TS EAMCET · Physics · Units and Dimensions
If \(V_0\) is the volume of a standard unit cell of germanium crystal containing \(N_0\) atoms, then the expression for the mass \(m\) of a volume \(V\) in terms of \(V_0, N_0, M_{\text {mol }}\) and \(N_A\) is [here, \(M\) is the molar mass of germanium and \(N_A\) is the Avogadro's constant]
- A \(M \frac{V}{V_0} \frac{N_A}{N_0}\)
- B \(\frac{N_{\mathrm{A}}}{N_0} \frac{V_0}{V} M\)
- C \(M \frac{V}{V_0} \frac{N_0}{N_A}\)
- D \(M \frac{V_0}{V} \frac{N_0}{N_A}\)
Answer & Solution
Correct Answer
(C) \(M \frac{V}{V_0} \frac{N_0}{N_A}\)
Step-by-step Solution
Detailed explanation
Number of unit cells in volume \(V\) \(=\frac{\text { Total volume }}{\text { Volume of a unit cell }}=\frac{V}{V_0}\) Number of atoms in volume \(V\) \(=\) Number of unit cells \(\times\) Number of atoms in 1 unit cell \(=\frac{V}{V_0} \times N_0\) Number of moles in volume…
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