An X-ray tube is operated at 1.24 million volts. The shortest wavelength of the produced photon will be:

CONCEPT:

• The energy of the given X-ray tube is equal to

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

         The minimum wavelength is defined as the shortest wavelength as it is denoted as \(\lambda_{min}\), therefore

           \(E = \frac{hc}{\lambda_{min}}\)

Here we have \(\lambda _{min}\) as the minimum wavelength, h is the Planck's constant, and c is the velocity.

CALCULATION:

As we know, the energy of the photon is written as;

\(E = \frac{hc}{\lambda_{min}}\) ----(1)

Also, the energy is written as;

E = eV -----(2)

Here V is the voltage applied and e is the charge of an electron.

Now, from equation (1) and equation (2) we have minimum wavelength,

\(\lambda_{min}=\frac{hc}{eV}\)

⇒ \(\lambda_{min}=\frac{6.626\times 10^{-34}\times 3 \times 10^{8}}{1.60 \times 10^{-19}\times 1.24 \times 10^{6}}\)

⇒\(\lambda_{min}=10^{-3} nm\)

Hence, option 3) is the correct answer.