The Hydrogen Atom: the Bohr Model in Detail

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]

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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 Schrödinger equation appropriate for this problem. ~r = xi + yj + zk As is common in these atomic problems, we be- come conflicted between the Cartesian coördinate 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 coördinates, which appear to be d~r somehow “cleaner”, although much of this is in the eye of ~v =x ˙î +y ˙ˆj +z ˙kˆ = = v î + v ˆj + v kˆ dt x y z the reader. Let’s recall what spherical polar coördinates 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 î 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~ = î(AyBz−AzBy)+ˆj(AzBx−AxBz)+k(AxBy−AyBz) which we repeat (reversed) as y Note that ϕ = tan−1   x î 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 coördinates 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 conserved, 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 coördinate 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. We thus have L~ = ~r ⊗ ~p = |~r||~p| sin π/2 |∆~v| |∆~r| ∆t = ∆t Bohr’s postulate consisted of quantizing angular momen- v r tum, i.e., which is, in the limit ∆t → 0 becomes π ` = n¯h = ~r × ~p = mvr = |~r||~p| sin d|~v| |~a| d|~r| |~v| 2 dt = = dt = v v r r where n is an integer (greater than or equal to 1), andh ¯ h i.e., (which is 2π ) is a constant, meaning that only integral multiples of ¯h were legal values of `. Note that we have v|~v| v2 used the fact that in circular motion, the angle between |~a| = = r r ~r and ~p is π/2, so the sine evaluates to one, and we get a simplified form for the angular momentum. which is a classic formula. All we need now is to remem- In the Coulomb problem of two oppositely charged par- ber that F=ma, i.e., Newton’s Second Law, and apply it ticles interacting (as in our problem), the force is given to our case. by the charge on one particle times the charge on the Remember that the angular momentum is conserved other divided by the square of the distance of separation. in this kind of “central force” model, and therefore, the This corresponds to vector representing the angular momentum, L~ , is a constant. And this means that L~ has constant length and Ze × e F = − direction. But if the direction is fixed, then it is perpen- r2 dicular to the motion at the instant, and at all future and past instants, i.e., it is perpendicular to the motion ~ where F is collinear with ~r or, more formally always. That means that the motion must take place in ~ Ze × e × rˆ Ze × e × ~r a plane, the plane to which L is perpendicular to. F~ = − = − (6) So, in that plane, we can construct a local x-y r2 r3 coördinate system which allows us to discuss the (cir- which is the same thing. Here, we have used the charge on cular) motion in this plane. If we write the electron (e) and the atomic number of the nucleus (Z) x = r cos ωt to force the nuclear charge, Ze, and the electron charge, e, to be multiplied together in Coulomb’s expression. y = r sin ωt This force is equal and opposite to the centripetal and takes the time derivative of these two equations, “force” which arises from the circular motion. which are describing a particle moving on a circle, we It is clear from the figure that the two triangles, one obtain involving changing r, the other involved in changing v, are similar, with the same included angle θ. From Figure x˙ =r ˙ cos ωt − rω sin ωt 2 part (e) one sees that y˙ =r ˙ sin ωt + rω cos ωt θ x/2 sin = where ω is constant, but r = r(t). We obtain, after a 2 y second differentiation: which, for θ → 0 (Taylor expansion for sin γ → γ) be- x¨ =r ¨cos ωt − 2rω ˙ sin ωt − rω2 cos ωt comes y¨ =r ¨sin ωt + 2rω ˙ cos ωt − rω2 sin ωt θ x/2 = which, by Newton’s Second Law should be the force, i.e. 2 y (see Equation 6), r cos ωt cos ωt which for the “v” (velocity) triangle (d) becomes m x¨ = F (x) = −Ze2 = −Ze2 e r3 r2 θ |~v|/2 r sin ωt sin ωt = m y¨ = F (x) = −Ze2 = −Ze2 2 v e r3 r2 4

(where we have employed the Coulomb’s Law force pre- which we can substitute into Equation 11 to obtain viously derived) which yields 2 2 2 2 n h¯ m v Ze me m2r2 e = = e 2 r cos ωt 2 r r2 r −(Ze /me) =r ¨cos ωt − 2rω ˙ sin ωt − rω cos ωt r3 Simpliﬁcation leads to r sin ωt −(Ze2/m ) =r ¨sin ωt + 2rω ˙ cos ωt − rω2 sin ωt 2 2 2 e r3 Ze n ¯h 2 = 3 r mer so, multiplying the ﬁrst of these by cos ωt and the second which, solving for r gives us by sin ωt and adding gives us n2¯h2 2 r = Ze 2 2 Ze me me r¨ − rω = − 2 r which we re-write as This led to the notion that if one re-wrote this as n2 ¯h2 r = (15) Z e2m Ze2 e m r¨ = m rω2 − → 0 e e r2 isolating the quantum number and the atomic number h¯2 for future use. The quantity 2 is known as the Bohr e me then the r.h.s could be forced to zero if the “centrifu- radius, a0. gal force” (centripetal acceleration) balanced the grav- The total energy itational (or electrostatic) attraction. Remember that 2 2 1 Ze ω = v , so rω2 = v . We then have E = m v2 − r r total 2 e r m v2 Ze2 now becomes (using Equation 14) e − = 0 (7) r r2 1 n2¯h2 Ze2 Etotal = me 2 2 − as the balance between the centripetal “force” on the one 2 mer r hand, and the Coulomb force on the other. which becomes, after substituting for r (using Equation We now have all three of the Bohr starting equations, 15),

