The wave number of the radiation absorbed in the j=0 -> j=1 transition of carbon

Question

The wave number of the radiation absorbed in the j=0 -> j=1 transition of carbon monoxide has been measured as 3.842 cm^-1. Convert this wave number to joules. Calculate the wavelength of the photon. In what region of the electromagnetic spectrum is this photon? If a molecule in the J=1 state loses energy and rotates with the energy given by J=0, what is the energy, in joules, of the emitted photon? What is the moment of inertia of the CO molecule? If the absorption of this photon is from a molecule comprised of 12C and 16O, what is the bond length of the CO molecule?

Explanation / Answer

wavenumber = 3.842 cm-1 = 384.2 m-1

now Energy in joules = h x wavenumber x c = 6.625 x 10^-34 x 384.2 x 3 x10^8 = 7.636 x 10^-23 J

wavelength = 1/wavenumber = 1/384.2 = 0.0026028 m = 2.603 x 10^ -3 m

Region is IR ( Infra red)

for CO , wavenumber absorbed for J = 0 to 1 is wavenumber = B J' ( J'+1) - B ( J) ( J+1)

3.842 = B 1(1+1) - B ( 0)(0+1)

3.842 = 2B

B = 1.921 cm-1 is rotational constant

enegry of photon releases when molecules deexcites from J = 1 to 0 is equal energy absrobed when molecule excites to J to 0 to 1 = 3.842 cm-1 = energy of photon emitted

we had B = h/8 x (3.14^2) x I where I = moment of inertia

384.2 = 6.625 x 10^-34 / ( 8x3.14^2 x I)

I = 2.186 x 10^ -38 Kg m^2 is moment of inertia

we had I = u x r^2 where u = reduced mass = ( mC x mO ) / ( mC + mO) = ( 12 x 16) / ( 12+16) = 6.85714 amu = 6.85714 x 1.67 x 10^ -27 = 1.145 x 10^ -26 kg

now we find r

2.186 x 10^-38 = 1.145 x 10^-26 x r^2

r = 1.91 x 10^ -12 m = 1.91 pm

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