Calculating Bit-Density of a CD using AFM Images

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AFM images of a CD revealed 224 bits in a 900µm² area, prompting a calculation of the bit-density for the entire disc. The total area of the CD was calculated as 10,746µm², with an inner inactive area of 1,661µm², resulting in a total active area of 9,085µm². The calculation of total bits using the formula (224 bits/900µm²) * 9,085µm² yields approximately 2,261 bits, which is inconsistent with the expected storage capacity of around 650MB for a CD. Concerns were raised about the accuracy of the radius measurement of 58.5µm, suggesting it may be incorrect. The discussion highlights the need to verify measurements and calculations to resolve the discrepancy in expected storage capacity.
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We took AFM images of a CD in lab. Manual counting provided that there were 224 bits in the 900µm^2 image I have. Based on this I want to determine the bit-density of the whole CD.

The total active area for the disc is the total area minus the inner inactive are of the disc. The total area was calculated using a radius of 58.5µm and was found to be 10,746 µm^2. The inner inactive area of the disc was found to be 1,661µm^2. Therefore, the total active area of the disk was 9,085µm^2.

Shouldn't the next step be to take (224 bits/900µm^2)*9,085µm^2 to get the total number of bits? For this I get 2,261.156 bits, but it can't be right because it would mean that the whole CD had only 0.00215640640258789MB of storage! I know that CDs have ~650MB of storage. What did I do wrong? Please?
 
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Radius of 58.5µm doesn't look correct to me. Are you sure you mean µm?
 
the area looks waaaay off!
 
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