2 1 n2¯h2 Ze2 1 2 Ze E = m v − (8) Etotal = me 2 − 2 2 total e 2 2 n h¯ 2 r 2 2 n h¯ 2 m 2 Z e me e Z e me ` = n¯h = mevr (9) 2 2 All it takes now is some elementary manipulations to mev Ze = (10) obtain r r2 2 2 2 4 1 n ¯h Z e me Etotal = me − 2 2 2 2 2 2 n2 h¯ n ¯h m 2 II. PART 3 e Z e me or The relevant Bohr equations are: 1 Z2e4m Z2e4m E = e − e total 2 2 2 2 2 m v2 Ze2 n ¯h n ¯h e = (11) r r2 which is 2 Z2e4m 1 2 Ze e E = m v − (12) Etotal = − total 2 e r 2n2¯h2 mevr = n¯h (13) which is the infamous Bohr result. We have for v from Equation 14, and their combination leads, irresistibly, to the Balmer n¯h formula (and others). What a bargain! v = First, we re-arrange Equation 13 to solve for v, i.e., mer and we have an expression for r from Equation 15, so n¯h v = n¯h v = mer n2h¯2 me 2 Ze me so, squaring, we have which is, n2¯h2 Ze2 2 v = v = 2 2 (14) mer n¯h 5

−27 III. PART 4 n2 ( 6.627×10 erg sec)2 r = 2π Z (4.8 × 10−10statcoulomb)29.1094 × 10−28grams We start with the total electronic energy in the Bohr model, and seek to use an appropriate unit system which and it behooves us to check the units here also using the will work easily in our case. The formula for the total same scheme. We have energy is (erg sec)2 r → Z2e4m (dynes−1/2cms)2grams E = − e total 2 2 2n ¯h which is where Z is the atomic number (an integer) e is the charge dyne2 cm2 sec2 on the electron, me is the mass of an electron, n is a non- r → 2 h dynes cm grams zero integer, and ¯h is 2π . The normal unit of charge is the Coulomb, but we will which ﬁnally becomes use instead the statcoulomb, since it is appropriate in the 2 2 gm cm 2 cgs system, i.e., dyne sec n 2 sec r → = sec = cm gm Z gm 1statcoulomb = 1dyne1/2cm which means that two charges (of one statcoulomb each) separated by one centimeter would exert a force of 1 dyne (gram − cm/sec2) (direction set by the charges’ signs. It turns out that the charge on the electron, in statcoulombs, is 4.8 × 10−10statcoulombs = e, (which corresponds to 1.6×10−19Coulomb). The mass of the electron, −28 −27 me is 9.1094 × 10 grams. h= 6.627 × 10 erg − sec, h and ¯h = 2π , so Z2(4.8 × 10−10)4(9.1094 × 10−28) Etotal = − 2 6.627×10−27 2 2n ( 2π ) where the units are

Z2(dynes1/2cm)4grams E = − (16) total 2n2(erg seconds)2 which is dynes2cm4grams E → total (dyne cm seconds)2 which is dynes2cm4grams E → total cm 2 (grams sec2 cm sec) which is

gram cm 2 4 sec2 cm grams Etotal → cm 2 (grams ( sec ) cm) which ﬁnally is

g cm2 E → = dyne cm = ergs total sec2 We know that

n2 ¯h2 r = 2 (17) Z e me 6

z D ϑ r = cos ϑ a r = sin ϑ r C

x O y = cos ϕ a a y = sin ϕ A ϕ B a x

FIG. 1: If we know x and y, then we can compute a. From z and a we can compute ϑ, while from the ratio of y and x we can compute ϕ. 7

v(t+ ∆ t) v v(t) r(t+∆ t) ϑ r r(t)

(a) (b)

v(t+ ∆ t) v(t) ∆ r(t) r(t+∆ t) ϑ r(t) x/2

FIG. 2: Construction of the centripetal “force”.

IV. ACKNOWLEDGMENTS herency, attempt to remove the last little errors (there were several) and place it in a “permanent” location, if This material was originally composed for interactive such can be said to exist. reading on the World Wide Web, but given the immi- The material is dedicated to all my students who over nent demise of the site which housed the material, given the years have either hated, loved, or were indiﬀerent to, my retirement, it seems worthwhile to re-edit it into come.