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  • The differential scattering cross section is given by <img alt="\mathrm{D(\theta)=a^2+b^2 \cos ^2 \theta}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BD%28%5Ctheta%29%3Da%5E2a

The differential scattering cross section is given by \mathrm{D(\theta)=a^2+b^2 \cos ^2 \theta} if the intensity of incoming particle is I what will be intensity of scatteried particle.

Option: 1

\mathrm{4 \pi\left(a^2+\frac{b^2}{3}\right)}


Option: 2

\mathrm{4 \pi\left(\frac{a^2+b^2}{3}\right)}


Option: 3

\mathrm{4 \pi I\left(a^2+\frac{b^2}{3}\right)}


Option: 4

\mathrm{4 \pi I\left(a^2-\frac{b^2}{3}\right)}


Answers (1)

best_answer

We know that - 
\mathrm{D(\theta)=\frac{b}{\sin \theta}\left|\frac{d b}{d \theta}\right|}
& \mathrm{\sigma=\iint D(\theta) \sin\theta\ d \theta\ d \phi}
\mathrm{\therefore \quad \sigma=\iint\left(a^2+b^2 \cos ^2 \theta\right)\ \sin \theta\ d \theta\ d \phi}
\mathrm{=\int a^2 \sin \theta\ d \theta\ d \phi +\int_{0}^{\pi} b^2 \cos ^2 \theta \sin \theta\ d \theta\ d \phi}
\mathrm{\begin{aligned} & =4 \pi a^2+2 \pi b^2 \times \frac{2}{3} \\ & \sigma=4 \pi a^2+4 \pi \frac{b^2}{3} \end{aligned}}
Intensity is given by \mathrm{=4\pi\ I \left ( a^{2}+\frac{b^{2}}{3} \right )}
 

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Gunjita

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