Question

Iridium has a prominent x-ray emission line at 75.8 keV.

HINT

(a)

What is the minimum speed (in m/s) of an incident electron that could produce this emission line? (Hint: Recall the expression for relativistic kinetic energy given in Topic 26.)

m/s

(b)

What is the wavelength (in m) of a 75.8 keV x-ray photon?

m

Answer 1

a) for the speed of electron needed

kinetic energy of electron = 75.8 KeV

total energy - rest energy = 75.8 *10^3 * e

9.11 *10^-31 * (3 *10^8)^2 * (1/sqrt(1 - (v/(3*10^8))^2) - 1) = 75.8 *10^3 * 1.602 *10^-19

solving for v

v = 1.47*10^8 m/s

the speed of electron is 1.47*10⁸ m/s

b)

for the photon

energy of photon = h * c/wavelength

75.8 *10^3 * 1.602 * 10^-19 = 6.626 *10^-34 * 3 *10^8/wavelength

wavelength = 1.64 *10^-11 m

the wavelength of the x ray is 1.64 *10^-11 m