Question

(a) What is the shortest-wavelength x-ray radiation that can be generated in an x-ray tube with an applied voltage of 50.0kV ?

(b) Calculate the photon energy in eV.

(c) Explain the relationship of the photon energy to the applied voltage.

Short answer

a) The wavelength with the smallest length is computed at 2.48165 x 10^-11m.

b)The photon energy is measured in electron volts eV is 50 keV.

c)The photon energy is proportional to applied voltage E = qV.

Step-by-step solution

Determine the formulas:

Consider the formula for the minimum wavelength is:

\[\begin{array}{l}{\bf{h}}{{\bf{\nu }}_{{\bf{max}}}}{\bf{ = }}\frac{{{\bf{hc}}}}{{{{\bf{\lambda }}_{{\bf{min}}}}}}\\{{\bf{\lambda }}_{{\bf{min}}}}{\bf{ = }}\frac{{{\bf{hc}}}}{{{\bf{eV}}}}\end{array}\]

Here, e is the charge on electron, V is theapplied voltage:

Determine the values of the wavelength:

(a)

Substitute the values and solve for the wavelength as:

\[\begin{array}{c}{\lambda _{\min }} = \frac{{\left( {6.626 \times {{10}^{ - 34}}J \cdot s} \right)\left( {3 \times {{10}^3}\;\;\frac{{\rm{m}}}{s}} \right)}}{{\left( {1.602 \times {{10}^{ - 19}}{\rm{C}}} \right)\left( {50.0 \times {{10}^3}\;{\rm{V}}} \right)}}\\{\lambda _{\min }} = 2.48165 \times {10^{ - 11}}\;{\rm{m}}\end{array}\]

As a result, the calculated wavelength with the shortest length is2.48165 x 10^-11m.

Calculate the photon energy in eV

b)

Calculating to find energy in photon.

Photon energy is given:

\[E = qV\]

Here,E is the photon energy, q is the charge and V is the voltage.

Consider the values as:

\[\begin{array}{l}q = 1.602 \times {10^{ - 19}}{\rm{C}}\\V = 50.0 \times {10^3}\;{\rm{V}}\end{array}\]

The photon energy is calculated as

\[\begin{array}{c}E = \left( {1.602 \times {{10}^{ - 19}}{\rm{C}}} \right)\left( {50.0 \times {{10}^3}\;{\rm{V}}} \right)\\E = 8.01 \times {10^{ - 15}}\;{\rm{J}}\\E = 8.01 \times {10^{ - 15}}\;{\rm{J}} \times \frac{{{\rm{leV}}}}{{1.602 \times {{10}^{ - {1_J}}}}}\\E = 50\;{\rm{keV}}\end{array}\]

As a result, the photon energy is measured in eV is 50 keV.

Step 4: Explaining the relationship of the photon energy to the applied voltage.

(c)

The relation between the photon energy and applied voltage is given\[E = qV\].

Photon energy and applied voltage are related as follows

\[E = qV\]

Here,

\[E = \]Photon energy

\[\begin{array}{l}q = {\rm{Charge }}\\V = {\rm{Voltage}}\end{array}\]

Hence, photon energy is related to applied voltage as E = qV.