Question

The graph which depicts the result of Rutherford's gold foil experiment with $\alpha -$particle is

$\theta$: Scattering angle

$Y$: Number of scattered $\alpha -$ particles detected

(Plots are schematic and not to scale)

• A.

  

• B.

  

• C.

  

• D.

  

Answer 1

The correct option is B

Step 1: Given data:

$\theta$: Scattering angle

$Y$: Number of scattered $\alpha -$ particles detected

Step 2: Formula Used:

According to Rutherford's observations from the gold foil experiment-

$N\left(\theta \right)=\frac{K}{{\sin }^4{\displaystyle \frac{\theta }{4}}}$

Where, $K$ is constant,

$N\left(\theta \right)$ is no. of alpha particles scattered through an angle $\theta$

Step 3: Computing the graph between $Yand\theta$:

From the scattering formula,

$N\left(\theta \right)=\frac{K}{{\sin }^4{\displaystyle \frac{\theta }{4}}}$

Here, $Y$: number of scattered $\alpha -$ particles detected

$Y\alpha \frac{1}{{\left(\sin {\displaystyle \frac{\theta }{4}}\right)}^4}$

When,

$$\begin{gathered} \theta =2\pi \\ \frac{\theta }{4}=\frac{\pi }{2} \end{gathered}$$

Since $0to2\pi$ function ${\sin }^4\frac{\theta }{4}$increases so $\frac{1}{{\sin }^4{\displaystyle \frac{\theta }{4}}}$decreases

So, At $\theta \to 0,\sin 0=0,\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${\sin }^4\theta /4$}\right.=\infty$

Thus the graph will decrease exponentially.

Note that this figure which depicts the result of Rutherford's gold foil experiment with $\alpha -$particles will be

Therefore the $\alpha -$particle is$Y\alpha \frac{1}{{\left(\sin {\displaystyle \frac{\theta }{2}}\right)}^4}$.

Hence,option B is the correct answer.