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Q16E

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Modern Physics
Found in: Page 134
Modern Physics

Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087

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Short Answer

A brag diffraction experiment is conducted using a beam of electrons accelerated through a1.0 kV potential difference. (a) If the spacing between atomic planes in the crystal is 0.1 nm , at what angles with respect to the planes will diffraction maximum be observed? (b) If a beam of X-rays products diffraction maxima at the same angles as the electron beam, what is theX -ray photon energy?

(a) The angles for the maximum is

θ1=11.2°θ2=22.83°θ3=35.59° θ4=50.89° θ5=75.93°

(b) The photon energy of the x-ray is

Ep=31.88 keV

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Formula of Bragg’s Law.

So, we'll use the relationship shown in Example 4.3directly, linking the particle's wavelength (λ)to the potential for rapid growth (V). can use Bragg's law to find the equivalent values of once have the wavelength, such that,

λ=h22mqV………………..(1)

Step 2: Calculation of wavelength(a)

Substitute the given data in equation (1), and we get,

λ=(6.63×10-34 J⋅s)22×9.1×10-31 kg×1.6×10-19 C×1000 V

λ=1.5×10-21 m2=3.89×10-11 m

2dsin(θ)=nλsin(θ)=(λ2d)nsin(θ)=(3.89×10-11 m2×0.1×10-9 m)×nθ=sin-1(0.194n)

θ1=11.2°θ2=22.83°θ3=35.59° θ4=50.89° θ5=75.93°

Keep in mind that the value ofn=6 isn't allowed because arcsine's parameter exceeds1 , but used the integer n instead of m for the integer counter so it doesn't get mixed up with the unit of measurement (m or meter).

Step 3:Wavelength of Electrons and Photons.

(b)

To achieve the same maxima, both the electron and the photons must have the same wavelength value.

Ep=hcλ

Step 4: Substitution.

Ep=(4.14×10−23 eV⋅s×3×108 m/s3.89×10-11 m) =31.88 keV

Therefore, the energy of the photon isEp=31.88 keV .

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