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

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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**:
S_{7}P_{5}A_{51} (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-m^{2}/s^{2};

Mass of electron = 9.109 x 10^{-28} g = 9.109 x 10^{-28} /10^{3} 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} /10^{3})

So, v = (6.626 x 10^{-34})/(6.476 x 10^{-41})

So, v = (6.626/6.476) x 10^{-34} x 10^{41}

So, v = 1.02 x 10^{7} m/sec.

So, speed of electron will be 1.02 x 10^{7} 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.

Math

The *point* **.** is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.

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