Question

In Rutherford experiment the number of alpha-particle scattered through an angle of 60 is 12 per minute,the find the number of alpha particle scattered through an angle of 90 per minute by the same nucleus.
quick plzzzzzz...............

Answer 1

according to Rutherford scattering experiment, $N(\theta)\propto\frac{Z^2}{sin^4\left(\frac{\theta}{2}\right)}$

where $N(\theta)$ denotes number of alpha - particle scattered through an angle of $\theta$.

case1: $N(\theta)$ = 12, $(\theta)$ = 60°

so, $12=k\frac{Z^2}{sin^4(30^{\circ})}$

case2 : $(\theta)$=90°

so, $N(\theta)=k\frac{Z^2}{sin^4(45^{\circ})}$

from case 1 and 2 ,

$\frac{12}{N(\theta)}=\frac{sin^4(45)}{sin^4(30)}$

or, 12/N = (1/√2)⁴/(1/2)⁴ = (√2)⁴

or, N = 3

hence, number of alpha particles scattered through an angle of 90° per minute by the same nucleus is 3

Answer 2

Answer:

3

Explanation:

N1/N2=[sin^4 teta2/2]/sin^4teta1/2

        =[sin^4 90/2] /[sin^4 60/2]

          =sin^4 45degrees /sin^4 30 degrees

          =[1/4]/[1/16]

12/N2= 4

N2=3