Question
When photons pass through matter, the intensity $I$ of the beam (measured in watts per square meter) decreases exponentially according to$$I=I_{0} e^{-\mu x}$$where $I$ is the intensity of the beam that just passed through a thickness $x$ of material and $I_{0}$ is the intensity of the incident beam. The constant $\mu$ is known as the linear absorption coefficient, and its value depends on the absorbing material and the wavelength of the photon beam. This wavelength (or energy) dependence allows us to filter out unwanted wavelengths from a broad-spectrum $\mathrm{x}-$ ray beam. (a) Two x-ray beams of wavelengths $\lambda_{1}$ and $\lambda_{2}$ and equal incident intensities pass through the same metal plate. Show that the ratio of the emergent beam intensities is$$\frac{I_{2}}{I_{1}}=e^{-\left(\mu_{2}-\mu_{1}\right) x}$$(b) Compute the ratio of intensities emerging from an aluminum plate 1.00 $\mathrm{mm}$ thick if the incident beam contains equal intensities of 50 $\mathrm{pm}$ and 100 $\mathrm{pm}$ x-rays. The values of $\mu$ for aluminum at these two wavelengths are $\mu_{1}=5.4 \mathrm{cm}^{-1}$ at 50 $\mathrm{pm}$ and $\mu_{2}=41.0 \mathrm{cm}^{-1}$ at $100 \mathrm{pm} .$ (c) Repeat part $(\mathrm{b})$ for an aluminum plate 10.0 $\mathrm{mm}$ thick.
Step 1
This is given by the equation $I=I_{0} e^{-\mu x}$, where $I$ is the intensity of the beam after passing through a thickness $x$ of material, $I_{0}$ is the initial intensity of the beam, and $\mu$ is the linear absorption coefficient. Show more…
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When photons pass through matter, the intensity $I$ of the beam (measured in watts per square meter) decreases exponentially according to $$ I=I_{0} e^{-\mu x} $$ where $I$ is the intensity of the beam that just passed through a thickness $x$ of material and $I_{0}$ is the intensity of the incident beam. The constant $\mu$ is known as the linear absorption coefficient, and its value depends on the absorbing material and the wavelength of the photon beam. This wavelength (or energy) dependence allows us to filter out unwanted wavelengths from a broad-spectrum x-ray beam. (a) Two x-ray beams of wavelengths $\lambda_{1}$ and $\lambda_{2}$ and equal incident intensities pass through the same metal plate. Show that the ratio of the emergent beam intensities is $$ \frac{I_{2}}{I_{1}}=e^{-\left(\mu_{2}-\mu_{1}\right) x} $$ (b) Compute the ratio of intensities emerging from an aluminum plate 1.00 $\mathrm{mm}$ thick if the incident beam contains equal intensities of 50 $\mathrm{pm}$ and 100 $\mathrm{pm}$ x-rays. The values of $\mu$ for aluminum at these two wavelengths are $\mu_{1}=5.40 \mathrm{cm}^{-1}$ at 50 $\mathrm{pm}$ and $\mu_{2}=41.0 \mathrm{cm}^{-1}$ at 100 $\mathrm{pm}$ . (c) Repeat part (b) for an aluminum plate 10.0 $\mathrm{mm}$ thick.
When photons pass through matter, the intensity $I$ of the beam (measured in watts per square meter) decreases exponentially according to $$I=I_{0} e^{-\mu x}$$ where $I$ is the intensity of the beam that just passed through a thickness $x$ of material and $I_{0}$ is the intensity of the incident beam. The constant $\mu$ is known as the linear absorption coefficient, and its value depends on the absorbing material and the wavelength of the photon beam. This wavelength (or energy) dependence allows us to filter out unwanted wavelengths from a broad-spectrum x-ray beam. (a) Two x-ray beams of wavelengths $\lambda_{1}$ and $\lambda_{2}$ and equal incident intensities pass through the same metal plate. Show that the ratio of the emergent beam intensities is $$\frac{I_{2}}{I_{1}}=e^{-\left(\mu_{y}-\mu_{1}\right) x}$$ (b) Compute the ratio of intensities emerging from an aluminum plate $1.00 \mathrm{~mm}$ thick if the incident beam contains equal intensities of $50 \mathrm{pm}$ and $100 \mathrm{pm} \mathrm{x}$ -rays. The values of $\mu$ for aluminum at these two wavelengths are $\mu_{1}=5.40 \mathrm{~cm}^{-1}$ at $50 \mathrm{pm}$ and $\mu_{2}=41.0 \mathrm{~cm}^{-1}$ at $100 \mathrm{pm} .$ (c) Repeat part (b) for an aluminum plate $10.0 \mathrm{~mm}$ thick.
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