Categories: Physics, Quantum mechanics.

Hydrogen atom

The quantum-mechanical calculation of the hydrogen atom is, in my opinion, the single most important model in all of physics: miraculously, it is possible to find closed-form solutions for the wave function of an electron in a proton’s potential well. The results are highly educational, and also qualitatively tell us a lot about all other chemical elements.

We start from the time-independent Schrödinger equation, where μ\mu is the reduced mass of the electron-proton system, and VV is the proton’s Coulomb potential:

Eψ=22μ2ψ+Vψ\begin{aligned} E \psi = - \frac{\hbar^2}{2 \mu} \nabla^2 \psi + V \psi \end{aligned}

In spherical coordinates (r,θ,φ)(r, \theta, \varphi) it becomes as follows, where VV only depends on rr:

Eψ=22μ(2ψr2+2rψr+1r22ψθ2+1r2tanθψθ+1r2sin2θ2ψφ2)+Vψ\begin{aligned} E \psi = - \frac{\hbar^2}{2 \mu} \bigg( \pdvn{2}{\psi}{r} + \frac{2}{r} \pdv{\psi}{r} + \frac{1}{r^2} \pdvn{2}{\psi}{\theta} + \frac{1}{r^2 \tan{\theta}} \pdv{\psi}{\theta} + \frac{1}{r^2 \sin^2{\theta}} \pdvn{2}{\psi}{\varphi} \bigg) + V \psi \end{aligned}

We will use the method of separation of variables by making the following ansatz, such that the Schrödinger equation takes the form below:

ψ(r,θ,φ)=R(r)Y(θ,φ)\begin{aligned} \psi(r, \theta, \varphi) = R(r) \: Y(\theta, \varphi) \end{aligned} ERY=22μ(RY+2RYr++RYθθr2+RYθr2tanθ+RYφφr2sin2θ)+VRY\begin{aligned} E R Y &= - \frac{\hbar^2}{2 \mu} \bigg( R'' Y + \frac{2 R' Y}{r} + + \frac{R Y_{\theta\theta}}{r^2} + \frac{R Y_\theta}{r^2 \tan{\theta}} + \frac{R Y_{\varphi\varphi}}{r^2 \sin^2{\theta}} \bigg) + V R Y \end{aligned}

After multiplying by 2μr2/(2RY)- 2 \mu r^2 / (\hbar^2 R Y), each term depends on rr or (θ,φ)(\theta, \varphi), but not both:

0=(r2RR+2rRR2μ2r2V+2μ2r2E)+(YθθY+1tanθYθY+1sin2θYφφY)\begin{aligned} 0 &= \bigg( r^2 \frac{R''}{R} + 2 r \frac{R'}{R} - \frac{2 \mu}{\hbar^2} r^2 V + \frac{2 \mu}{\hbar^2} r^2 E \bigg) + \bigg( \frac{Y_{\theta\theta}}{Y} + \frac{1}{\tan{\theta}} \frac{Y_\theta}{Y} + \frac{1}{\sin^2{\theta}} \frac{Y_{\varphi\varphi}}{Y} \bigg) \end{aligned}

Since these two groups are independent, this equation can only hold if there exists a separation constant CC such that:

C=r2RR+2rRR2μ2r2(VE)=YθθY1tanθYθY1sin2θYφφY\begin{aligned} C &= r^2 \frac{R''}{R} + 2 r \frac{R'}{R} - \frac{2 \mu}{\hbar^2} r^2 ( V - E ) \\ &= - \frac{Y_{\theta\theta}}{Y} - \frac{1}{\tan{\theta}} \frac{Y_\theta}{Y} - \frac{1}{\sin^2{\theta}} \frac{Y_{\varphi\varphi}}{Y} \end{aligned}

Now we have two simpler equations than the one we started with. We multiply them by RR and YY respectively, and define C(+1)C \equiv \ell (\ell + 1) to help us later (\ell is unknown for now). The results are the radial equation and the angular equation:

(+1)R=r2R+2rR2μ2r2(VE)R(+1)Y=Yθθ+1tanθYθ+1sin2θYφφ\begin{aligned} \boxed{ \begin{aligned} \ell (\ell + 1) R &= r^2 R'' + 2 r R' - \frac{2 \mu}{\hbar^2} r^2 (V - E) R \\ - \ell (\ell + 1) Y &= Y_{\theta\theta} + \frac{1}{\tan{\theta}} Y_\theta + \frac{1}{\sin^2\theta} Y_{\varphi\varphi} \end{aligned} } \end{aligned}

Note that this calculation has not really been specific to hydrogen so far: it is applicable to all spherically symmetric quantum systems.

