An x ray tube has an applied voltage of 100 kV.

(a) What is the most energetic x-ray photon it can produce? Express your answer in electron volts and joules.

(b) Find the wavelength of such an X–ray.

Consider the formula for the energy of X ray photons as follows:

\[{\bf{E = }}\frac{{{\bf{hc}}}}{{\bf{\lambda }}}\]

Here,

λ = Wavelength

E = Energy of the x-ray photons

h = Planck's constant

c = Speed of light

Consider the formula for the energy in terms of the voltage as follows:

\[{\bf{E = qV}}\]

Here,

q = Charge

V = Voltage applied

a)

Consider the applied voltage of 100 kV.

Substituting the values, energy is calculated as

\[\begin{array}{c}E = \left( {1.602 \times {{10}^{ - 19}}\;{\rm{C}}} \right)\left( {100\;{\rm{kV}} \times \frac{{100\;{\rm{V}}}}{{1\;{\rm{kV}}}}} \right)\\E = 1.602 \times {10^{ - 14}}\;\;{\rm{J}}\\E = 1.602 \times {10^{ - 14}}\;\;{\rm{J}} \times \frac{{1\;{\rm{eV}}}}{{1.6012 \times {{10}^{ - 10}}\;{\rm{J}}}}\\E = {10^5}\;{\rm{eV}}\end{array}\]

Hence, the energy is measured in joules 1.60 x 10^-14 J and the energy in electron volts is 10⁵eV.

b)

Rearrange the formula for the energy in terms of the wavelength as:

\[E = \frac{{hc}}{\lambda }\]

Substituting the values in above equation and solve as:

\[\begin{array}{l}\lambda = \frac{{\left( {6.626 \times {{10}^{ - 34}}J \cdot s} \right)\left( {3 \times {{10}^8}\;\frac{{\rm{m}}}{{\rm{s}}}} \right)}}{{\left( {1.602 \times {{10}^{ - 14}}J} \right)}}\\\lambda = 1.24 \times {10^{ - 11}}\;{\rm{m}}\end{array}\]

Hence,the wavelength of x-ray is 1.24 x 10^-11 m.