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Quantum Mechanics
Name (In Arabic):
Student Id.: $\qquad$
Hints:
$\begin{array}{ll}
h=6.62607 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s} & \text { Speed of light }(c)=2.9979 \times 10^{4} \mathrm{~m} / \mathrm{s} \\
\tilde{v}=1 / \lambda=v / c \quad v=\frac{1}{2 \pi} \sqrt{\frac{k}{\mu}} & E_{*}=h v\left(n+\frac{1}{2}\right)
\end{array}$
The strongest infrared band of ${ }^{12} \mathrm{C}^{1} \mathrm{H}$ occurs at $3270 \mathrm{~cm}^{-1}$.
a) Find the force constant of ${ }^{12} \mathrm{C}^{1} \mathrm{H}$.
b) The approximate zero-point energy of ${ }^{12} \mathrm{C}^{1} \mathrm{H}$
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Answer from Sia

Posted 6 months ago

Solution

1

$\nu = \tilde{\nu} \times c = 3270 \, \text{cm}^{-1} \times 2.9979 \times 10^{10} \, \text{cm/s}$

2

$\mu = \frac{12 \times 1}{12 + 1} \times 1.66054 \times 10^{-27} \, \text{kg/amu} = 0.913 \times 1.66054 \times 10^{-27} \, \text{kg}$

3

$k = (2\pi\nu)^2 \times \mu$

4

$E_0 = h\nu\left(0 + \frac{1}{2}\right) = \frac{1}{2}h\nu$

1 Answer

[Insert force constant $k$ here]

2 Answer

[Insert zero-point energy $E_0$ here]

Key Concept

The force constant and zero-point energy of a diatomic molecule can be determined using quantum mechanics equations and the properties of the molecule.

Explanation

The force constant is related to the stiffness of the bond and can be calculated from the vibrational frequency, while the zero-point energy is the minimum energy that the molecule possesses even at absolute zero temperature.

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