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the j=3 to 4 transition for a diatomic molecule occurs at 0.50cm⁻¹ what is the wavenumber for the J=6 to 7 transition for this molecule. Assume the molecule is a rigid motor.

User Keishawn
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Final answer:

To calculate the wavenumber for the J=6 to 7 transition, we need to determine the energy difference between the two states. This can be done using the rotational energy formula for diatomic molecules and the relation between energy and wavenumber. By equating the expressions for energy difference, we can solve for the rotational constant and then calculate the wavenumber.

Step-by-step explanation:

The wavenumber for the J=6 to 7 transition can be obtained by considering the energy difference between the J=6 and J=7 states. Assuming the molecule is a rigid rotor, the energy for a diatomic molecule can be calculated using the formula:



E = B * J(J+1), where B is the rotational constant.



Since the transition is from J=6 to J=7, the energy difference can be calculated as:



∆E = E(J=7) - E(J=6) = B * 7(7+1) - B * 6(6+1)



Given that the transition occurs at 0.50 cm⁻¹, we can use the relation between energy and wavenumber:



∆E = hc∆ν, where h is Planck's constant and c is the speed of light.



By equating the two expressions for ∆E, we can solve for B:



B = (hc∆ν) / [(7(7+1) - 6(6+1)) * hc]



Once B is known, we can calculate the wavenumber for the J=6 to 7 transition using the formula:



∆ν = B * (7(7+1) - 6(6+1))

User Tom Warner
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Final answer:

The wavenumber for the J=6 to 7 transition of a diatomic molecule can be found by multiplying the wavenumber for the J=3 to 4 transition by Planck's constant times the speed of light.

Step-by-step explanation:

To find the wavenumber for the J=6 to 7 transition, we need to understand the relationship between wavenumber and energy in a diatomic molecule. Wavenumber is defined as the reciprocal of wavelength and is given by the formula wavenumber = E / hc, where E is the energy difference between the two states and hc is Planck's constant times the speed of light.

In this case, we are given the wavenumber for the J=3 to 4 transition as 0.50 cm⁻¹. To find the wavenumber for the J=6 to 7 transition, we can use the fact that the energy difference between two states is equal to the product of the wavenumber and Planck's constant times the speed of light. Therefore, wavenumber6-7 = 0.50 cm⁻¹ x hc = 0.50 x 3 x 10^10 cm/s x 6.626 x 10^(-34) J×s = 9.936 x 10^(-25) J.

User Vivek Salve
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