MCQ
An $X-$ray machine has an accelerating potential difference of $25,000$ volts. By calculation the shortest wavelength will be obtained as $\left(\mathrm{h}=6.63 \times 10^{-34} \mathrm{~J}-\mathrm{sec} ; \mathrm{e}=1.6 \times 10^{-19}\right.$ coulomb $)$
- A$0.25 \mathring A$
- ✓$0.50 \mathring A$
- C$1.00 \mathring A$
- D$2.50 \mathring A$
