$(a)$ An $X$-ray tube produces a continuous spectrum of radiation with its short wavelength end at $0.45\,\mathring{A}$. What is the maximum energy of a photon in the radiation?
$(b)$ From your answer to $(a)$,guess what order of accelerating voltage (for electrons) is required in such a tube?

(N/A) The wavelength produced by an $X$-ray tube is $\lambda = 0.45\,\mathring{A} = 0.45 \times 10^{-10}\,\text{m}$.
Planck's constant,$h = 6.626 \times 10^{-34}\,\text{J s}$.
Speed of light,$c = 3 \times 10^{8}\,\text{m/s}$.
The maximum energy of a photon is given by $E = \frac{hc}{\lambda}$.
Converting to electron-volts $(eV)$: $E = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{0.45 \times 10^{-10} \times 1.602 \times 10^{-19}}\,\text{eV} \approx 27.6 \times 10^{3}\,\text{eV} = 27.6\,\text{keV}$.
Therefore,the maximum energy of an $X$-ray photon is $27.6\,\text{keV}$.
$(b)$ The accelerating voltage provides kinetic energy to the electrons to produce $X$-rays. To obtain an $X$-ray photon of $27.6\,\text{keV}$,the incident electrons must possess at least $27.6\,\text{keV}$ of kinetic energy.
Hence,an accelerating voltage of the order of $30\,\text{kV}$ is required for producing these $X$-rays.