University of Connecticut OpenCommons@UConn Chemistry Education Materials Department of Chemistry June 2006 The yH drogen Atom: The Bohr Model in Detail Carl W. David University of Connecticut, [email protected] Follow this and additional works at: https://opencommons.uconn.edu/chem_educ Recommended Citation David, Carl W., "The yH drogen Atom: The Bohr Model in Detail" (2006). Chemistry Education Materials. 3. https://opencommons.uconn.edu/chem_educ/3 The Hydrogen Atom; The Bohr Model in Detail C. W. David∗ Department of Chemistry University of Connecticut Storrs, Connecticut 06269-3060 (Dated: June 1, 2006) When studying the hydrogen atom’s electron, the pro- distance of separation. Since r = px2 + y2 + z2, we have totypical atomic problem, one should keep in mind that that most of our mental impressions about atoms and their (Ze)(e) (Ze)(e) electronic structure will be gleaned from this problem. PE = − = − r p 2 2 2 That means that paying attention to it is tantamount x + y + z to laying an adequate foundation for continuing study of where we are using the electron charge, e, and the nuclear higher atomic number atoms, and ultimately molecules, charge of an atomic number Z, to form either an ion the stuff of Chemistry. (Z>1) or the neutral H-atom (Z=1). The minus sign We start with elementary Bohr theory, not because we is there because the nucleus is positive and the electron can do it better than others, but just so that we can carries a negative charge, and the numerator (as written) review the material in this appropriate place. If, to be- obliterates that fact. Actually, there are other reasons, gin, we place the proton (or nucleus) at the origin, and but let’s leave it at that for the time being. the electron somewhere in space, and if we assume (tem- We now have that the energy of the H-atom is porarily) that the nucleus is infinitely massive, so that it 1 (Ze)(e) 1 Ze2 doesn’t ever move, then the total energy of the two par- E = m v2 − = m v2 − (1) ticle system is contained in the energy of the electron. It total 2 e r 2 e r has two forms of energy, as usual, potential and kinetic. We note in passing that this last expression will form the Assuming the electron is located at ~r at time t, i.e., at basis for constructing the Hamiltonian and ultimately the ˆ ˆ ˆ Schr¨odinger equation appropriate for this problem. ~r = xi + yj + zk As is common in these atomic problems, we be- come conflicted between the Cartesian co¨ordinate sys- (where x = x(t), y = y(t), and z = z(t)) then the (clas- tem which is the appropriate one for starting the analy- sical) velocity of the electron would be sis, and spherical polar co¨ordinates, which appear to be d~r somehow “cleaner”, although much of this is in the eye of ~v =x ˙ˆi +y ˙ˆj +z ˙kˆ = = v ˆi + v ˆj + v kˆ dt x y z the reader. Let’s recall what spherical polar co¨ordinates are, an alternative scheme for locating a particle in space, and the kinetic energy is (by definition) using two angles and the distance from the origin. The definitions are 1 KE = m ~v • ~v 2 e x = r sin ϑ cos ϕ y = r sin ϑ sin ϕ where me is the mass of the electron itself. This dot product produces z = r cos ϑ (2) 1 which is the traditional set used by physicists and KE = m v2 + v2 + v2 2 e x y z chemists (mathematicians sometimes exchange ϑ and ϕ and it is essential when starting out to keep clear which which is just definitions you are using). Squaring and adding these together, we recover the definition of the radius, i.e., 1 KE = m v2 2 e x2 + y2 + z2 = r2 where v is the scalar magnitude of the velocity vector where we make copious use of cos2 + sin2 = 1. q 2 2 2 From Figure 1 we have from the O-A-B triangle, that (sometimes called c), i.e., v = vx + vy + vz . x y cos ϕ = a and sin ϕ = a , while from triangle O-B-C we The potential energy of the electron is given by a have sin ϑ = r so, eliminating “a” from these gives us the Coulomb’s law, i.e., it corresponds to the charge on the transformation equations we desire. For triangle O-C-D nucleus times the charge on the electron divided by the we get r cos ϑ = z, independent of “a”. Forming y r sin ϑ sin ϕ = ∗Electronic address: [email protected] x r sin ϑ cos ϕ Typeset by REVTEX 2 we obtain The other definition, which is caveat free, is −1 y tan = ϕ ˆi ˆj kˆ x ~ ~ A ⊗ B ≡ det Ax Ay Az (5) and from the z-equation we obtain Bx By Bz z z cos−1 = cos−1 = ϑ which expands to px2 + y2 + z2 r ˆ A~⊗B~ = ˆi(AyBz−AzBy)+ˆj(AzBx−AxBz)+k(AxBy−AyBz) which we repeat (reversed) as y Note that ϕ = tan−1 x ˆi ˆj kˆ −1 z −1 z ˆ ˆ ˆ ϑ = cos p = cos (3) i ⊗ j ≡ det 1 0 0 → k x2 + y2 + z2 r 0 1 0 p r = x2 + y2 + z2 as expected. In our case, we have, employing Equation 4 These three equations (Equations 4) tell us, given values and Equation 5, we obtain: for x, y, and z, (on the r.h.s.) how to compute r, ϑ ~ ˆ ˆ ˆ and ϕ. The first set of equations (Equations 2) gave L ≡ ~r ⊗ ~p = i(ypz − zpy) + j(zpx − xpz) + k(xpy − ypz) us the opposite, i.e., knowing the values of the spherical polar co¨ordinates of the electron (on the r.h.s.), we could which means that we can work with the definition of L~ calculate the x, y, and z values of that same position. and know that we can compute its components as needed. The original Bohr calculation for the electron’s energy So, let’s learn first what the time derivative of L~ is. involved using circular orbits for the path of the electron Thus around the nucleus (analogous to the Earth orbiting the dL~ d~r d~p Sun (elliptical orbits, hint, hint)). So we need to think = ⊗ ~p + ~r ⊗ about circular orbits for a bit. dt dt dt In a circular orbit, r would be constant, wouldn’t it? The first term is zero, since the time derivative of ~r is That’s the meaning of circular (or spherical, if we just proportional (through the mass) to ~p, and, we remember thought the electron stayed on the surface of a sphere). that the cross product of a vector on itself (or anything To continue this discussion then, we need to recall the an- parallel to itself) is zero. This leaves gular momentum vector from elementary physics. When a particle is located at the point P(x,y,z), with instanta- dL~ d~p neous velocity v in the x-direction, v in the y-direction, = ~r ⊗ x y dt dt and vz in the z-direction, i.e., has a velocity vector but the time derivative of the momentum is the force ˆ ˆ ˆ ~v = vxi + vyj + vzk (Newton’s second law), so dx (where vx = dt (and is itself a function of time, i.e., dL~ v (t)), with equivalent terms for the instantaneous values = ~r ⊗ F~ ≡ ~τ x dt of y and z), and a momentum vector where ~τ is the torque exerted by the force at ~r. If ~r is ˆ ˆ ˆ ~p = mvxi + mvyj + mvzk parallel to F~ then the torque is zero, and, lo and behold, the angular momentum has a zero time derivative. And a then the angular momentum is defined as the cross prod- zero time derivative means that the angular momentum uct of the radius vector on the momentum vector, i.e., is constant, and that means we have discovered one of L~ = ~r ⊗ ~p (4) the infamous “constants of the motion”. where we remember two different definitions of the cross product. One can define it as I. PART 2 |A~ ⊗ B~ | ≡ |A~||B~ | sin θ Having established that the angular momentum is con- served, i.e., is time independent, in central force prob- where θ is the angle between A~ and B~ and the vertical lems such as the hydrogen atom, i.e., problems in which lines indicate “magnitude” of the enclosed vector. The the force vector is collinear with the radius vector, i.e., added caveat is that the direction of the new vector, the ~r collinear to F~ , we turn now to the Bohr calculation cross product, is perpendicular to the plane of A~ and B~ , of the H atom’s electronic energy in the infinite nuclear and follows a right-hand rule analogous to ˆi ⊗ ˆj → kˆ. mass approximation. 3 We have a co¨ordinate system with the nucleus at the while for the “r” (radius) triangle (c) we have origin and an electron circumnavigating the nucleus at θ |~r|/2 ~r(t). If, in that circumnavigation, the path is a circle = (the Kepler problem uses ellipses), then we know that 2 r the instantaneous velocity vector ~v(t) of the electron at so, setting the two equal to each other we have the point ~r(t) is perpendicular to the instantaneous force |∆~v| |∆~r| (Coulomb) between the proton (nucleus) and the elec- = tron, and therefore that the cross product becomes v r all in time ∆t.