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Question

# While being X-rayed, a person absorbs $3.2 \times 10^{-3} \mathrm{J}$ of energy. Determine the number of 40,000-eV X-ray photons absorbed during the exam.

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The energy of the X-ray photon is given as

\begin{align*} E_{ph} & = 40,000\ \mathrm{eV} \\ & = (40,000) \cdot \left(1.6\times 10^{-19}\right) \ \ \mathrm{J} \\ & = 6.4\times 10^{-15} \ \mathrm{J} \end{align*}

The energy absorbed during the the X-ray exam is given as

\begin{align*} E_{ab} & = 3.2\times 10^{-3} \ \mathrm{J} \end{align*}

Therefore, the number of absorbed photon are estimated as

\begin{align*} E_{ab} & = n \cdot E_{ph} \\ n & = \dfrac{E_{ab}}{E_{ph}} \\ & = \dfrac{(3.2\times 10^{-3})}{6.4\times 10^{-15} } \\ & = 5.0 \times 10^{11} \ \mathrm{photons} \\ & \hspace*{-3mm} \boxed{n = 5.0 \times 10^{11} \ \mathrm{photons} } \end{align*}

Therefore, $\displaystyle {n = 5.0 \times 10^{11} \ \mathrm{photons} }$ were absorbed during the said X-ray exam.

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