Atoms....

Let the radius of the nuclei (in meters) be $r$ . To find $r$ , we do the following calculations :

Step 1 : Number of atoms in the mono-layer( $n$ ) = $\dfrac{\text{Weight in grams}}{\text{Gram Atomic Weight}} \times N_{A} = \dfrac{6 \times 10^{-10}}{124} \times 6.023 \times 10^{23} = 2.91435 \times 10^{12}$ .

Step 2 :

Probability of back-scattering = $\dfrac{\text{Number of Rebounds}}{\text{Total number of hits}} = \dfrac{\text{Rebound count rate}}{\text{Hit count rate}} = \dfrac{2}{132000}$ .

Since the particles rebound when and only when they hit one of the nuclei, we can compute this probability in another way :

Probability of back-scattering = $\dfrac{\text{Area occupied by the nuclei}}{\text{Total area of the mono-layer}}$

$= \dfrac{(\pi r^{2}) \times n }{\text{Area of the mono-layer}}$

$=\dfrac{22 \times r^{2} \times 2.91435 \times 10^{12} }{7 \times 9 \times 10^{-4}}$ .

Step 3 :

Now, we equate the final answers of both ways in which we computed the same probability :

$\dfrac{22 \times r^{2} \times 2.91435 \times 10^{12} }{7 \times 9 \times 10^{-4}} = \dfrac{2}{132000}$ $\Rightarrow \boxed{r = 3.8584755... \times 10^{-11} \quad m }$ .

Hence, the required answer is $\boxed{(-11)}$ .

I did using the following three steps :

(a) Calculate the total number of atoms of the element in the sample (call it N).

(b) Since the radiation is uniform, calculate the area of which the alpha particles actually rebound.

(c) Now, this area is occupied collectively by N atoms. Calculated the area occupied by one element.

Hints for steps:

(a) Apply formula : N = (moles of element in sample)(Avogadro's number).

(b) 132,000 alpha particles strike .0009 square meters of area. How much area will 2 particles strike?

(c) N atoms occupy the above found area. How much area would 1 atom occupy? Apply standard area formula for circle and you have the answer.