The energy levels of the 1D harmonic oscillator are given...

The energy levels of the 1D harmonic oscillator are given by: En = (n + 1/2) ??, n = 0, 1, 2, 3, ... The CO molecule has a (reduced) mass of mCO = 1.139 x 10^-26 kg. Assuming a force constant of kCO = 1860 N/m, what is: a) The angular frequency, ?, of the ground state CO bond vibration? b) The energy separation between the ground and first excited vibrational states? The N2 molecule has a (reduced) mass of mN2 = 1.163 x 10^-26 kg. Assuming a force constant of kN2 = 2287 N/m : c) How much higher is the angular frequency of the N2 ground state in comparison to CO? The (unnormalized) 1st excited state of the harmonic oscillator is given by: ?_1^HO(x) = xe^(-ax^2) Where a = m?/2?. At what values of x is the probability distribution associated with this wavefunction: d) Zero? e) A maximum?

Transcript

00:04 In this question, given that the mass of carbon monoxide is 1 .139 multiplied by 10 raised to power minus 26 kg.

00:16 The value of kco is given as 1860 n per m and the value of mass of nitrogen is given as 1 .163 multiplied by m.

00:33 10 range to power minus 26 kg and the value of k and do is given as 2287 n per m in the part a we have to calculate the value of w in the part b we have to calculate the value of delta e in the part c we have to calculate the value of angular frequency of nitrogen in the part d and e we have to calculate the value of x so solving for the part a so here the expression for mu is 1 divided by 2 pi multiply by under root of k divided by mu.

01:48 Now here, 2 pi mu is equal to w.

01:56 Now where w is the angular velocity, mu is the frequency, k is the constant, and m is the reduced, mass.

02:34 Now we can write here, therefore the expression will be w is equal to under root of k divided by mu.

02:49 So here we put the values.

02:51 So under root of 1860 n per m divided by 1 .139, multiply by 10 raised to power minus 26 kg.

03:11 And from here we can write w is equal to under root of 1633 .01 multiply by 10 raised to power 26 kg m per s square per m divided by kg.

03:43 So, here we can say that this value is equal to under root of s minus raised to power minus 2.

04:02 Or we can say that s raised to power minus 1.

04:07 So, therefore, w is equal to 40 .41 multiplied by 10 raised to power 13 .13.

04:22 S raised to power minus 1 or we can say that 40 .41 multiply by 10 raised to power 13 hertz so this is the value of the angular velocity now the part b here the expression we will write as e n is equal to n plus 1 divided by 2 multiply by h w so here the delta e, which is energy separation, is equal to e1 minus e 0.

05:10 So here the e 0 is equal to 1 divided by 2, hw for n is equal to 0.

05:22 And e1 is equal to 3 divided by 2, hw for n is equal to 1.

05:33 So here we will write as delta e is equal to 3 divided by 2, hw minus 1 divided by 2, hw.

05:48 So that will be equal to hw...

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