Question

Quantum mechanics is used to describe the vibrational motion of molecules, but analysis using classical physics gives some useful insight. In a classical model the vibrational motion can be treated as SHM of the atoms connected by a spring. The two atoms in a diatomic molecule vibrate about their center of mass, but in the molecule HI, where one atom is much more massive than the other, we can treat the hydrogen atom as oscillating in SHM while the iodine atom remains at rest. (a) A classical estimate of the vibrational frequency is f=7×1013 Hz. The mass of a hydrogen atom differs little from the mass of a proton. If the HI molecule is modeled as two atoms connected by a spring, what is the force constant of the spring? (b) The vibrational energy of the molecule is measured to he about 5×10−20 J. In the classical model, what is the maximum speed of the H atom during its SHM? (c) What is the amplitude of the vibrational motion? How does your result compare to the equilibrium distance between the two atoms in the HI molecule, which is about 1.6×10−10 m?

Answer 1

Answer:

a)  k = 1.93 10² N / m , b) v = 0.773 10⁴ m / s , c)  A = 3.6 10¹² m

Explanation:

a) As the iodine atom is still, there are no forces on it, so we can work on the hydrogen atom only with mass m = 1.67355 10⁻²⁷ kg, the angular velocity is

           w = √ k / m

           k = m w²

 Angular velocity and frequency are related.

           w = 2π f

           k = 4π² m f²

Let's calculate

           k = 4π²  1.67355 10⁻²⁷ (7 10¹³)²

           k = 1.93 10² N / m

b) The energy of the atom is conserved, in the approximation of a mass-spring system, the energy ranges from kinetic to power, when the elongation is zero all energy is kinetic

             Em = K = ½ m v²

             v = √ 2Em / m

             v = √ (2 5 10⁻²⁰ / 1.67355 10⁻²⁷)

             v = √ (0.5975 10⁸)

            v = 0.773 10⁴ m / s

c) At the point of maximum elongation all energy is potential

               Em = U = ½ k A²

               A = √ 2 Em / k

               A = √ (2 5 10⁻²⁰ / 0.773 10⁴)

               A = √ (12,937 10⁻²⁴)

              A = 3.6 10¹² m

We see that the classic result is much smaller by one factor.

              A / A_true = 3.6 10⁻¹² / 1.6 10⁻¹⁰

             A / A_true = 2.25 10⁻²

Answer 2

Answer:

The spring force constant is $k=1.93 \times 10^{2} \;\rm N/m$.

The maximum speed of H-atom is 7738.23 m/s.

The amplitude of vibrational motion is $2.27 \times 10^{-11} \;\rm m$.

Explanation:

Given data:

Vibrational frequency is, $f=7 \times 10^{13} \;\rm Hz$.

Vibrational energy is, $E=5 \times 10^{-20} \;\rm J$.

Equilibrium distance between two atoms is, $d = 1.6 \times 10^{-10} \;\rm m$.

(a)

The angular frequency is,

$\omega = \sqrt{\dfrac{k}{m} }$ .................................................................(1)

Here, k is the force constant of spring and m is the mass of hydrogen atom. And its value is $1.67 \times 10^{-27} \;\rm kg$.

Angular frequency is also expressed as,

$\omega = 2 \pi f$.......................................................(2)

Comparing equation (1) and (2) as,

$2 \pi f= \sqrt{\dfrac{k}{m} } \\(2 \pi \times 7 \times 10^{13})^{2}= \dfrac{k}{1.67 \times 10^{-27}}\\k=1.93 \times 10^{2} \;\rm N/m$

Thus, the spring force constant is $k=1.93 \times 10^{2} \;\rm N/m$.

(b)

The vibrational energy provides the necessary kinetic energy to the molecule. Then,

Vibrational Energy = Kinetic energy

$E= KE\\E= \dfrac{1}{2}mv^{2}\\5 \times 10^{-20}= \dfrac{1}{2} \times 1.67 \times 10^{-27} \times v^{2}\\v = 7738.23 \;\rm m/s$

Thus, the maximum speed of H-atom is 7738.23 m/s.

(c)

The Amplitude (A) is calculated using the expression,

$E= \dfrac{1}{2}kA^{2}\\5 \times 10^{-20}= \dfrac{1}{2} \times (1.93 \times 10^{2}) \times A^{2}\\A= 2.27 \times 10^{-11} \;\rm m$

Thus, the amplitude of vibrational motion is $2.27 \times 10^{-11} \;\rm m$.

Compare with equilibrium distance as,

$\dfrac{A}{d} =\dfrac{2.27 \times 10^{-11} \;\rm m}{1.6 \times 10^{-10} \;\rm m}\\\dfrac{A}{d} =0.142$

And it is 0.142 times the equilibrium distance between the atoms.

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