Angular equation

Let us keep this generality, by keeping VV unspecified for now, In that case, the radial equation cannot be solved yet, but the angular one can. We separate the variables again:

Y(θ,φ)=Θ(θ)Φ(φ)\begin{aligned} Y(\theta, \varphi) = \Theta(\theta) \: \Phi(\varphi) \end{aligned}

Insert this into the equation and multiply by sin2θ/(ΘΦ)\sin^2\theta / (\Theta \Phi) to get a clean separation:

sin2θΘΘ+sinθcosθΘΘ+(+1)sin2θ=ΦΦ\begin{aligned} \sin^2{\theta} \frac{\Theta''}{\Theta} + \sin{\theta} \cos{\theta} \frac{\Theta'}{\Theta} + \ell (\ell + 1) \sin^2{\theta} = - \frac{\Phi''}{\Phi} \end{aligned}

Each term depends on θ\theta or φ\varphi but not both, so there exists a constant m2m^2 such that:

m2=sin2θΘΘ+sinθcosθΘΘ+(+1)sin2θ=ΦΦ\begin{aligned} m^2 &= \sin^2{\theta} \frac{\Theta''}{\Theta} + \sin{\theta} \cos{\theta} \frac{\Theta'}{\Theta} + \ell (\ell + 1) \sin^2{\theta} \\ &= - \frac{\Phi''}{\Phi} \end{aligned}

These are two distinct equations; multiplying by Θ\Theta and Φ\Phi respectively yields:

m2Θ=sin2θΘ+sinθcosθΘ+(+1)sin2θΘm2Φ=Φ\begin{aligned} \boxed{ \begin{aligned} m^2 \Theta &= \sin^2{\theta} \:\Theta'' + \sin{\theta} \cos{\theta} \:\Theta' + \ell (\ell + 1) \sin^2{\theta} \:\Theta \\ - m^2 \Phi &= \Phi'' \end{aligned} } \end{aligned}

Clearly the latter is the simplest, so we start there. It is an eigenvalue problem for m2m^2, but it looks like a harmonic oscillator equation, so the solutions are easily found to be:

Φ(φ)=eimφ\begin{aligned} \boxed{ \Phi(\varphi) = e^{i m \varphi} } \end{aligned}

Because the coordinate φ\varphi is only defined in the interval [0,2π][0, 2\pi], we demand periodic boundary conditions Φ(0)=Φ(2mπ)\Phi(0) = \Phi(2 m \pi), which tells us that mm is an integer.

The other equation, for Θ\Theta, needs a bit more work. We write it out like so:

0=d2Θdθ2+cosθsinθdΘdθ+((+1)m2sin2θ)Θ\begin{aligned} 0 &= \dvn{2}{\Theta}{\theta} + \frac{\cos{\theta}}{\sin{\theta}} \dv{\Theta}{\theta} + \Big( \ell (\ell + 1) - \frac{m^2}{\sin^2{\theta}} \Big) \Theta \end{aligned}

And then perform a change of variables θξ\theta \to \xi where ξcosθ\xi \equiv \cos{\theta}, leading to:

0=ddθ(sinθdΘdξ)cosθdΘdξ+((+1)m2sin2θ)Θ=sin2θd2Θdξ22cosθdΘdξ+((+1)m2sin2θ)Θ=(1ξ2)d2Θdξ22ξdΘdξ+((+1)m21ξ2)Θ\begin{aligned} 0 &= - \dv{}{\theta} \bigg( \sin{\theta} \dv{\Theta}{\xi} \bigg) - \cos{\theta} \dv{\Theta}{\xi} + \bigg( \ell (\ell + 1) - \frac{m^2}{\sin^2{\theta}} \bigg) \Theta \\ &= \sin^2{\theta} \dvn{2}{\Theta}{\xi} - 2 \cos{\theta} \dv{\Theta}{\xi} + \bigg( \ell (\ell + 1) - \frac{m^2}{\sin^2{\theta}} \bigg) \Theta \\ &= (1 - \xi^2) \dvn{2}{\Theta}{\xi} - 2 \xi \dv{\Theta}{\xi} + \bigg( \ell (\ell + 1) - \frac{m^2}{1 - \xi^2} \bigg) \Theta \end{aligned}

