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Chapter 5: Q52E (page 190)

To a good approximation. the hydrogen chloride molecule, HCI, behaves vibrationally as a quantum harmonic ascillator of spring constant 480N/mand with effective osciltating mass just that of the lighter atom, hydrogen If it were in its ground vibtational state, what wave. Iength photon would be just right to bump this molecule. up to its next-higher vibrational energy state?.

Short Answer

Expert verified

The wavelength位=3.51.10-6m

Step by step solution

01

Given Data

k=480Nm

02

Concept of the wavelength

In order to obtain the wavelength, need to calculate for the change in energy.鈭�Ebetween E0 and E1 of the harmonic oscillator.

En=n+12h蝇鈭�E=E1-E0=1+12h蝇-0+12h蝇=h蝇

Where 鈭�E is change in energy between E0 and E1 of the harmonic oscillator, h is the plack鈥檚 constant and 蝇is the wavelength.

03

Calculate the value of the wavelength by using the formula

The angular frequency, 蝇, to the mass and spring constant:

蝇=km鈭�E=hkm=1.055路10-34路4801.67路10-27=5.65路10-20J

04

Evaluate the value of the wavelength

Now the change in energy, 鈭�E, as:

鈭�E=hc位位=hc鈭�E=6.62路10-343路1085.65路10-20位=3.15路10-6m

Thus, the wavelength位=3.15路10-6m

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Most popular questions from this chapter

It is possible to take the finite well wave functions further than (21) without approximation, eliminating all but one normalization constant C . First, use the continuity/smoothness conditions to eliminate A, B , andG in favor of Cin (21). Then make the change of variables z=x-L/2 and use the trigonometric relations

sin(a+b)=sinacosb+cosasinband

cos(a+b)=cosacosb-sinasinbon the

functions in region I, -L/2<z<L/2. The change of variables shifts the problem so that it is symmetric about z=0, which requires that the probability density be symmetric and thus that 蠄(z)be either an odd or even function of z. By comparing the region II and region III functions, argue that this in turn demands that (伪/k)sinkL+coskL must be either +1 (even) or -1 (odd). Next, show that these conditions can be expressed, respectively, as 伪k=tankL2 and 伪k=-cotkL2. Finally, plug these separately back into the region I solutions and show that

蠄(z)=C脳{e伪(z+L/2)鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌�z<L/2coskzcoskL2鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌�-L/2<z<L/2e-伪(z-L/2)鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌�z>L/2


or

蠄(z)=C脳{e伪(z+L/2)鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌�z<L/2-sinkzsinkL2鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌�-L/2<z<L/2e-伪(z-L/2)鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌�z>L/2

Note that Cis now a standard multiplicative normalization constant. Setting the integral of |蠄(z)|2 over all space to 1 would give it in terms of kand 伪 , but because we can鈥檛 solve (22) exactly for k(or E), neither can we obtain an exact value for C.

In several bound systems, the quantum-mechanically allowed energies depend on a single quantum number we found in section 5.5 that the energy levels in an infinite well are given by, En=a1n2wheren=1,2,3.....andis a constant. (Actually, we known whata1is but it would only distract us here.) section 5.7 showed that for a harmonic oscillator, they areEn=a2(n鈭�12), wheren=1,2,3.....(using ann鈭�12with n strictly positive is equivalent towith n non negative.) finally, for a hydrogen atom, a bound system that we study in chapter 7,En=鈭�a3n2, wheren=1,2,3.....consider particles making downwards transition between the quantized energy levels, each transition producing a photon, for each of these three systems, is there a minimum photon wavelength? A maximum ? it might be helpful to make sketches of the relative heights of the energy levels in each case.

There are mathematical solutions to the Schr枚dinger equation for the finite well for any energy, and in fact. They can be made smooth everywhere. Guided by A Closer Look: Solving the Finite Well. Show this as follows:

(a) Don't throw out any mathematical solutions. That is in region Il (x<0), assume that (Ce+ax+De-ax), and in region III (x>L), assume that蠄(x)=Fe+ax+Ge-ax. Write the smoothness conditions.

(b) In Section 5.6. the smoothness conditions were combined to eliminate A,Band Gin favor of C. In the remaining equation. Ccanceled. leaving an equation involving only kand 伪, solvable for only certain values of E. Why can't this be done here?

(c) Our solution is smooth. What is still wrong with it physically?

(d) Show that

localid="1660137122940" D=12(B-k伪A)andF=12e-伪L[(A-Bk伪)sin(kL)+(Ak伪+B)cos(kL)]

and that setting these offending coefficients to 0 reproduces quantization condition (5-22).

Calculate the uncertainty in the particle鈥檚 momentum.

Determine the expectation value of the momentum of the particle. Explain.
See all solutions

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