Question

An electron, traveling at a speed of 5.90 × 10 6 5.90×106 m/s, strikes the target of an X-ray tube. Upon impact, the electron decelerates to three-quarters of its original speed, with an X-ray photon being emitted in the process. What is the wavelength of the photon?

Answer 1

Answer:

$2.84\cdot 10^{-8} m$

Explanation:

Due to the law of conservation of energy, the energy of the emitted X-ray photon is equal to the energy lost by the electron.

The initial kinetic energy of the electron is:

$K_i = \frac{1}{2}mv_i^2 = \frac{1}{2}(9.11\cdot 10^{-31}kg)(5.90\cdot 10^6 m/s)^2=1.59\cdot 10^{-17}J$

The electrons decelerates to 3/4 of its speed, so the new speed is

$v_f = \frac{3}{4}v_i = \frac{3}{4}(5.90\cdot 10^6 m/s)=4.425\cdot 10^6 m/s$

So the final kinetic energy is

$K_f = \frac{1}{2}mv_f^2=\frac{1}{2}(9.11 \cdot 10^{-31} kg)(4.425\cdot 10^6 m/s)^2=8.9\cdot 10^{-18} J$

So, the energy lost by the electron, which is equal to the energy of the emitted photon, is

$E=K_i - K_f =1.59\cdot 10^{-17} J-8.9\cdot 10^{-18} J=7\cdot 10^{-18} J$

The wavelength of the photon is related to its energy by

$\lambda=\frac{hc}{E}$

where h is the Planck constant and c the speed of light. Substituting E, we find

$\lambda=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{7\cdot 10^{-18} J}=2.84\cdot 10^{-8} m$