This result can be recognized as Legendre’s generalized equation, a known eigenvalue problem for (+1)\ell (\ell + 1), which has solutions when \ell is a non-negative integer. Those solutions are called the associated Legendre polynomials Pm(x)P_\ell^m(x) of degree \ell and order mm. For a given \ell, there exist 2+12 \ell + 1 such “polynomials” (they actually contain square roots too) indexed by the integer mm in the range [,][-\ell, \ell], so e.g. for =2\ell = 2 there is m=2,1,0,1,2m = -2, -1, 0, 1, 2. We now have:

Ym(θ,φ)Pm(cosθ)eimφ\begin{aligned} Y_\ell^m(\theta, \varphi) \propto P_\ell^m(\cos{\theta}) \: e^{i m \varphi} \end{aligned}

We are still missing a constant factor, found by imposing the normalization condition:

02π0πYm2sinθdθdφ=1\begin{aligned} \int_0^{2 \pi} \int_0^\pi |Y_\ell^m|^2 \sin{\theta} \dd{\theta} \dd{\varphi} = 1 \end{aligned}

Calculating the normalization constant (not shown here) leads to the following definition of the so-called spherical harmonics YmY_\ell^m of degree \ell and order mm:

Ym(θ,φ)=(1)m(2+1)(m)!4π(+m)!Pm(cosθ)eimφ\begin{aligned} \boxed{ Y_\ell^m(\theta, \varphi) = (-1)^m \sqrt{\frac{(2 \ell + 1) (\ell - m)!}{4 \pi (\ell + m)!}} \: P_\ell^m(\cos{\theta}) \: e^{i m \varphi} } \end{aligned}

These are important functions: the wave function of any spherically symmetric quantum system is a superposition of YmY_\ell^m with rr-dependent coefficients. And, as befits a (component of a) wave function, they form an orthonormal basis, specifically:

02π0πYmYmsinθdθdφ=δδmm\begin{aligned} \int_0^{2 \pi} \int_0^\pi Y_\ell^m \:Y_{\ell'}^{m'} \:\sin\theta \:d\theta \:d\varphi = \delta_{\ell\ell'} \delta_{mm'} \end{aligned}

Radial equation

With the angular part solved, we now turn to the radial part. Introducing u(r)=rR(r)u(r) = r R(r), such that the derivatives of R(r)R(r) become:

R=ruur2R=r2u2ru+2ur3\begin{aligned} R' = \frac{r u' - u}{r^2} \qquad\qquad R'' = \frac{r^2 u'' - 2 r u' + 2 u}{r^3} \end{aligned}

Inserting this into the radial equation and cancelling some of the terms:

(+1)ur=r2u2ru+2ur+2ru2ur2μ2(VE)ru=ru2μ2(VE)ru\begin{aligned} \ell (\ell + 1) \frac{u}{r} &= \frac{r^2 u'' - 2 r u' + 2 u}{r} + \frac{2 r u' - 2 u}{r} - \frac{2 \mu}{\hbar^2} (V - E) r u \\ &= r u'' - \frac{2 \mu}{\hbar^2} (V - E) r u \end{aligned}

After multiplying by 2/(2μr)\hbar^2 / (2 \mu r) and rearranging, this turns into:

Eu=22μu+(V+22μ(+1)r2)u\begin{aligned} E u = - \frac{\hbar^2}{2 \mu} u'' + \bigg( V + \frac{\hbar^2}{2 \mu} \frac{\ell (\ell + 1)}{r^2} \bigg) u \end{aligned}

Here it is useful to define an effective potential Veff(r)V_{\mathrm{eff}}(r) as below. Keep in mind that \ell is known after solving the angular equation:

VeffV+22μ(+1)r2\begin{aligned} V_{\mathrm{eff}} \equiv V + \frac{\hbar^2}{2 \mu} \frac{\ell (\ell + 1)}{r^2} \end{aligned}

This yields a relation of the same form as the time-independent Schrödinger equation, just with VV replaced by VeffV_{\mathrm{eff}}. This is the “true” radial equation, an eigenvalue problem for EE:

Eu=22μu+Veffu\begin{aligned} \boxed{ E u = - \frac{\hbar^2}{2 \mu} u'' + V_{\mathrm{eff}} u } \end{aligned}

