Question

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? (\(N = 6.02 \times 10^{23} \, \text{mol}^{-1}\))

• A. 4.04
• B. 2.93
• C. 1.93
• D. 3.04

Hint:

- For a simple cubic (SC) lattice: Number of atoms per unit cell \(Z=1\). Edge length \(a=2r\), where \(r\) is atomic radius. - Density \( \rho = \frac{Z \times M}{N_A \times V_{cell}} \), where M is molar mass (g/mol), \(N_A\) is Avogadro's number, \(V_{cell}=a^3\). - So, \( \rho = \frac{Z M}{N_A (2r)^3} = \frac{Z M}{8 N_A r^3} \). (Here r is in cm if \(\rho\) is in g/cm\(^3\)). - To get \(r\) in Angstroms directly, if \(r_{\text{\AA}}\) is radius in \(\text{\AA}\), then \(r_{cm} = r_{\text{\AA}} \times 10^{-8}\). \( \rho = \frac{Z M}{N_A 8 (r_{\text{\AA}} \times 10^{-8})^3} = \frac{Z M \times 10^{24}}{8 N_A r_{\text{\AA}}^3} \). So, \( r_{\text{\AA}}^3 = \frac{Z M \times 10^{24}}{8 N_A \rho} \).

Answer 1

The Correct Option is C

For a simple cubic (SC) lattice: Number of atoms per unit cell \( Z = 1 \).
Relationship between edge length \(a\) of the unit cell and atomic radius \(r\): \( a = 2r \).
Density \( \rho \) of the unit cell is given by: \[ \rho = \frac{Z \times M}{N_A \times a^3} \] where \(M\) is the atomic weight (in g/mol), \(N_A\) is Avogadro's number, and \(a\) is the edge length.
Given: Atomic weight \( M = 250 \) g/mol (since 250 u per atom implies 250 g per mole).
Density \( \rho = 7.
2 \, \text{g cm}^{-3} \).
Avogadro's number \( N_A = 6.
02 \times 10^{23} \, \text{mol}^{-1} \).
(Given as N in question) \(Z=1\).
Substitute these values into the density formula to find \(a^3\): \[ 7.
2 = \frac{1 \times 250}{(6.
02 \times 10^{23}) \times a^3} \] \[ a^3 = \frac{250}{7.
2 \times 6.
02 \times 10^{23}} \] \[ a^3 = \frac{250}{43.
344 \times 10^{23}} = \frac{250}{4.
3344 \times 10^{24}} \] \[ a^3 \approx \frac{250}{4.
33} \times 10^{-24} \approx 57.
7 \times 10^{-24} \, \text{cm}^3 = 5.
77 \times 10^{-23} \, \text{cm}^3 \] Let's use \(N_A \approx 6 \times 10^{23}\) for a quicker estimate: \( a^3 \approx \frac{250}{7.
2 \times 6 \times 10^{23}} = \frac{250}{43.
2 \times 10^{23}} \approx 5.
787 \times 10^{-24} \, \text{cm}^3 \).
\( a = (5.
787 \times 10^{-24})^{1/3} \, \text{cm} = (57.
87 \times 10^{-25})^{1/3} \).
No, \( (5.
787)^{1/3} \times 10^{-8} \, \text{cm} \).
\( (5.
787)^{1/3} \): \( 1^3=1, 2^3=8 \).
So it's between 1 and 2, closer to 2.
\( 1.
7^3 \approx 4.
913 \), \( 1.
8^3 \approx 5.
832 \).
So \( (5.
787)^{1/3} \approx 1.
8 \).
More precisely, \( (5.
787037.
.
.
)^{1/3} \approx 1.
795 \).
Let's use \( (250 / (7.
2 \times 6.
02 \times 10^{23})) \).
Using \(6.
02 \times 10^{23}\): \( a^3 = \frac{250}{7.
2 \times 6.
02 \times 10^{23}} = \frac{250}{43.
344 \times 10^{23}} = 5.
7678 \times 10^{-24} \, \text{cm}^3 \).
\( a = (5.
7678 \times 10^{-24})^{1/3} \, \text{cm} = (5.
7678)^{1/3} \times 10^{-8} \, \text{cm} \).
\( (5.
7678)^{1/3} \approx 1.
7935 \).
So \( a \approx 1.
7935 \times 10^{-8} \, \text{cm} \).
Convert cm to Angstroms (\( \text{\AA} \)): \( 1 \, \text{cm} = 10^8 \, \text{\AA} \).
