# I don't expect anybody to solve this - Part (2)

**Classical Mechanics**Level 5

**Step 1** : You carefully cut out a \( 3 \) cm by \( 3 \) cm mono-layer of atoms from it, and weigh it accurately and precisely to get \( 0.6 \) nano grams. You can safely assume that the mono-layer looks similar to the figure above.

Note that this is **not** the actual mono-layer you have, but its lattice is similar to the above showed picture. We also assume that the atoms are spherical. The yellow dots represent the nuclei of the atoms.

**Step 2**: You put aside the mono-layer, and take a sample of radioactive radium bromide, which emits \( \alpha \) radiation. With the help of a Geiger counter ( or simply, a Detector ), you measure the count rate of this beam of \( \alpha \) particles to be \( 132000 \) hits min\(^{-1}\). The setup is something like this :

**Step 3**: Without further delay, you subject the mono-layer you have with the same radiation uniformly throughout its surface. You also place a Geiger counter beside the emitter to measure the count rate of rebounded \( \alpha \) particles, which you found out to be \( 2 \) hits min\(^{-1}\). The setup is something like this:

Now, given this information, you calculate the radius of nuclei of the atoms in the mono-layer to be some \( \lambda \times 10^{\beta} \) metres, where \( 1 < \lambda < 10 \). Enter \( \beta \) as your answer.

**Details and Assumptions**

Avogadro Number is taken as \( 6.023 \times 10^{23} \).

Pi is taken as \( \frac{22}{7} \).

\( \alpha \) particles are assumed to be point sized, when compared to large sized atoms in the mono-layer.

Step 2 and Step 3 are performed for the same amount of time.

###### This problem is original, and is inspired from MIT 5.111 Principles of Chemical Science Lecture-2, where the students perform an in-class imitation of the scattering experiment and calculate the radius of the "ping-pong ball" nuclei.

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