Now, finally, we specialize for the hydrogen atom. Coulomb’s law tells us the attractive force F(r)F(r) between the electron and the proton, which we integrate to find the potential energy V(r)V(r):

F(r)=q24πε0r2    V(r)=q24πε0r\begin{aligned} F(r) = \frac{q^2}{4 \pi \varepsilon_0 r^2} \qquad \implies \qquad V(r) = - \frac{q^2}{4 \pi \varepsilon_0 r} \end{aligned}

Where q<0q < 0 is the electron’s charge, and ε0\varepsilon_0 is the permittivity of free space. Note that V<0V < 0, so there is a natural distinction between bound states E<0E < 0 (where the electron is trapped in the proton’s well), and scattering states E>0E > 0 (where the electron is free). The true radial equation, after dividing by EE, is now given by:

u=22μEu+(22μE(+1)r22μ2μq24πε0rE)u\begin{aligned} u = - \frac{\hbar^2}{2 \mu E} u'' + \bigg( \frac{\hbar^2}{2 \mu E} \frac{\ell (\ell + 1)}{r^2} - \frac{\hbar^2 \mu}{\hbar^2 \mu} \frac{q^2}{4 \pi \varepsilon_0 r E} \bigg) u \end{aligned}

For brevity, let us introduce new constants κ\kappa and ρ0\rho_0, defined as follows:

κ2μEρ0μq22πε02κ\begin{aligned} \kappa \equiv \frac{\sqrt{-2 \mu E}}{\hbar} \qquad\qquad \rho_0 \equiv \frac{\mu q^2}{2 \pi \varepsilon_0 \hbar^2 \kappa} \end{aligned}

Where E<0E < 0, as we are interested in bound states. Now the radial equation has become:

0=1κ2u+(ρ0κr(+1)κ2r21)u\begin{aligned} 0 = \frac{1}{\kappa^2} u'' + \Big( \frac{\rho_0}{\kappa r} - \frac{\ell (\ell + 1)}{\kappa^2 r^2} - 1 \Big) u \end{aligned}

To clean this up further, we switch to the dimensionless variable ρκr\rho \equiv \kappa r, yielding:

0=u+(ρ0ρ(+1)ρ21)u\begin{aligned} 0 = u'' + \Big( \frac{\rho_0}{\rho} - \frac{\ell (\ell + 1)}{\rho^2} - 1 \Big) u \end{aligned}

We then choose the following ansatz for u(ρ)u(\rho), where v(2ρ)v(2 \rho) is unknown:

u(ρ)=w(2ρ)ρ+1eρ\begin{aligned} u(\rho) &= w(2 \rho) \: \rho^{\ell + 1} \: e^{- \rho} \end{aligned}

For reference, we also calculate its first and second derivatives:

u(ρ)=(2ρw+(+1ρ)w)ρeρu(ρ)=(4ρ2w+4(+1ρ)ρw+(ρ22ρ(+1)+(+1))w)ρ1eρ\begin{aligned} u'(\rho) &= \Big( 2 \rho w' + (\ell + 1 - \rho) w \Big) \rho^\ell \: e^{- \rho} \\ u''(\rho) &= \bigg( 4 \rho^2 w'' + 4 (\ell + 1 - \rho) \rho w' + \big( \rho^2 - 2 \rho (\ell + 1) + \ell (\ell + 1) \big) w \bigg) \rho^{\ell-1} \: e^{- \rho} \end{aligned}

Inserting this into the radial equation and dividing out all common factors gives:

0=4ρw+4(+1ρ)w+(ρ02(+1))w\begin{aligned} 0 &= 4 \rho w'' + 4 (\ell + 1 - \rho) w' + \big( \rho_0 - 2 (\ell + 1) \big) w \end{aligned}

Let us rearrange this to put it in a more suggestive form. Keep in mind that w=w(2ρ)w = w(2 \rho):

0=(2ρ)w+((2+1)+1(2ρ))w+(ρ021)w\begin{aligned} 0 &= (2 \rho) w'' + \Big( (2 \ell + 1) + 1 - (2 \rho) \Big) w' + \Big( \frac{\rho_0}{2} - \ell - 1 \Big) w \end{aligned}

