2 Louis de Broglie relation

CP100X year 2012/2013

C. Crespos, F. Castet and L. Truflandier Universit´ e Bordeaux 1

File: CorrTDCP100X, Revision v1.1 (as of 18/10/2013)

This document summarizes the various relations practical examples introduced during the courses. It gives you the numerical values and the essential materials. Physical and chemical interpretations are only accessible during the tutorials. All the fundamental constants and conversion of energy units can be found in “Report HUKI 1-89”, ISSN 0784-065. The other very useful links are:

• http://physics.nist.gov/cuu/Constants/index.html

• http://www.webelements.com

Version Information: Revision: v1.1 Date: 18/10/2013 File: CorrTDCP100X,

1 Planck–Einstein relation:

• Planck relation, with ν = wave frequency [s⁻¹= Hz];λ= wave length [m]; ¯ν = wave number [m⁻¹] E[J] =h ν=h c

λ=h cν¯ (1)

• IfE is given for 1 mole then

E[J.mol⁻¹] =h νN, with N[mol⁻¹] = Avogadro constant (2)

2 Louis de Broglie relation

• de Broglie relation for wave–particle

λ= h

p = h[J.s⁻¹]

m[kg]v[m.s⁻¹] (3)

3 Bohr model

Let’s consider a system composed of a nucleus with a charge Ze (where Z is the number of protons), and a single electron of charge −e with a massm. We assume that the electron is moving with a constant velocity

~v around the nucleus at a distance~rand describes a circular orbit. The system is stable and we consider that the orbit of the electron is stationary.

a) As a result, following the classical mechanics laws for a system at dynamical equilibrium, we write, md~v

dt =F~Coulomb+F~velocity= 0, (4)

where F~_Coulomb describes the electrostatic interaction and F~_velocity the centrifugal force. If we take the Cartesian frame centered on the nuclei and thex–axis collinear to the forces, we have

F~_Coulomb = − 1 4π0

Ze²

r² ~i, with r≡ |~r| (5)

F~velocity = mv²

r ~i, with v≡ |~v| (6)

and~iis the unit vector along thex–axis.

1

b) Then, from 4 and 5, we get

v²(r) = 1 4π0

Ze²

mr (7)

c) Form the classical expression of the kinetic energyT = ¹₂mv²we have, T(r) =1

2 1 4π0

Ze²

r (8)

d) From the relation of the electrostatic force to the potential, i.e.F~ =−∇V~ , we found that V(r) = − 1

4π0

Ze²

r (9)

and E_T(r) = T(r) +V(r) =−1 2

1 4π0

Ze²

r (10)

Notice: the ratioV /T =−2 is a constant independent ofr; this is the classical representation of the virial theo- rem which states that the time–average of the kinetic and potential energies of a system of particles interacting by Coulomb forces are related byhTi=−¹₂hVi. From relation 9 we observe that for each value ofrthere must exists a value ofET. This is in contradiction with the observation of well defined emission rays in the spectra for the hydrogen atom. Moreover, the total energy of the system is inversely proportional to the distance r, that is

E_T(r→0)−→ −∞ (11)

In other words, to reach the most stable configuration, the electron must fall on the nucleus. Using the rela- tion 7, this means that for a distance nucleus–electron below 3×10⁻⁵ ˚A,¹ the velocity of the particle is higher than the celerityc. This result is forbidden by the laws of physics even within the relativistic generalization.

a) To circumvent these contradictions and respond to the experimental observations, Bohr has proposed 3 postulates:

1. The electron occupy stationary states which can be describe by classical mechanics as a stationary orbits.

This has been already used in the introduction of this exercise.

2. The passing between different stationary states is followed by the emission/absorption of homogeneous radiations as defined in the Planck’s theory.

3. Quantification of the electronic angular (or kinetic) momentum defines the allowed stationary states. We recall that the kinetic momentumL~ is given by

L~ =~p×~r and |L|~ =|~r| |~p|sinθ (12) For a velocity vector perpendicular to the position vector, i.e.θ=^π₂, we obtain

L = pr (13)

mvr = n~, with n∈N⁺ (14)

where the second equality arises from the quantification of|~L|.

b) From the relation 14 we can express the velocity of the electron in function ofn, v= h

2πm n

r; v²= h² 4π²m²

n²

r² (15)

By combining relation 15 with 7, we obtain the following equalities v = e²

20h Z

n, (16)

1

r = 1

4π0

4π²me² h²

Z

n², (17)

r = ₀h² πme²

n²

Z =a₀n²

Z, (18)

with a0 = 0h²

πme². (19)

a₀(∼= 0.5292 ˚A) is called the Bohr radius (you can easily check thata₀ has the dimension of a distance).

