Physics Ph 1205

Md. Kamrul Hasan Reza

Department of Physics

Khulna University of Engineering & Technology Khulna-9203, Bangladesh

Tel.: +880-41-769468~75 Ext. 587(O), 588 (R)

e-mail: mkhreza@phy.kuet.ac.bd, mkhreza1@gmail.com Website : www.kuet.ac.bd/phy/reza/

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Welcome to my Class

Physics Ph 1205

09:15 AM

February 28, 2021

COVID-19 Precautions

Don’t be afraid

Be aware of the pandemic

Use appropriate outfits if you compelled to go out

Try to maintain proper diet

Do not forget to exercise (at least one hour) regularly

Try to follow the guidelines of WHO and Bangladesh Government

 Try to stay at home

Bohr Correspondence Principle

The greater the quantum number, the closer quantum physics approaches classical physics

According to electromagnetic theory, an electron moving in a circular orbit radiates EM waves whose frequencies are equal to its frequency of revolution and to harmonics (that is, integral multiples) of that frequency. In a hydrogen atom the electron’s speed is

∴ 𝒗 = 𝒆

𝟒𝝅𝝐_𝒐𝒎𝒓

Fig.6: Force balance in the hydrogen atom

𝑭_𝑪 = 𝒎𝒗^𝟐 𝒓 𝐹_𝑒 = 1

4𝜋𝜀_𝑜 𝑒² 𝑟²

𝐹_𝑒 = 𝑭_𝑪 The condition for orbit stability

The centripetal force,

The electrostatic force, …….(1)

…….(2)

…….(3)

∴ 𝒗 = 𝒆

𝟒𝝅𝝐_𝒐𝒎𝒓

⟹ 𝒎𝒗^𝟐 𝒓 =

𝟏 𝟒𝝅𝝐_𝒐

𝒆^𝟐 𝒓^𝟐

Condition for orbit stability

n λ = 2 π r

n n=1, 2, 3, …..

Using

𝝀 = 𝒉 𝒎𝒗

…….(4)

…….(5)

…….(6)

Hence the frequency of revolution f of the electron is

The radius r_n of a stable orbit is given in terms of its quantum number n by Eq. (7)

= 𝒆

2𝝅 4𝜋𝜖_𝑜𝑚𝑟³

∴ 𝒓_𝒏= 𝒏²𝒉²𝝐_𝒐

𝝅𝒎𝒆² …….(7)

and so the frequency of revolution is

…….(8)

a hydrogen atom dropping from the n_i^th energy level to the n_f^th energy level emits a photon whose frequency is

Let us write n for the initial quantum number n_i and n – p (where p = 1, 2, 3, . . .) for the final quantum number n_f.

…….(9)

With this substitution

When n_i and n_f are both very large, n is much greater than p, and

= 𝑚𝑒⁴ 8𝜖_𝑜²ℎ³

2𝑛𝑝 − 𝑝² 𝑛²(𝑛 − 𝑝)²

2𝑛𝑝 − 𝑝² ≈ 2𝑛𝑝

…….(10)

When p = 1, the frequency of the radiation is exactly the same as the frequency of rotation f of the orbital electron given in Eq. (8). Multiples of this frequency are radiated when p = 2, 3, 4, . . . .

Hence both quantum and classical pictures of the hydrogen atom make the same predictions in the limit of very large quantum numbers.

Nuclear Motion and Reduced mass

The nuclear mass affects the wavelengths of spectral lines

Fig. 1: Both the electron and nucleus of a hydrogen atom revolve around a common center of mass.

In developing the theory of hydrogen atom, its nucleus remains stationary while the orbital electron revolves around it. This is not an unreasonable assumption: the proton mass is 1836 times greater than the electron mass and the electrostatic force

We begin in expressing in a different way the condition that a stable orbit in an atom consist of an integral number of electron de Broglie wavelengths.

(1)

This condition states that

n λ = 2 π r

Since the de Broglie wavelength λ is given by

We may equivalently write

(2)

(3)

(4)

In terms of the angular velocity of the electron, the quantization rule is stated

Since

If the nucleus has a mass M that is not infinite, both it and its orbital electron revolve around a common center of mass. If we think of the nucleus and electron in a hydrogen atom as being at opposite ends of a massless rod r long, in effect constituting a lopsided dumbbell, the center of mass , which is r_N distant from the nucleus and r_E distant from the electron, may be found from the requirement

(5)

(6)

m r_E = M r_N Where

r = r_E + r_N

Total angular momentum = electron angular momentum +

nuclear angular momentum The total angular momentum of the hydrogen atom is the sum of the angular momentum of the electron and that of the nucleus

The angular velocity ω is the same for both particles

(7)

(8)

According to Bohr’s first postulate, the total angular momentum of the atom must be an integral multiple of ℏ, and so

(9)

From eqns. (7) & (8) we find that

(10)

(11)

And so eq. (9) becomes

(12)

The condition for force balance must also be generalized to take into account nuclear motion. Its proper expression is

Substituting r_E from eq. (10 ) we find

(13)

(14)

If we let

We see that eqns. (12) & (14) become respectively

(15)

(16)

(17)

From eqns. (16) & (17) we can obtain an expression for the energy levels of the hydrogen atom.

The result is

Here m’ replaces the electron mass m. The quantity m’ is known as reduced mass. Due to the motion of the nucleus, all the energy levels of the hydrogen are changed by the fraction

(18)

An increase of 0.055 percent since the energies E_n, being smaller in absolute value, are therefore less negative.

The value of the Rydberg number, R to eight significant figures without correcting for nuclear motion is 1.0973731x10⁷m^-1; the correction lowers it to 1.0967758x10⁷m^-1.

Because of the greater nuclear mass, the spectral lines of deuterium are all shifted slightly to wavelengths shorter than the corresponding one of ordinary hydrogen.

Thus the H_α line of deuterium, which arises from a transition from the n=3 to the n=2 energy level, occurs at a wavelength of 6561 A, whereas the H_α line of hydrogen occurs at 6563 A. This difference in wavelength was responsible for the identification of deuterium in 1932 by the American chemist Harold Urey.