This can be recognized as Laguerre’s generalized equation, a well-known eigenvalue problem for λ(ρ0/2 ⁣ ⁣ ⁣ ⁣1)\lambda \equiv (\rho_0 / 2 \!-\! \ell \!-\! 1). It has solutions when λ\lambda is a non-negative integer, in other words for ρ0=2n\rho_0 = 2n with n=1,2,3,...n = 1, 2, 3,..., which also tells us that \ell cannot be larger than n1n - 1. Then the solutions are the so-called associated Laguerre polynomials Ln12+1(2ρ)L_{n - \ell - 1}^{2 \ell + 1}(2 \rho), therefore:

u(ρ)ρ+1eρLn12+1(2ρ)\begin{aligned} u(\rho) \propto \rho^{\ell + 1} \: e^{-\rho} \: L_{n - \ell - 1}^{2 \ell + 1}(2 \rho) \end{aligned}

We are still missing a constant factor, found by imposing the normalization condition:

0R2r2dr=1\begin{aligned} \int_0^\infty R^2 \: r^2 \dd{r} = 1 \end{aligned}

Calculating the normalization constant (not shown here) leads to this radial solution Rn(r)R_{n\ell}(r):

Rn(r)=(n1)!2n(n+)!(2κ)3/2(2κr)eκrLn12+1(2κr)\begin{aligned} R_{n \ell}(r) = \sqrt{\frac{(n - \ell - 1)!}{2 n (n + \ell)!}} \: (2 \kappa)^{3/2} \: (2 \kappa r)^\ell \: e^{-\kappa r} \: L_{n - \ell - 1}^{2 \ell + 1}(2 \kappa r) \end{aligned}

Meanwhile, by isolating the definitions of κ\kappa and ρ0\rho_0 for EE, we find the eigenenergies to be:

E=22μκ2=22μ(μq22πε02ρ0)2\begin{aligned} E = - \frac{\hbar^2}{2 \mu} \kappa^2 = - \frac{\hbar^2}{2 \mu} \bigg( \frac{\mu q^2}{2 \pi \varepsilon_0 \hbar^2 \rho_0} \bigg)^2 \end{aligned}

Since ρ0=2n\rho_0 = 2 n, these allowed Bohr energies EnE_n of the electron are as follows:

En=1n2μ22(q24πε0)2\begin{aligned} \boxed{ E_n = - \frac{1}{n^2} \frac{\mu}{2 \hbar^2} \bigg( \frac{q^2}{4 \pi \varepsilon_0} \bigg)^2 } \end{aligned}

At this point, it is customary to also define the reduced Bohr radius a0a_0^*, given by:

a01nκ=4πε02μq25.295×1011m=0.5295A˚\begin{aligned} \boxed{ a_0^* \equiv \frac{1}{n \kappa} = \frac{4 \pi \varepsilon_0 \hbar^2}{\mu q^2} } \approx 5.295 \times 10^{-11} \:\mathrm{m} = 0.5295 \:\mathrm{\AA} \end{aligned}

The non-reduced Bohr radius a0a_0 simply uses the electron’s raw mass mem_e instead of μ\mu. Roughly speaking, a0a_0^* is the most probable electron-proton distance after a measurement of the electron’s position while it is in its ground state. This is often to used to write Rn(r)R_{n \ell}(r) as:

Rn(r)=(n1)!2n(n+)!(2na0)3/2(2rna0)er/na0Ln12+1(2rna0)\begin{aligned} \boxed{ R_{n \ell}(r) = \sqrt{\frac{(n - \ell - 1)!}{2 n (n + \ell)!}} \bigg( \frac{2}{n a_0^*} \bigg)^{3/2} \bigg( \frac{2 r}{n a_0^*} \bigg)^\ell e^{- r / n a_0^*} \: L_{n - \ell - 1}^{2 \ell + 1}\Big(\frac{2 r}{n a_0^*}\Big) } \end{aligned}

Quantum numbers

Multiplying the angular and radial parts together, we thus arrive at the following expression for the full wave function ψnm\psi_{n \ell m}:

ψnm(r,θ,φ)=Rn(r)Ym(θ,φ)\begin{aligned} \boxed{ \psi_{n \ell m}(r, \theta, \varphi) = R_{n \ell}(r) \: Y_\ell^m(\theta, \varphi) } \end{aligned}

The indices nn, \ell, and mm are the quantum numbers, which describe the state of the electron. There is also a fourth not shown here, the spin quantum number, which is +1/2+1/2 or 1/2-1/2 for spin-up or spin-down electrons respectively.

