/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q39P Calculate the ratio of the wavel... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Calculate the ratio of the wavelength of theKα line for niobium (Nb) to that for gallium (Ga) .Take needed data from the periodic table of Appendix G.

Short Answer

Expert verified

The ratio of the wavelength of the Kα line for niobium (Nb) to that for gallium (Ga) is 0.563.

Step by step solution

01

The given data

The periodic table of Appendix G is given with elements niobium and gallium.

02

Understanding the concept of wavelength:

Moseley's law states that "the square root of the frequency of an atom's emitted x-ray is proportionate to its atomic number."

Formula:

According to the Moseley’s law, we can get the relation of frequency or wavelength to atomic number as:

f∞Z-12OR cλ(Z-1)2 ….. (1)

03

Calculation of the ratio of the wavelengths:

The atomic number of Niobium, ZNb=41

The atomic number of Gallium, ZGa=31

Using the relation of wavelength and atomic number, the ratio of the wavelengths of the line of the elements niobium and gallium is as follows:

λNbλGa=ZGa-12ZNb-12

Substitute known values in the above equation.

λNbλGa=31-1241-12=302402=916=0.563

Hence, the value of the required ratio of the wavelengths is 0.563 .

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The x-ray spectrum of Fig. 40-13 is for 35.0 keV electrons striking a molybdenum ( Z = 42 ) target. If you substitute a silver ( Z = 47 ) target for the molybdenum target, will

(a) λmin,

(b) the wavelength for therole="math" localid="1661495146456" Kα line, and

(c) the wavelength for theKβ line increase, decrease, or remain unchanged?

What is the wavelength associated with a photon that will induce a transition of an electron spin from parallel to anti-parallel orientation in a magnetic field of magnitude 0.200 T? Assume that l=0.

In Fig. 40-13, the x-rays shown are produced when 35.0 keV electrons strike a molybdenum (Z = 42) target. If the accelerating potential is maintained at this value but a silver (Z = 47) target is used instead, what values of (a)λmin, (b) the wavelength of the Kαline, and (c) the wavelength of the Kβ line result? The K,L and M atomic x-ray levels for silver (compare Fig. 40-15) are 25.51, 3.56 and 0.53 keV.

A tungsten (Z=74) target is bombarded by electrons in an x-ray tube. The K,L and M energy levels for tungsten (compare Fig. 40-15) have the energies 69.5 keV,11.3 keV, and 2.30 keV respectively. (a) What is the minimum value of the accelerating potential that will permit the production of the characteristickα andkβ lines of tungsten? (b) For this same accelerating potential, what is λmin? What are the (c) kαand (d)kβ wavelengths?

Two of the three electrons in a lithium atom have quantum numbers (n,I,mI,ms)of (1,0,0,+12)and (1,0,0,-12). What quantum numbers are possible for the third electron if the atom is (a) in the ground state and (b) in the first excited state?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.