Solid State JEE Main previous year questions with solutions

5 solved JEE Main questions on Solid State, free to read — no sign-in needed. The full chapter has 85 questions; sign in to attempt the remaining 80 in the exam simulator.

1. Q1JEE Main 2023 (15 Apr, Shift 1)
   Which of the following expressions is correct in case of a $CsCl$ unit cell (edge length ‘$a$’)$?$
   1. A.${r}_{{Cs}^{+}}+{r}_{{Cl}^{-}}=\dfrac{a}{2}$
   2. B.${r}_{{Cs}^{+}}+{r}_{{Cl}^{-}}=\dfrac{\sqrt{3}}{2}a$
   3. C.${r}_{{Cs}^{+}}+{r}_{{Cl}^{-}}=\dfrac{a}{\sqrt{2}}$
   4. D.${r}_{{Cs}^{+}}+{r}_{{Cl}^{-}}=a$
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   Answer: (B)

   $CsCl$ has body centered type structure in which ${Cs}^{+}$ occupies at corner of a cube and ${Cl}^{–}$ occupies the centre of the cube. $2{r}_{{Cs}^{+}}+2{r}_{{Cl}^{-}}=\sqrt{3}a$ (where a is the edge length of the cube) ${r}_{{Cs}^{+}}+{r}_{Cr}=\dfrac{\sqrt{3}}{2}a$
2. Q2JEE Main 2023 (13 Apr, Shift 2)
   Sodium metal crystallises in a body centred cubic lattice with unit cell edge length of $4Å$. The radius of sodium atom is $\underline{\phantom{0000}}\times {10}^{-1}Å$. (Nearest integer)
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   Answer: 17

   Edge length of the unit cell $(a)=4x{10}^{-8}cm$ Radius of sodium atom $(r)=?$ Since, sodium crystallises in body centered cubic arrangement. The radius of the atom and the edge length of the cubic unit cell are related as follows, For bcc structure, $4r=\sqrt{3}a\Rightarrow r=\dfrac{\sqrt{3}}{4}a$ $r=\dfrac{1.732\times 4\times {10}^{-8}}{4} r=1.732\times {10}^{-8}cm r=17.3Å$.
3. Q3JEE Main 2023 (11 Apr, Shift 1)
   An atomic substance A of molar mass $12g{mol}^{–1}$has a cubic crystal structure with edge length of $300pm$. The no. of atoms present in one unit cell of A is (Nearest integer) Given the density of A is $3.0gm{m}^{-1}$ and ${NA}_{A}=6.02\times {10}^{23}{mol}^{-1}$
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   Answer: 4

   The density of crystal can be calculated as follows, $\rho =\dfrac{Z\times M}{{N}_{A}\times {a}^{3}}$ $\rho =$density of the unit cell $Z=$Effective atomic number $M=$Molar mass ${N}_{A}=$Avogadro's number $a=$Edge length of the unit cell $3=\dfrac{Z\times 12}{6.02\times {10}^{23}\times (3{)}^{3}\times {10}^{-24}}$ $Z=\dfrac{6.02\times {10}^{-1}\times 27}{4}=4$
4. Q4JEE Main 2023 (10 Apr, Shift 2)
   The correct relationships between unit cell edge length ‘$a$’ and radius of sphere ‘$r$’ for face-centred and body-centred cubic structures respectively are:
   1. A.$2\sqrt{2}r=a$ and $\sqrt{3}r=4a$
   2. B.$r=2\sqrt{2}a$ and $4r=\sqrt{3}a$
   3. C.$r=2\sqrt{2}a$ and $\sqrt{3}r=4a$
   4. D.$2\sqrt{2}r=a$ and $4r=\sqrt{3}a$
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   Answer: (D)

   A face-centered cubic (fcc) unit cell contains a component in the center of each face in addition to those at the corners of the cube. The atoms touch each other along the face diagonal. For $FCC:a\sqrt{2}=4r$ $a=2\sqrt{2}r$ A body-centered cubic (bcc) unit cell contains a component in the center of the cube in addition to those at the corners of the cube. The atoms touch each other along the body diagonal. For $BCC:a\sqrt{3}=4r$
5. Q5JEE Main 2023 (06 Apr, Shift 2)
   Number of crystal systems from the following where body centred unit cell can be found, is _____. Cubic, tetragonal, orthorhombic, hexagonal, rhombohedral, monoclinic, triclinic
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   Answer: 3

   Bravais Lattice refers to the 14 different 3-dimensional configurations into which atoms can be arranged in crystals. The smallest group of symmetrically aligned atoms which can be repeated in an array to make up the entire crystal is called a unit cell. Crystal systems where body centred unit cell can be found are Cubic, orthorhombic and tetragonal Hence, correct answer is $3$