Please explain the answers. Real molecules have to be treated by quantum, not cl

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Please explain the answers.

Real molecules have to be treated by quantum, not classical,mechanics. Although the classical treatment gives a correct description of the parameters involved and of the potential energy, there are differences. In the classical treatment, any energy is allowed; in the quantum treatment, only particular energy levels are allowed, as shown in the figure at the right.
The lowest level permitted (the ground state, E0) is not at the bottom of the well but a little up from it. For a simple harmonic oscillator, the spacing
between neighboring allowed levels is E = hf. According to our study of thermodynamics, we know that the relative probability of a molecule being
excited by an amount E = E – E0 in a medium at temperature, T, has the form

2. (10 pts) Real molecules have to be treated by quantum, not classical, mechanics. Although the classical treatment gives a correct description of the parameters involved and of the potential energy, there are differences. In the classical treatment, any energy is allowed; in the quantum treatment, only particular energy levels are allowed, as shown in the figure at the right. The lowest level permitted (the ground state, Eo is not at the bottom of the well but a little up from it. For a simple harmonic oscillator, the spacing between neighboring allowed levels is AE hf. According to our study of thermodynamics, we know that the relative probability of a molecule being excited by an amount AE E-Eo in a medium at temperature, T, has the form POE) -AE POEo) where E is a parameter that has units of energy and depends on T 2.1 (5 pts) If we are using AE to represent the excitation energy of a single diatomic molecule, and the molecule is in a human body at normal body temperature of 37 oC, what value should we take for E? A. 3.1 meV B. 25 meV C. 26 meV D. 0.03 kJ E. 2.4 kJ F. 2.5 kJ

Explanation / Answer

2.1 . For thermal energy, = kBT, where kB is the Boltzman constant and T is the temperature.

=> = 1.38×10-23 × (273.15 + 37) × 6.24 × 1018 eV= 0.0267 eV = 26meV (option C)

2.2. Ground state energy E0 = 0.11eV, first excited state E1 = E0 + hf = 0.11 + 0.22 = 0.33 eV, T=37C

-E/ = -(0.33-0.11) /0.0267 = -8.24

Probability to find in first excited state P = exp(-8.24) = 2.6×10-4 .~2×10-4 (option C)

3.1. E0 = 110 meV, E2 = 110+2×220=550 meV, E= E2-E0 = 550-110 meV = 440 meV

=> =hc/E = (6.63 × 10-34 × 3× 108)/(440 × 10-3 /6.24 × 1018) = 2.8 × 10-6 m (option B)

3.2. Energy is inversely proportional to wavelength. When the wavelength is half, the energy is double. This wavelength can excite from E0 to E4. When the wavelength is double, the energy is half. This wavelength can excite from E0 to E1. This can happen will other wavelengths too, however whose energy is multiple of 220 meV. (option D)

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