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MHT CET · Chemistry · Solid State

What is the number of unit cells present in \(3.9 \mathrm{~g}\) of potassium if it crystallizes in BCC Structure?

  1. A \(\frac{\mathrm{N}_{\mathrm{A}}}{10}\)
  2. B \(\mathrm{N}_{\mathrm{A}} \times 10\)
  3. C \(2 \mathrm{~N}_{\mathrm{A}}\)
  4. D \(\frac{\mathrm{N}_{\mathrm{A}}}{20}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{\mathrm{N}_{\mathrm{A}}}{20}\)

Step-by-step Solution

Detailed explanation

\(\text { Number of atoms }=\frac{\text {Mass}}{\text {Atomicmass}} \times \mathrm{N}_{\mathrm{A}}=\) \(\frac{3.9}{39} \times \mathrm{N}_{\mathrm{A}}=0.1 \mathrm{~N}_{\mathrm{A}} \)
\( \text { In BCC unit cell, } \mathrm{n}=2 \)
\( \therefore \text { Number of unit cells }=\frac{0.1 \mathrm{~N}_{\mathrm{A}}}{2}=\frac{\mathrm{N}_{\mathrm{A}}}{20}\)