de Broglie Wavelength Of An Electron As Determinant Of Velocity Of An Electron
TECTechnics Classroom   TECTechnics Overview

de Broglie Wavelength Of An Electron As Determinant Of Velocity Of An Electron

Assume the particle illustrated in Figure 22.13 is an electron moving with velocity v:
(a) How fast would the electron be moving in order to have a wavelength of 0.711 Å?
(b) Under certain conditions, the element molybdenum emits light with characteristic wavelength of 0.711 Å. What region of the electromagnetic spectrum do the emitted light belong?
(c) Name an important use for the light emitted by molybdenum.

The strings: S7P5A51 (Change - Physical Change)

The math:
Pj Problem of Interest is of type change (physical change). Velocity problems are change problems because it measures rate of change of distance with respect to time.

(a) The de Broglie equation that relates the wavelength of a particle with the velocity of a particle is:
λ = h/(mv) --------------(1)
Where λ is wavelength in meters; h is Plank's constant in Joules-sec; m is mass of particle in kilogram; v is velocity of particle in meter/sec.
So, from equation (1) we have:
v = h/(λm)
Plank's constant, h = 6.626 x 10-34 J-s = 6.626 x 10-34 kg-m2/s2;
Mass of electron = 9.109 x 10-28 g = 9.109 x 10-28 /103 kg.
Wavelength, λ = 0.711 Å = 0.711 x 10-10 m.
So, v = (6.626 x 10-34) /[(0.711 x 10-10) x (9.109 x 10-28 /103)
So, v = (6.626 x 10-34)/(6.476 x 10-41)
So, v = (6.626/6.476) x 10-34 x 1041
So, v = 1.02 x 107 m/sec.
So, speed of electron will be 1.02 x 107 m/sec, if it is to have a wavelength of 0.711 Å.

(b) If the conditions are appropriate, molybdenum emits X rays.

(c) X rays emitted by molybdenum are used in diffraction experiments to determine the structure of molecules.

Blessed are they that have not seen, and yet have believed. John 20:29