So \( a \approx 1.
7935 \, \text{\AA} \).
For a simple cubic lattice, \( a = 2r \).
So, radius \( r = a/2 \).
\[ r = \frac{1.
7935 \, \text{\AA}}{2} \approx 0.
89675 \, \text{\AA} \] This does not match any option closely.
Let me recheck calculation or formula usage.
Options are 1.
93, 2.
93, 3.
04, 4.
04.
My \(a\) is around their \(r\).
Maybe the \(a\) I found is actually \(2r\).
Is the atomic weight definition different? No, 250u is 250 g/mol.
Let's check my \(a^3\) value again.
\( a^3 = \frac{Z M}{N_A \rho} = \frac{1 \times 250 \text{ g/mol}}{(6.
02 \times 10^{23} \text{ mol}^{-1}) \times (7.
2 \text{ g/cm}^3)} \) \( = \frac{250}{43.
344 \times 10^{23}} \text{ cm}^3 = 5.
7678 \times 10^{-24} \text{ cm}^3 \).
This is correct.
\( a = (5.
7678)^{1/3} \times 10^{-8} \text{ cm} \).
\( 1.
5^3 = 3.
375 \), \( 1.
6^3 = 4.
096 \), \( 1.
7^3 = 4.
913 \), \( 1.
8^3 = 5.
832 \).
So \( (5.
7678)^{1/3} \) is slightly less than 1.
8.
About 1.
7935.
\( a = 1.
7935 \times 10^{-8} \text{ cm} = 1.
7935 \times 10^{-10} \text{ m} = 1.
7935 \, \text{\AA} \).
Then \( r = a/2 = 1.
7935 / 2 = 0.
89675 \, \text{\AA} \).
Let's check if one of the options for \(r\) gives the density.
Option (3): \( r = 1.
93 \, \text{\AA} = 1.
93 \times 10^{-8} \, \text{cm} \).
Then \( a = 2r = 2 \times 1.
93 \times 10^{-8} \, \text{cm} = 3.
86 \times 10^{-8} \, \text{cm} \).
\( a^3 = (3.
86 \times 10^{-8})^3 = (3.
86)^3 \times 10^{-24} \, \text{cm}^3 \).
\( 3.
86^3 \approx (3.
9)^3 \).
\( 3.
9^2 = 15.
21 \).
\( 15.
21 \times 3.
9 \approx 15 \times 4 = 60 \).
\( 3.
86^3 = 57.
512 \).
\( a^3 = 57.
512 \times 10^{-24} \, \text{cm}^3 \).
Density \( \rho = \frac{1 \times 250}{(6.
02 \times 10^{23}) \times (57.
512 \times 10^{-24})} = \frac{250}{6.
02 \times 57.
512 \times 10^{-1}} \) \( = \frac{2500}{6.
02 \times 57.
512} = \frac{2500}{346.
22} \approx \frac{2500}{346} \approx 7.
219.
.
.
\, \text{g/cm}^3 \).
This is very close to 7.
2 g/cm\(^3\).
So \(r=1.
93 \, \text{\AA}\) is likely the correct answer.
My calculation of \(a\) was correct, but the value of \( (5.
7678)^{1/3} \) might have been interpreted as the final radius \(r\) by mistake, or there's a factor of 2 error.
Let's retrace calculation for \(a\): \(a^3 = 5.
7678 \times 10^{-24} \text{ cm}^3\).
My prior calculation of \(a\) was \(1.
7935 \times 10^{-8} \text{ cm}\).
This is \(a\).
Then \(r = a/2 = 0.
89675 \times 10^{-8} \text{ cm} = 0.
89675 \text{ \AA}\).
What if \(a=r\) for simple cubic? No, for simple cubic, atoms touch along the edge, so \(a=2r\).
What if the problem uses \(N = 6.
0 \times 10^{23}\) instead of \(6.
02\)? \( a^3 = \frac{250}{7.
2 \times 6 \times 10^{23}} = \frac{250}{43.
2 \times 10^{23}} = 5.
787 \times 10^{-24} \, \text{cm}^3 \).
\( a = (5.
787)^{1/3} \times 10^{-8} \, \text{cm} \approx 1.
795 \times 10^{-8} \, \text{cm} = 1.
795 \, \text{\AA} \).
\( r = a/2 \approx 1.
795/2 = 0.
8975 \, \text{\AA} \).
Still around \(0.
9 \, \text{\AA}\).
The options are: 1.
93, 2.
93, 3.
