In obtaining the results for hydrogen-like atoms in Section...

In obtaining the results for hydrogen-like atoms in Section $8.5,$ the atomic nucleus was assumed to be immobile due to its much larger mass compared with that of the electron. If this assumption is relaxed, the results remain valid if the electron mass $m$ is replaced everywhere by the reduced mass $\mu$ of the electron-nucleus combination: $$\mu=\frac{m M}{m+M}$$ Here $M$ is the nuclear mass. (a) Making this replacement in Equation $8.38,$ show that a more general expression for the allowed energies of a one-electron atom with atomic number $Z$ is $$E_{n}=-\frac{\mu k^{2} e^{4}}{2 \hbar^{2}}\left\{\frac{Z^{2}}{n^{2}}\right\} $$ (b) The wavelength for the $n=3$ to $n=2$ transition of the hydrogen atom is $656.3 \mathrm{nm}$ (visible red light). What is the wavelength of this same transition in singly ionized helium? In positronium? (Note: Positronium is an "atom" consisting of a bound positron-electron pair. A positron is a positively charged electron.)

In obtaining the results for hydrogen-like atoms in Section $8.5,$ the atomic nucleus was assumed to be immobile due to its much larger mass compared with that of the electron. If this assumption is relaxed, the results remain valid if the electron mass $m$ is replaced everywhere by the reduced mass $\mu$ of the electron-nucleus combination:
$$\mu=\frac{m M}{m+M}$$
Here $M$ is the nuclear mass. (a) Making this replacement in Equation $8.38,$ show that a more general expression for the allowed energies of a one-electron atom with atomic number $Z$ is
$$E_{n}=-\frac{\mu k^{2} e^{4}}{2 \hbar^{2}}\left\{\frac{Z^{2}}{n^{2}}\right\} $$
(b) The wavelength for the $n=3$ to $n=2$ transition of the hydrogen atom is $656.3 \mathrm{nm}$ (visible red light). What is the wavelength of this same transition in singly ionized helium? In positronium? (Note: Positronium is an "atom" consisting of a bound positron-electron pair. A positron is a positively charged electron.)

Key Concepts

Reduced Mass

In the context of two-body problems such as an electron orbiting a nucleus, the reduced mass is a mathematical construct that simplifies the dynamics by replacing the two-body system with a single particle of mass ? moving in an effective potential. It is defined as ? = (mM)/(m + M), where m is the electron mass and M is the nuclear mass, and accounts for the fact that the nucleus is not completely stationary. This concept is crucial when precise energy levels are calculated, as the smaller but finite movement of the nucleus affects the energy spectrum of the atom.

Hydrogen-like Atom Energy Levels

The energy levels of hydrogen-like atoms are derived from solving the Schrödinger equation for a one-electron system with a Coulomb potential. These levels depend on the atomic number Z, the principal quantum number n, and the mass of the orbiting particle (usually the electron or the reduced mass when corrections are needed). The general expression E_n = - (? k^2 e^4 Z^2)/(2 ?^2 n^2) demonstrates that the energy is quantized and scales with Z^2/n^2. This concept forms the basis for understanding the spectral properties and stability of atoms with one electron.

Spectral Transitions and Wavelength Calculation

Spectral transitions occur when an electron moves between energy levels, emitting or absorbing a photon with energy equal to the difference between these levels. The wavelength of the spectral line, such as that for transitions between specific levels (e.g., n = 3 to n = 2), can be determined by equating this energy difference to the energy of a photon (E = hc/?). This relationship allows the calculation of distinct wavelengths for different systems, including variations due to changes in reduced mass in systems like helium ion and positronium, which in turn affect the observed spectral lines.

Transcript

00:01 In this question, we need to show that the allowed energy is of a one electron atom, with atomic number z is given as such, whereby the mass of electron is replaced by the reduced mass of the system.

00:15 Okay, and then in part b, we need to find a wavelength of the transition from n equals to 3 to n equals to 2 in he plus ion and positronium.

00:24 Okay, so solution.

00:32 Okay, in part a, we are supposed to start from equal to equal to 2.

00:35 Equation 8 .38.

00:42 So this equation is e n equals to minus k e square over to a not c squared over n square and then we'll replace a not okay so a not is equal to h bar square over mass of electron times k the coolum constant times charge of electron square so, e .n would be equal to minus m .e, k square, e to the 4, divided by 2, h bar square, c square over n square.

01:27 Okay.

01:29 Then we replace the mass of electron with the reduced mass.

01:43 Okay, m is equal to small m times big m divide by small m plus big m.

01:49 Then we have en is equal to minus new k square e to the 4 divide 2 hbath square times c square over n square shown if we are done with part a now we move on to parts b so we are given that for hydrogen atom the wavelength length for n equals to 3 to n equals to 2 transition is equal to 656 .3 nanometer which means that so so e3 minus e2 is equal to hc over lambda is equal to so, new, k square, e square over 2 hbath square, c square, 1 over 3 square, minus 1 over 2 square, with the negative sign.

03:32 Okay, so i'm just going to, this is the general formula, okay, so for hydrogen atom, okay, z is equals to 1, mu is equal to the mass of electron, okay? so delta e equals to hc over lambda, is equal to me, k square, e to the 4, derived by 2h bar square, 1 over 2 square, minus 1 over 3 square.

04:11 So this is the delta e for each atom...

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