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Chapter 41: Q. 29 (page 1207)

A hydrogen atom in its fourth excited state emits a photon with a wavelength of 1282nm. What is the atom’s maximum possible orbital angular momentum (as a multiple of h) after the emission?

Short Answer

Expert verified

Maximum possible orbital angular momentum is 3h2π.

Step by step solution

01

:  Given information

Fourth excited state means n=5

Wavelength, λ=1282nm=1282×10-9m

02

:  Simplification

Using Formula , 1λ=Rz21x2-1n2, where λis the wavelength , Ris rydberg constant , xis the final orbital of electron , nis the initial orbital of electron , zis atomic no

Now for hydrogen z=1

Now , Putting all the values in the formula we get 11282×10-9=1.01×1071x2-152

Solving it we get x=3

It means n=5⇒n=3transition

Now , we know that orbital angular momentum is nh2π

Now after emmision we haven=3

So, the maximum angular momentum comes out to be 3h2π

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Most popular questions from this chapter

a. Draw a diagram similar to Figure 41.2 to show all the possible orientations of the angular momentum vector L→for the case l=3. Label each L→with the appropriate value of m.

b. What is the minimum angle between L→and the z-axis?

An electron accelerates through a 12.5V potential difference, starting from rest, and then collides with a hydrogen atom, exciting the atom to the highest energy level allowed. List all the possible quantum-jump transitions by which the excited atom could emit a photon and the wavelength (in nm) of each.

FIGUREP41.41shows the first few energy levels of the lithium atom. Make a table showing all the allowed transitions in the emission spectrum. For each transition, indicate

a. The wavelength, in nm.

b. Whether the transition is in the infrared, the visible, or the ultraviolet spectral region.

c. Whether or not the transition would be observed in the lithium absorption spectrum.

Does each of the configurations in FIGURE Q41.6represent a possible electron configuration of an element? If so,

(i) identify the element and

(ii) determine whether this is the ground state or an excited state. If not, why not?

The 1997 Nobel Prize in physics went to Steven Chu, Claude Cohen-Tannoudji, and William Phillips for their development of techniques to slow, stop, and “trap” atoms with laser light. To see how this works, consider a beam of rubidium atoms (mass 1.4x10-25kg) traveling at 500m/safter being evaporated out of an oven. A laser beam with a wavelength of 780nmis directed against the atoms. This is the wavelength of the 5s→5ptransition in rubidium, with 5sbeing the ground state, so the photons in the laser beam are easily absorbed by the atoms. After an average time of 15ns, an excited atom spontaneously emits a 780-nm-wavelength photon and returns to the ground state.

a. The energy-momentum-mass relationship of Einstein’s theory of relativity is E2=p2c2+m2c4. A photon is massless, so the momentum of a photon is p=Ephoton/c. Assume that the atoms are traveling in the positive x-direction and the laser beam in the negative x-direction. What is the initial momentum of an atom leaving the oven? What is the momentum of a photon of light?

b. The total momentum of the atom and the photon must be conserved in the absorption processes. As a consequence, how many photons must be absorbed to bring the atom to a halt?

NOTE Momentum is also conserved in the emission processes. However, spontaneously emitted photons are emitted in random directions. Averaged over many absorption/emission cycles, the net recoil of the atom due to emission is zero and can be ignored.

c. Assume that the laser beam is so intense that a ground-state atom absorbs a photon instantly. How much time is required to stop the atoms?

d. Use Newton’s second law in the form F=∆p/∆tto calculate the force exerted on the atoms by the photons. From this, calculate the atoms’ acceleration as they slow.

e. Over what distance is the beam of atoms brought to a halt?
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