Answer:
[tex]2.84\cdot 10^{-8} m[/tex]
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:
[tex]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[/tex]
The electrons decelerates to 3/4 of its speed, so the new speed is
[tex]v_f = \frac{3}{4}v_i = \frac{3}{4}(5.90\cdot 10^6 m/s)=4.425\cdot 10^6 m/s[/tex]
So the final kinetic energy is
[tex]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[/tex]
So, the energy lost by the electron, which is equal to the energy of the emitted photon, is
[tex]E=K_i - K_f =1.59\cdot 10^{-17} J-8.9\cdot 10^{-18} J=7\cdot 10^{-18} J[/tex]
The wavelength of the photon is related to its energy by
[tex]\lambda=\frac{hc}{E}[/tex]
where h is the Planck constant and c the speed of light. Substituting E, we find
[tex]\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[/tex]
