Question

In Rutherford’s experiment the number of $\alpha$ particles scattered through an angle ${60^ \circ }$ is $112$ per minute. Find the number of $\alpha$ particles scattered through an angle of ${90^ \circ }$ per minute by the same nucleus?
(A) $28$ per minute.
(B) $112$ per minute
(C) $12.5$ per minute
(D) $7$ per minute

Answer 1

Hint : The given question can be solved by the concepts of Rutherford’s atomic experiments from where he published the scattering formula establishing an inversely proportional relationship between the number of particles scattered and scattering angle.

Formula Used: In this question, the following formulae will be used,
$\Rightarrow N(\theta ) = \dfrac{K}{{{{\sin }^4}\left( {\dfrac{\theta }{2}} \right)}}$ where $N(\theta )$ is the number of $\alpha$ particles scattered through an angle $\theta$ and $K$ is a constant.

Complete step by step answer
Ernest Rutherford in his $\alpha - {\text{particle}}$ scattering experiment shot a beam of $\alpha - {\text{particles}}$ at an extremely thin gold foil. It was observed that a major fraction of the $\alpha - {\text{particles}}$ passed through the sheet without any deflection while some were deflected by both small and large angles (about ${180^ \circ }$ ).
According to Rutherford’s scattering formula, $N(\theta ) \propto \dfrac{1}{{{{\sin }^4}\dfrac{\theta }{2}}}$ where $N(\theta )$ is the number of $\alpha$ particles scattered through an angle $\theta$ where $\theta$ is the scattering angle.
It can be written as, $N(\theta ) = \dfrac{K}{{{{\sin }^4}\left( {\dfrac{\theta }{2}} \right)}}$ where $K$ is a constant.
We know that 112 particles can be scattered through an angle of ${60^ \circ }$ in one minute.
Assigning the values, $N(\theta ) = 112$ and $\theta = {60^ \circ }$ we get the equation,
$\Rightarrow 112 = \dfrac{K}{{{{\sin }^4}\left( {\dfrac{{{{60}^ \circ }}}{2}} \right)}}$
$\Rightarrow \;112 = \dfrac{K}{{{{\sin }^4}({{30}^ \circ })}}$
Simplifying this equation,
$\Rightarrow K = 112 \times {\sin ^4}({30^{^ \circ }})$
We know that, $\sin {30^ \circ } = \dfrac{1}{2}$ . Therefore, ${\sin ^4}({30^ \circ }) = {\left( {\dfrac{1}{2}} \right)^4}$
$\Rightarrow {\sin ^4}{30^ \circ } = \dfrac{1}{{16}}$ .
From here, we get the value of $K$ .
$\Rightarrow K = 112 \times \dfrac{1}{{16}}$
$\Rightarrow K = 7$ .
Let $N'({90^ \circ })$ be the number of particles scattered through a scattering angle of ${90^ \circ }$ . Hence, $\theta = {90^ \circ }$ $\theta = {90^ \circ }$ .
According to the formula, $N'({90^ \circ }) = \dfrac{K}{{{{\sin }^4}\left( {\dfrac{{{{90}^ \circ }}}{2}} \right)}}$
$\Rightarrow N'({90^ \circ }) = \dfrac{K}{{{{\sin }^4}({{45}^ \circ })}}$
Here we substitute the value of $K$ that has been calculated.
$\Rightarrow N'({90^ \circ }) = \dfrac{7}{{{{\sin }^4}({{45}^ \circ })}}$
The value of $\sin {45^ \circ }$ is, $\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$ .
Thus, ${\sin ^4}{45^ \circ } = {\left( {\dfrac{1}{{\sqrt 2 }}} \right)^4} = \dfrac{1}{4}$
We have to find the value of $N'({90^ \circ })$ .
Therefore, substituting the values, $K = 7{\text{ and si}}{{\text{n}}^4}({45^ \circ }) = \dfrac{1}{4}$ ,
We get the equation, $N'({90^ \circ }) = \dfrac{7}{{\dfrac{1}{4}}}$
$\Rightarrow N'({90^ \circ }) = 7 \times 4$
$\therefore$ $N'({90^ \circ }) = 28.$
The number of $\alpha - particles$ scattered through an angle of ${90^ \circ }$ per minute by the same nucleus is 28 per minute.
Hence, the correct option is option A.

Note
Another way to solve the problem has been given here. If $N(\theta ) \propto \dfrac{1}{{{{\sin }^4}\dfrac{\theta }{2}}}$ , then according to the information given in the question,
$112 \propto \dfrac{1}{{{{\sin }^4}\left( {\dfrac{{{{60}^ \circ }}}{2}} \right)}}$ where 112 is the number of $\alpha$ particles scattered through an angle of ${60^ \circ }$ per minute.
So, $N'({90^ \circ }) \propto \dfrac{1}{{{{\sin }^4}\left( {\dfrac{{{{90}^ \circ }}}{2}} \right)}}$ where $N'({90^ \circ })$ is the number of $\alpha$ particles scattered through an angle of ${90^ \circ }$ per minute.
From the above equations, it can be written that,
$\Rightarrow \dfrac{{N'({{90}^ \circ })}}{{112}}\alpha \dfrac{{{{\sin }^4}\left( {\dfrac{{60}}{2}} \right)}}{{{{\sin }^4}\left( {\dfrac{{{{90}^ \circ }}}{2}} \right)}}$
$\therefore N'({90^ \circ }) = 28.$
The number of $\alpha - particles$ scattered through an angle of ${90^ \circ }$ per minute by the same nucleus is 28 per minute.