Question

In Rutherford's experiment the number of $\alpha$ particles scattered through an angle ${60}^o$ is $112$ per minute, then the number of $\alpha$ particles scattered through an angle of ${90}^0$ per minute by the same nucleus is:

• A. $28$ per minute
• B. $112$ per minute
• C. $12.5$ per minute
• D. $7$ per minute

Answer 1

The correct option is A $28$ per minute

In Rutherford's experiment the number of particles scattered through an angle $\theta$ per minute is given by:
$N\propto \frac{Z^2}{Si{n}^4\left(\theta /2\right)}$

According to the question: the number of particles scattered through an angle ${60}^o$ is 112 per minute. So,
$112\propto \frac{Z^2}{Si{n}^4\left(60/2\right)}$------ (1)

And, lets assume the number of particles scattered through an angle ${90}^o$ is N' per minute
$N^{\prime }\propto \frac{Z^2}{Si{n}^4\left(90/2\right)}$------(2)

Using (1) and (2),
$\frac{N^{\prime }}{112}\propto \frac{Si{n}^4\left(60/2\right)}{Si{n}^4\left(90/2\right)}$

$N^{\prime }=28$