The principal quantum number nn, often called the shell number, gives the energy level (shell) of the electron, because the other numbers do not appear in EnE_n’s formula. Since En=E1/n2E_n = E_1 / n^2, the energy differences decrease with increasing nn, so electrons in higher shells can be excited more easily (i.e. they need less energy to get excited).

The azimuthal quantum number \ell gives the subshell of shell nn in which the electron is located. It takes integer values from 00 to n1n - 1 inclusive, with 00, 11, 22, and 33 respectively also called the ss, pp, dd, and ff subshells. The electron’s total angular momentum is given by (+1)\hbar \sqrt{\ell (\ell + 1)}.

The magnetic quantum number mm splits the electrons in each subshell into orbitals, and takes integer values from -\ell to \ell. The zz-component of the electron’s angular momentum is m\hbar m.

The total degeneracy of each energy level nn can be calculated as the sum of an arithmetic series, and is found to be n2n^2 excluding spin (or 2n22 n^2 with spin).

Unsurprisingly, all these wave functions form an orthonormal basis (although not a complete one unless the scattering states with E>0E > 0 are included):

02π0π0ψnmψnmr2sinθdrdθdφ=δnnδδmm\begin{aligned} \int_0^{2 \pi} \int_0^\pi \int_0^\infty \psi_{n \ell m}^* \: \psi_{n' \ell' m'} \: r^2 \sin{\theta} \dd{r} \dd{\theta} \dd{\varphi} = \delta_{nn'} \delta_{\ell\ell'} \delta_{mm'} \end{aligned}

When an excited electron drops from a state with energy EiE_i to a lower level EfE_f, it emits a photon with energy ω\hbar \omega, where ω\omega is the angular frequency of the resulting electromagnetic wave:

ω=EiEf=E1(1ni21nf2)\begin{aligned} \hbar \omega = E_i - E_f = E_1 \bigg( \frac{1}{n_i^2} - \frac{1}{n_f^2} \bigg) \end{aligned}

The corresponding vacuum wavelength is λ0=2πc/ω\lambda_0 = 2 \pi c / \omega, leading to the Rydberg formula, which was discovered empirically before the hydrogen atom had been solved:

1λ0=ω2πc=RH(1ni21nf2)\begin{aligned} \boxed{ \frac{1}{\lambda_0} = \frac{\omega}{2 \pi c} = R_\mathrm{H} \bigg( \frac{1}{n_i^2} - \frac{1}{n_f^2} \bigg) } \end{aligned}

Quantum mechanics then successfully gave a theoretical value to the experimentally determined Rydberg constant RHR_\mathrm{H} (or RR_\infty if the raw electron mass mem_e is used):

RH=E12πc=μ4π3c(q24πε0)21.097×107m1\begin{aligned} \boxed{ R_\mathrm{H} = \frac{|E_1|}{2 \pi \hbar c} = \frac{\mu}{4 \pi \hbar^3 c} \bigg( \frac{q^2}{4 \pi \varepsilon_0} \bigg)^2 } \approx 1.097 \times 10^7 \:\mathrm{m}^{-1} \end{aligned}

The transitions from excited states to the ground state nf=1n_f = 1 correspond to ultraviolet spectral lines known as the Lyman series. Similarly, transitions to nf=2n_f = 2 give visible lines known as the Balmer series, and transitions to nf=3n_f = 3 explain the Paschen series of infrared lines.

The Rydberg constant is not to be confused with the Rydberg energy Ry\mathrm{Ry}, which is the ionization energy of ground-state hydrogen, and is sometimes used as a unit in calculations:

Ry2πcRH=E1=μ22(q24πε0)213.61eV\begin{aligned} \mathrm{Ry} \equiv 2 \pi \hbar c R_\mathrm{H} = |E_1| = \frac{\mu}{2 \hbar^2} \bigg( \frac{q^2}{4 \pi \varepsilon_0} \bigg)^2 \approx 13.61 \:\mathrm{eV} \end{aligned}

The point is that the hydrogen atom’s solution gave clear explanations for known experimental data, and settled the mystery of what an atom actually looks like. While other elements’ atoms generally do not have such closed-form solutions (because they have more than one electron), their orbitals are qualitatively very similar. In short, this model is the foundation of our modern understanding of atoms.


  1. D.J. Griffiths, D.F. Schroeter, Introduction to quantum mechanics, 3rd edition, Cambridge.
  2. R. Shankar, Principles of quantum mechanics, 2nd edition, Springer.