Question:
A metal crystallizes in simple cubic lattice. The volume of one unit cell is \( 6.4 \times 10^{-7} \, \text{pm}^3 \). What is the radius of the metal atom in pm?
A metal crystallizes in simple cubic lattice. The volume of one unit cell is \( 6.4 \times 10^{-7} \, \text{pm}^3 \). What is the radius of the metal atom in pm?
Show Hint
In simple cubic lattices, the edge length of the unit cell is twice the radius of the atoms. Use this to relate volume and atomic radius.
Updated On: Jun 6, 2025
- 100
- 200
- 300
- 400
Hide Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
For a simple cubic lattice, the relationship between the unit cell volume \( V_{\text{unit}} \) and the atomic radius \( r \) is:
\[
V_{\text{unit}} = (2r)^3
\]
Given that the volume is \( 6.4 \times 10^{-7} \, \text{pm}^3 \), solving for \( r \) gives us a radius of 200 pm.
Was this answer helpful?
0
0
Top Questions on The solid state
- A solution is prepared by adding 124 g of ethylene glycol (molar mass = 62 g mol\(^{-1}\)) to \( x \) g of water to get a 10 m solution. What is the value of \( x \) (in g)?
- AP EAPCET - 2025
- Chemistry
- The solid state
- The composition of a sample of wustite is \( \text{Fe}_{0.93} \text{O}_{1.00} \). Percentage of iron in the form of \( \text{Fe}^{3+} \) ion is nearly:
- AP EAPCET - 2025
- Chemistry
- The solid state
- An element (atomic weight = 250 u) crystallises in a simple cubic lattice. If the density of the unit cell is \( 7.2 \, \text{g cm}^{-3} \), what is the radius (in \( \text{\AA} \)) of the atom of the element? (\(N = 6.02 \times 10^{23} \, \text{mol}^{-1}\))
- AP EAPCET - 2025
- Chemistry
- The solid state
- The composition of a sample of wustite is \( \text{Fe}_{0.93} \text{O}_{1.00} \). Percentage of iron in the form of \( \text{Fe}^{3+} \) ion is nearly:
- AP EAPCET - 2025
- Chemistry
- The solid state
- An element (atomic weight = 250 u) crystallises in a simple cubic lattice. If the density of the unit cell is \( 7.2 \, \text{g cm}^{-3} \), what is the radius (in \( \text{\AA} \)) of the atom of the element? (\(N = 6.02 \times 10^{23} \, \text{mol}^{-1}\))
- AP EAPCET - 2025
- Chemistry
- The solid state
View More Questions
Questions Asked in AP EAPCET exam
- If the binding energy per nucleon of deuteron (\( ^1\mathrm{H}^2 \)) is 1.15 MeV and an \(\alpha\)-particle has a binding energy of 7.1 MeV per nucleon, then the energy released per nucleon in the given reaction is \[ ^1\mathrm{H}^2 + ^1\mathrm{H}^2 \rightarrow ^2\mathrm{He}^4 + Q \]
- AP EAPCET - 2025
- Nuclear physics
- If the function \( f \) defined by \[ f(x) = \begin{cases} \dfrac{1 - \cos 4x}{x^2}, & x<0 \\ a, & x = 0 \\ \dfrac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}, & x>0 \end{cases} \] is continuous at \( x = 0 \), then \( a = \)
- AP EAPCET - 2025
- Limits and Exponential Functions
If \( \vec{u}, \vec{v}, \vec{w} \) are non-coplanar vectors and \( p, q \) are real numbers, then the equality:
\[ [3\vec{u} \quad p\vec{v} \quad p\vec{w}] - [p\vec{v} \quad \vec{w} \quad q\vec{u}] - [2\vec{w} \quad q\vec{v} \quad q\vec{u}] = 0 \]
holds for:
- AP EAPCET - 2025
- Geometry and Vectors
Statement-I: In the interval \( [0, 2\pi] \), the number of common solutions of the equations
\[ 2\sin^2\theta - \cos 2\theta = 0 \]
and
\[ 2\cos^2\theta - 3\sin\theta = 0 \]
is two.
Statement-II: The number of solutions of
\[ 2\cos^2\theta - 3\sin\theta = 0 \]
in \( [0, \pi] \) is two.
- AP EAPCET - 2025
- Trigonometric Identities
If \( A \) and \( B \) are acute angles satisfying
\[ 3\cos^2 A + 2\cos^2 B = 4 \]
and
\[ \frac{3 \sin A}{\sin B} = \frac{2 \cos B}{\cos A}, \]
Then \( A + 2B = \ ? \)
- AP EAPCET - 2025
- Trigonometric Identities
View More Questions
