Ratio of energy difference between the ground state of helium to that of hydrogen. Ratio of energy difference between the ground state of sodium to that of hydrogen:

Question

Want to see the full answer?

Answer 1

Question

Chapter 10, Problem 18P

To determine

Ratio of energy difference between the ground state of helium to that of hydrogen.

Ratio of energy difference between the ground state of sodium to that of hydrogen:

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Photons released by nuclear decays tend to be in the MeV range, and atomic nuclei are a few femtometers (10-15 m) across. If a single proton trapped in an inescapable rectangular box releases a 1.3 MeV photon when dropping from the n = 2 to the n = 1 state, how wide is the box, in femtometers? You should find that this quick and dirty estimate is remarkably close to the real size of a nucleus! The proton mass is about 1.7 x 10-27 kg. 1 MeV = 1.6 x 10-13 J. Planck's constant is approximately h = 6.6 x 10-34 J s.

(b) Calculate the half width in nanometers for Doppler broadening of the 4s S 4p transition for atomic nickel at 361.939 nm (3619.39 Å) at a temperature of 20,000 K in both wavelength and frequency units. (e) Calculate the speed that an iron atom undergoing the 4s S 4p transition at 385.9911 nm (3859.911 Å) would have if the resulting line appeared at the rest wavelength for the same transition in nickel. (f) Compute the fraction of a sample of iron atoms at 10,000 K that would have the velocity calculatedin (e). (g) Create a spreadsheet to calculate the Doppler half width DlD in nanometers for the nickel and iron lines cited in (b) and (e) from 3000–10,000 K. (h) Consult the paper by Gornushkin et al. (note 10) and list the four sources of pressure broadening that they describe. Explain in detail how two of these sources originate in sample atoms.

7.129 A technique called photoelectron spectroscopy is usedto measure the ionization energy of atoms. A gaseoussample is irradiated with UV light, and electrons areejected from the valence shell. The kinetic energies ofthe ejected electrons are measured. Because the energyof the UV photon and the kinetic energy of the ejectedelectron are known, we can write where ν is the frequency of the UV light, and m and uare the mass and velocity of the electron, respectively.In one experiment the kinetic energy of the ejectedelectron from potassium is found to be 5.34 × 10−19 Jusing a UV source of wavelength 162 nm. Calculate theionization energy of potassium. How can you be surethat this ionization energy corresponds to the electron inthe valence shell (i.e., the most loosely held electron)?