Q158 / 160 · AP EAMCET 2008 · Physics

An X-ray tube produces a continuous spectrum of radiation with its shortest wavelength of $45 \times 10^{-2} Ã…$. The maximum energy of a photon in the radiation in $\mathrm{eV}$ is $\left(h=6.62 \times 10^{-34} \mathrm{~J}-\mathrm{s}, c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)$

  • Correct answer
    A
    27,500
  • B
    22,500
  • C
    17,500
  • D
    12,500

Solution

$\begin{aligned} E & =\frac{h c}{\lambda} \\ & =\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{45 \times 10^{-12}} \\ & =\frac{0.44 \times 10^{-14}}{1.6 \times 10^{-19}} \\ & =0.275 \times 10^5 \mathrm{eV}\end{aligned}$

Hence, the correct answer is option: 'A'.

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