Question

At which of the following angles, the number of $\alpha -$particles scattering will be minimum when it is targeted at an atom?

• A. ${30}^{\circ }$
• B. ${50}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is D ${90}^{\circ }$

We know that, the number of $\alpha -$particles scattering at angle $\theta$ is given by,

$N\left(\theta \right)=\frac{K{Z}^2}{{\sin }^4\left(\theta /2\right)}$

So,

$N\left(\theta \right)\propto \frac{1}{{\sin }^4\left(\theta /2\right)}$

Value of $\sin \left(\theta /2\right)$ increases as $\theta /2$ increases for $0^{\circ }\le \theta /2\le {90}^{\circ }$.

Therefore, $N\left(\theta \right)$ will be minimum for maximum value of $\sin \left(\theta /2\right)$ which will be maximum for maximum value of $\theta /2$.

Hence, option $\left(D\right)$ is the correct answer.