Question

# Gamma rays of photon energy $$0.511 \mathrm{MeV}$$ are directed onto an aluminum target and are scattered in various directions by loosely bound electrons there. (a) What is the wavelength of the incident gamma rays?(b) What is the wavelength of gamma rays scattered at $$90.0^{\circ}$$ to the incident beam? (c) What is the photon energy of the rays scattered in this direction?

Solution

## (a) Using the value $$h c=1240 \mathrm{eV} \cdot \mathrm{nm}$$, we find$$\lambda=\dfrac{h c}{E}=\dfrac{1240 \mathrm{nm} \cdot \mathrm{eV}}{0.511 \mathrm{MeV}}=2.43 \times 10^{-3} \mathrm{nm}=2.43 \mathrm{pm}\\$$(b) Now, Eq. 38-11 leads to\begin{aligned}\lambda^{\prime} &=\lambda+\Delta \lambda=\lambda+\dfrac{h}{m_{e} C}(1-\cos \phi)=2.43 \mathrm{pm}+(2.43 \mathrm{pm})\left(1-\cos 90.0^{\circ}\right) \\&=4.86 \mathrm{pm}\end{aligned}\\(c) The scattered photons have energy equal to$$E^{\prime}=E\left(\dfrac{\lambda}{\lambda^{\prime}}\right)=(0.511 \mathrm{MeV})\left(\dfrac{2.43 \mathrm{pm}}{4.86 \mathrm{pm}}\right)=0.255 \mathrm{MeV}\\$$PhysicsNCERTStandard XII

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