Question

28 : The molecule 12c32S has been detected in
interstellar clouds using microwave
spectroscopy. Predict which rotational level in
12C32S will have the greatest population at a
temperature of 70 K. The masses of the two
atoms are mC = 12.00 u and mS= 31.972 u and
the equilibrium bond length of the molecule is
1.534 Å. Note: The Boltzmann constant, k, has
a value in wavenumbers of 0.69503 cm​

Answer 1

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Answer 2

Answer:

The rotational level in ¹²C³²S is 4.93 ≈ 5.

Explanation:

Data given,

The temperature at which the rotational level in ¹²C³²S has greatest population = 70 K

The atomic mass of carbon atom, $m_{c}$ = 12 u

The atomic mass of sulphur atom, $m_{s}$ = 31.972 u

The equilibrium bond length of the molecule = 1.534 Å

Convert Å into metre

• 1 Å = 10⁻¹⁰ m
• 1.534 Å = 1.534 × 10⁻¹⁰ m

The Boltzmann constant value in wavenumbers, K = 0.69503 cm⁻¹

Firstly, we have to calculate the rotational constant, B, by using the equation given below:

• $B = \frac{h}{8\pi ^{2} \mu r^{2} c}$

Here,

• B = The rotational constant
• h = The Planck's constant = 6.626 × 10⁻³⁴ Js
• μ = The reduced mass
• r = The equilibrium bond length = 1.534 × 10⁻¹⁰ m
• c = The speed of the light = 3 × 10¹⁰ cms⁻¹

Now, we have to calculate the reduced mass, μ, by using the equation given below:

• $\mu = \frac{m_{c} \times m_{s} }{m_{c} + m_{s} }$
• $\mu = \frac{12 \times 31.972 }{12 + 31.972 }$ = $\frac{383.664}{43.972}$ = 8.72 u

Convert u into Kg

• 1 u = 1.66 × 10⁻²⁷
• 8.72 u = 8.72 × 1.66 × 10⁻²⁷ = 14.47 × 10⁻²⁷ Kg

Now, after putting all the values in the equation of rotational constant, we get:

• $B = \frac{(6.626\times10^{-34}) }{8\times(3.14)^{2}\times(14.47\times10^{-27})\times(1.534\times10^{-10})^{2}\times(3\times10^{10}) }$
• $B = \frac{(6.626\times10^{-34}) }{8\times9.85\times14.47\times10^{-27}\times2.353\times10^{-20}\times(3\times10^{10}) }$
• $B = \frac{(6.626\times10^{3}) }{8048.9}$ = 0.823 cm⁻¹

Now,

• The rotational level ($J_{max}$) can be calculated by the equation given below:
• $J_{max} = \sqrt{\frac{KT}{2B} } -\frac{1}{2}$

Here,

• K = The Boltzmann constant = 0.69503 cm⁻¹
• T = Temperature = 70K
• B = Rotational constant = 0.823 cm⁻¹

Therefore,

• $J_{max} = \sqrt{\frac{0.69503\times70}{2\times 0.823} } -\frac{1}{2}$
• $J_{max} = \sqrt{\frac{48.65}{1.646} } -\frac{1}{2}$
• $J_{max} = 5.43 - 0.5$ = 4.93 ≈ 5

Hence, the rotational level in ¹²C³²S = 5.