MCQ
An $X$-ray machine has an accelerating potential difference of $25,000$ volts. By calculation the shortest wavelength will be obtained as $\left(h=6.62 \times 10^{-34} J-s e c ; \mathrm{e}=1.6 \times 10^{-19}\right.$ coulomb $)$
  • A
    $0.25 \mathring A$
  • $0.50 \mathring A$
  • C
    $1.00 \mathring A$
  • D
    $2.50 \mathring A$

Answer

Correct option: B.
$0.50 \mathring A$
$\lambda_{\text {min }}=\frac{h c}{e V}=\frac{12375}{V} \mathring A=0.495 \mathring A \ \approx 0.5 \mathring A$

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