment to measure potentials and distance), it is necessary to cause a separation of the ...


8

The value of the energy in the Bohr model is zero when the quantum number is infinity because that is the limiting value of the Coulombic potential at large distances, and because the electron is assumed bound to the nucleus (the atom is stable), which constrains the value of the total energy. The energy of a stationary hydrogen-like atom is described by ...


1

When exchange effects are not taken into consideration, the simplest possible description of a many-body wave function is simply the Hartree product, i.e. the product of individual-particle wave functions. This is reasonable for bosons, but electrons are fermions, which means that the exchange of any two must flip the sign of the wave function. Slater ...


5

Degeneracy refers to states having the same energy. If you have two or more coordinates that can be related by a symmetry operation then the system will contain degenerate states because swapping the coordinates associated with those degrees will result in a Hamiltonian with the same solutions. This is the case in both the hydrogen atom and in the particle ...


Top 50 recent answers are included