1Notice that nuclei present finite sizes comprise between 1 and 15 fm (10⁻⁵˚A).

c) By substituting the expression 17 of 1/rin 10, we obtain for the total energy En = −

1 4π0

2

2π²me⁴ h²

Z²

n² (20)

En = − me⁴ (4π₀~)²

1 2

Z²

n² (21)

E_n[eV] ∼= −13.6×Z²

n² (22)

From the fundamental Postulate given in 14, we found that the total energy of the hydrogenoid atom is quantified through thequantum number nfollowing 22. For the hydrogen atom (Z = 1), we found that the energy of the fundamental state, i.e. the state with the lower energy, is obtained for n = 1 and equal to

−13.6 eV. This also defined the ionization energyIH = +13.6 eV of hydrogen, i.e. the energy we need to furnish to the system to dissociate the electron from the nuclei.

IH = E_n→∞−En=1 (23)

= 13.6× 1

n²₁ − 1 n²_∞

(24) d) To compute the energy of transition ∆Efrom the orbitn(initial state) to the orbitn⁰ (final state) we use

∆E_n→n⁰ = En−En⁰ (25)

= 13.6×Z² 1

n⁰² − 1 n²

(26)

• Ifn < n⁰: this corresponds to the energy of absorption andE_n→n⁰ <0

• Ifn > n⁰: this corresponds to the energy of emission andE_n→n⁰ >0 e) IfZ= 1, we obtain

∆En→n⁰ = me⁴ (4π0~)²

1 2

1 n⁰² − 1

n²

n>n⁰

(27)

= me⁴ (20h)²

1 2

1 n⁰² − 1

n²

n>n⁰

(28)

= hν (29)

Then, using relation 1 we get, 1

λ = me⁴ (4π₀~)²

1 2hc

1 n⁰² − 1

n²

n>n⁰

(30)

= α²mc 4π~

1 n⁰² − 1

n²

n>n⁰

, withα= e²

4π₀~c (31)

= me⁴ 8²₀h³c

1 n⁰² − 1

n²

n>n⁰

(32)

= R_∞ 1

n⁰² − 1 n²

n>n⁰

, withR_∞= me⁴

8²₀h³c (33)

where α∼= 1/137.036 is the [dimensionless] fine-structure constant. In a first approximation, the constant R_∞∼= 109737 cm⁻¹ is taken as equal to the Rydberg constant² R_H∼= 109677 cm⁻¹.

2Indeed, we observe a deviation between the constantR∞deriving from the Bohr model and the measured Rydberg constant R_H. Within the Bohr approximation, we assumed that the mass of the nucleusMis huge with respect to the mass of the electron me, in other words, the center of mass of the system is centered on the nucleus. This statement leads to approximateR_HbyR∞, i.e. setting the [dimensionless] prefactorκM =µM/me= 1/(1 +^m_M^e) equal to 1 inR_H=κMR∞; withµM the reduced mass of the system nucleus + electron. For hydrogen atomκ_H∼= 0.99945, which is not a too crude approximation.

4 Exercices

EXO 1

1. After absorption of ∆E_n→n⁰ = hνabs, with n⁰ > n, we may use Eq. (33) to calculate the final state n⁰, considering that in this casen= 1, i.e. the fondamental state of the Hydrogen atom. We have:

−1 λ =RH

1 n⁰² − 1

n²

n⁰>n

. (34)

Afterstraigthforward algebra, we obtain:

n⁰ =

n²λRH

λR_H−n^{2 1/2}

=

1.066444 1.66444−1

^1/2

'4 (35)

If now we observe the emission phenomena ∆E_4→n⁰ =hνemi withn⁰< n, we have:

1 λ =RH

1 n⁰² − 1

n²

n⁰<n

, (36)

and

n⁰=

n²λR_H n²+λRH

^1/2

=

16×5.330302 16 + 5.330302

1/2

'2 (37)

2. For the excited state of hydrogrenn= 4, we must observe 6 different wave-lengths when the system is coming back to equilibrium. From Eq. (1), we observe that the smaller is the wave-number ¯ν, the smaller is ∆E_n→n⁰. By analysing Eq. (27), we found that this is achieved for the larger values ofnandn⁰, with the smaller ∆n such as|n−n⁰|= 1; that is in our case ∆E_4→3 (We can also check with a graphical representation).

¯

νmin= 1

λ_max =RH

1 3²− 1

4²

= 533152 m⁻¹= 5331.5 cm⁻¹ (38)

EXO 2

1. A hydrogen-like ion is any atomic nucleus with one electron, i.e. with any number of chargeZ. The ionization energy (IH) is the energy we need to furnish to the system to dissociate the electron from the nuclei; see Eq. (23).

2. see section (3) and Eq. (22).

