Question

Calculate the highest frequency X-rays produced by 8.00 · 104 eV electrons. _____ Hz

Answer 1

When an electron stops, it emits a photon with energy equal to the kinetic energy lost by the electron:
$E=\Delta K$
The energy of the photon is $E=hf$ where $h=6.63 \cdot 10^{-34} Js$ is the Planck constant and f is the frequency. Therefore, the maximum frequency of the emitted photon occurs when the loss of kinetic energy is maximum.

The maximum loss of kinetic energy of the electron occurs when the electron stops completely, so it loses all its energy:
$\Delta K = 8\cdot 10^4 eV$
Keeping in mind that $1 eV=1.6 \cdot 10^{-19}J$, we have
$\Delta K = 8 \cdot 10^4 eV \cdot 1.6 \cdot 10^{-19}J/eV = 1.28 \cdot 10^{-14}J$

And so, this corresponds to the energy of the emitted photon, E. Therefore, we can find the maximum frequency of the emitted photon:
$f= \frac{E}{h} = \frac{1.28 \cdot 10^{-14}J}{6.63 \cdot 10^{-34}Js}=1.93 \cdot 10^{19}Hz$