04, 4.
04.
These values seem to be \(a\) rather than \(r\), or my density formula is off by a factor.
Density \( \rho = \frac{Z \times (\text{Atomic weight}/N_A)}{a^3} \).
This is \( \frac{\text{mass of unit cell}}{\text{volume of unit cell}} \).
Mass of one atom \( = M/N_A = 250 / (6.
02 \times 10^{23}) \) g.
Mass of unit cell (Z=1 for SC) \( = 1 \times \frac{250}{6.
02 \times 10^{23}} \) g.
Volume of unit cell \( = a^3 \).
So \( \rho = \frac{M_{atom}}{a^3} = \frac{M}{N_A a^3} \) for Z=1.
This is correct.
Let's check the calculation using \(r=1.
93 \text{ \AA}\) which implies \(a=3.
86 \text{ \AA}\).
If \(r\) is radius (in \(\text{\AA}\)), then \(a = 2r \times 10^{-8}\) cm.
\( a^3 = (2r \times 10^{-8})^3 = 8r^3 \times 10^{-24} \text{ cm}^3 \).
\[ \rho = \frac{Z M}{N_A \cdot 8r^3 \times 10^{-24}} \] \[ r^3 = \frac{Z M}{8 N_A \rho \times 10^{-24}} \] (if r is in cm) If \(r\) is in \(\text{\AA}\), then \( r_{\text{cm}} = r_{\text{\AA}} \times 10^{-8} \).
\[ (r_{\text{\AA}} \times 10^{-8})^3 = \frac{Z M}{8 N_A \rho} \] \[ r_{\text{\AA}}^3 \times 10^{-24} = \frac{Z M}{8 N_A \rho} \] \[ r_{\text{\AA}}^3 = \frac{Z M}{8 N_A \rho} \times 10^{24} \] Substitute values: \(Z=1, M=250, N_A = 6.
02 \times 10^{23}, \rho = 7.
2 \).
\[ r_{\text{\AA}}^3 = \frac{1 \times 250}{8 \times (6.
02 \times 10^{23}) \times 7.
2} \times 10^{24} \] \[ r_{\text{\AA}}^3 = \frac{250}{8 \times 6.
02 \times 7.
2} \times \frac{10^{24}}{10^{23}} = \frac{250}{8 \times 6.
02 \times 7.
2} \times 10 \] \[ r_{\text{\AA}}^3 = \frac{2500}{8 \times 6.
02 \times 7.
2} = \frac{2500}{8 \times 43.
344} = \frac{2500}{346.
752} \] \[ r_{\text{\AA}}^3 \approx 7.
2097 \] Now, \( r_{\text{\AA}} = (7.
2097)^{1/3} \).
\( 1^3=1, 2^3=8 \).
So \(r\) is close to 2, slightly less.
\( 1.
9^3 = (2-0.
1)^3 = 8 - 3(4)(0.
1) + 3(2)(0.
01) - 0.
001 = 8 - 1.
2 + 0.
06 - 0.
001 = 6.
8 + 0.
059 = 6.
859 \).
\( 1.
93^3 \approx (1.
9)^3 \approx 6.
859 \).
Let's calculate \( (1.
93)^3 \): \( 1.
93 \times 1.
93 = 3.
7249 \).
\( 3.
7249 \times 1.
93 \approx 3.
72 \times 1.
93 \).
\( 3.
7249 \times 1.
93 = 7.
188057 \).
This is very close to \(7.
2097\).
So, \( r_{\text{\AA}} \approx 1.
93 \).
This matches option (3).
My initial placement of \(a=2r\) in the main density formula denominator was correct, but I solved for \(a\) first, then \(r\).
Direct formula for \(r\) is better.
My error was in \( a^3 = (2r)^3 = 8r^3 \).
Original: \( \rho = \frac{Z M}{N_A a^3} \).
If \(a=2r\), then \( \rho = \frac{Z M}{N_A (2r)^3} = \frac{Z M}{N_A 8r^3} \).
So \( r^3 = \frac{ZM}{8N_A\rho} \).
This formula is for \(r\) in cm.
\( r^3_{\text{cm}} = \frac{1 \times 250}{8 \times (6.
02 \times 10^{23}) \times 7.
2} = \frac{250}{346.
752 \times 10^{23}} = 0.
72097 \times 10^{-23} = 7.
2097 \times 10^{-24} \text{ cm}^3 \).
\( r_{\text{cm}} = (7.
2097)^{1/3} \times (10^{-24})^{1/3} = (7.
2097)^{1/3} \times 10^{-8} \text{ cm} \).
Since \(1 \text{ \AA} = 10^{-8} \text{ cm}\), \( r_{\text{\AA}} = (7.
2097)^{1/3} \approx 1.
931 \).
This is consistent.