Question

In Rutherford experiments of $\alpha -ray$ scattering the number of particles scattered at ${90}^0$ be $28$ per minute$.$ Then the number of particles scattered per minute by the same foil but at ${60}^0$ are

• A. $56$
• B. $112$
• C. $60$
• D. $120$

Answer 1

Answer: (B)

$112$

No. of particles scattered at angle is

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

Given:- the number of particles scattered at ${90}^o$ be $28$ per minute

$\Rightarrow$ $\frac{N_1}{N_2}=\frac{si{n}^4\left(\frac{{\theta }_2}{2}\right)}{si{n}^4\left(\frac{{\theta }_1}{2}\right)}$

$\Rightarrow$ $\frac{28}{N_2}=\frac{si{n}^4\left(\frac{{60}^0}{2}\right)}{si{n}^4\left(\frac{{90}^o}{2}\right)}$

$\Rightarrow$ $N_2=112$