Question

If $2000\alpha -$particles scattered at an angle of ${90}^{\circ }$ targeted at an atom, then the number of $\alpha -$particles scattering at ${60}^{\circ }$ will be -

• A. $2000$
• B. $4000$
• C. $6000$
• D. $8000$

Answer 1

The correct option is D $8000$

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,

$2000=\frac{K{Z}^2}{{\sin }^4\left({90}^{\circ }/2\right)}...\left(1\right)$

$N\left(\theta \right)=\frac{K{Z}^2}{{\sin }^4\left({60}^{\circ }/2\right)}...\left(2\right)$

From equations $\left(1\right)$ and $\left(2\right)$, we get,

$N\left(\theta \right)=8000$

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