What is the accelerating voltage of an x-ray tube that produces x rays with a shortest wavelength of \({\rm{0}}{\rm{.0103 nm}}\)?

A sinusoidal wave's frequency is defined as the number of complete oscillations made by any wave constituent per unit of time. One may comprehend that if a body is in periodic motion, it has completed one cycle after passing through a series of events or locations and returning to its original condition using the notion of frequency. As a result frequency is a quantity that describes the rate at which oscillations and vibrations occur.

The distance between the identical places between two successive waves is determined when the wavelength of a wave is measured.

The energy of x ray photons with a wavelength of \({\rm{\lambda = 0}}{\rm{.0103 nm}}\) is –

\(E = \frac{{hc}}{\lambda }\)

Substitute the values and solve as:

\(\begin{align}E &= \frac{{4.14 \times {{10}^{ - 15}}\;eV \cdot s \times 3 \cdot {{10}^8}\;\frac{m}{s}}}{{0.0103 \cdot {{10}^{ - 9}}\;\;m}}\\ &= 1.21 \cdot {10^5}\;eV\end{align}\)

Write the value of the energy in joules as:

\({\rm{E = 1}}{\rm{.94}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 14}}}}{\rm{\;J}}\)

The accelerating voltage of an x-ray tube that produces such photons is as follows:

\(V = \frac{E}{q}\)

Substitute the values and solve as:

\(\begin{align}V &= \frac{{1.94 \cdot {{10}^{ - 14}}\;J}}{{1.6 \cdot {{10}^{ - 19}}C}}\\ &= 1.21 \cdot {10^5}\;V\end{align}\)

Therefore, the value of voltage is obtained as \({\rm{1}}{\rm{.21}} \times {\rm{1}}{{\rm{0}}^{\rm{5}}}{\rm{\;V}}\).