Question

The distance between the cathode (filament) and the target in an X-ray tube is $15\text{m}$. If the cut-off wavelength is $30\text{pm}$, find the electric field between the cathode and the target.$(hc=1.24\times {10}^{-6}\text{eV})$

• A. $27.6\times {10}^{-19}\text{keV/m}$
• B. $27.6\text{keV/m}$
• C. $27.6\text{kV/m}$
• D. $27.6\times {10}^3\text{keV/m}$

Answer 1

The correct option is B $27.6\text{keV/m}$

Given, $d=1.5\text{m},\lambda =30\text{pm}=30\times {10}^{-12}\text{nm}$

$E=\frac{hc}{\lambda }=\frac{1.24\times {10}^{-6}}{30\times {10}^{-12}}=4.14\times {10}^4\text{eV}$

Electric field,

$E=\frac{V}{d}=\frac{4.14\times {10}^4}{1.5}=2.76\times {10}^4\text{eV/m}=27.6\text{keV/m}$

Hence, $\left(B\right)$ is the correct answer.

Why this question? Caution: Be careful with unit, as the answer is in $\text{keV/m}$ but most of the students mark the answer in $\text{kV/m}$.