3. For a hydrogen-like ion, we have:

IH = 13.6Z² (eV) (39)

Z = IH

13.6 1/2

= 217.6

13.6 1/2

(40) 4. Z = 4 ⇒ Be

5.

∆E_1→3 = 193.4 eV = 193.4×1.6022×10⁻¹⁹J (41)

= 3.1×10⁻¹⁷ J (42)

= 214.9×1.6022×10⁻¹⁹×10⁻³×6.0221×10²³kJ.mol⁻¹ (43)

= 18662 kJ.mol⁻¹ (44)

EXO 3

cf. section (3).

EXO 4

a. yes; if the energy levels involved are the same b. no; Lyman series

c. no; it depends also on Z;rn=−13.6^Z_n²2 ;Z = 3 for Li d. no; the spherical symmetry is associated withs-type orbitals

e. no; Be³⁺ is an hydrogen-like atom f. yes ;rn=a0n²

Z

g. no

h. no; only 4 times larger

EXO 5

1.

∆E_n→n⁰(eV) = En−En⁰ <0 (45)

= 13.6×Z² 1

n⁰² − 1 n²

(46) n⁰ =

13.6×n²Z² 13.6×Z²−∆E_n⁰_→n

^1/2

(47)

=

54.4 54.4−51.0

^1/2

= 4 (48)

2.

1

λ = R_H 1

n⁰² − 1 n²

n⁰<n

(49)

= RH

1 2² − 1

4²

(50)

λ = 16

3×1.0974×10⁷ = 486 nm (51)

3. ⇒visible

EXO 6

1. B−→^hν B⁴⁺ + 4e⁻

2. For a hydrogen-like atom such as B⁴⁺, with a number of chargeZ_B= 5, we have:

1

λ = R_B 1

n⁰²− 1 n²

n>n⁰

, (52)

withRB = RH×Z_B² (53)

ThenRB= 25×RH= 27.435 10⁷ m⁻¹

3. The lowest energetic transition for the Balmer series is: ∆E3→2. The associated wavelength is:

1

λ = 27.435×10⁷× 5

36 (54)

λ = 26 nm (55)

EXO 7

1. We compute the electronic velocity,v_e, from the kinetic energy (T) relation:

ve= r2T

m_e =

r2×1.10×1.6022×10⁻¹⁹

9.1094×10⁻³¹ = 622 049 m.s⁻¹ (56)

2. We need to compute the minimal energy (W0) needed to extract one electron from the cristal. Using the fact that the total energy (ET) is the sum of W0and T and the data of 1. we can deduce that:

W0 = ET−T (57)

with: ET =hc

λ= 4.414×10⁻¹⁹J = 2.75 eV (58)

and: T = 1.10×1.6022×10⁻¹⁹= 1.762×10⁻¹⁹J = 1.10 eV (59)

W0 = 2.652×10⁻¹⁹ J = 1.65 eV (60)

The associated wavelength is: λ₀= 749 nm. Notice thatλ₀> λ.

3. W0= 1.65 eV =^2.652×10_4.184⁻¹⁹ × N ×10⁻³= 38.2 kcal.mol⁻¹ 4. We haveλ⁰= 180 nm andT⁰ =E_T⁰ −W₀. Then,

E_T⁰ = hc

λ⁰ = 11.036×10⁻¹⁹ J = 6.89 eV (61)

T⁰ = 9.27410⁻¹⁹ J = 5.24 eV (62)

ve = 1 556 596 m.s⁻¹ (63)

EXO 8

1. Cs−→^hν B⁺+ e⁻ Photoelectric work function (´energie de seuil pho´eletrique):

W0 = h c λ0

(64)

= 6.6261×10⁻³⁴×2.9979×10⁸

0.66×10⁻⁶ (65)

= 30.1×10⁻²⁰J = 1.88 eV = 43.4 kcal.mol⁻¹ (66) 2. E_T =W₀+T =hν=hc/λ. IfE_T = 2W₀, thenT =W₀.

2W0 = 2×h c

λ₀ (67)

λ = λ0

2 (68)

ve =

r2W0

m_e =

r2×30.1×10⁻²⁰

9.1094×10⁻³¹ (69)

= 812 930 m.s⁻¹ (70)

3. cf. de Broglie relation in Eq. (3).

λ= h meve

= 6.6261×10⁻³⁴

9.1094×10⁻³¹×812930 = 8.9 ˚A (71)

EXO 9

Apply the de Broglie relation of Eq. (3) ; taking care about units conversion.

EXO 10

Apply Beer-Lamber relation ;T = 0.113

EXO 11

Apply Beer-Lamber relation ; solve a system of equations with two unknown ;CX= 1.15⁻⁴M;CY = 4.71⁻⁴M