Solid State Physics JEST 2012-2019
Solid State Physics JEST 2012-2019
Solid State Physics JEST 2012-2019
Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
Solid State Physics
JEST-2012
Q1. A beam of X-rays is incident on a BCC crystal. If the difference between the incident and
scattered wavevectors is K nxˆ kyˆ lzˆ where xˆ , yˆ , zˆ are the unit vectors of the
associated cubic lattice, the necessary condition for the scattered beam to give a Laue
maximum is
(a) h k l even (b) h k l
(c) h, k , l are all distinct (d) h k l odd
Ans.: (a)
1 1
Solution: In BCC basis 0, 0, 0, ,
1
,
2 2 2
F f S f e
2 i hun kVn ln
f e e 2 f 1 e
i h k l
n 1
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JEST-2013
Q4. A flat surface is covered with non-overlapping disks of same size. What is the largest
fraction of the area that can be covered?
3 5 6
(a) (b) (c) (d)
6 7 2 3
Ans.: (d)
Solution: In closed packed hexagonal lattice,
1 1 1
neff nC n f 1 ni 6 1 3 and a 2r
3 2 3
neff A 3 r 2
Now, largest fraction of area i.e., packing fraction
2r
3 3 2 3
6 a2 6
2
4 4
Q5. A metal suffers a structural phase transition from face-centered cubic FCC to the
simple cubic SC structure. It is observed that this phase transition does not involve any
change of volume. The nearest neighbor distances d fcc and d sc for the FCC and the SC
d 1
structures respectively are in the ratio fcc [Given 2 3 1.26 ]
d sc
(a) 1.029 (b) 1.122 (c) 1.374 (d) 1.130
Ans. : ()
Solution: Nearest neighbour in SC is a and C.N 6
a
Nearest neighbour in FCC is and C.N 12
2
a
d fcc 2 1 1 0.707
d sc a 2 1.414
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JEST-2014
Q6. Circular discs of radius 1 m each are placed on a plane so as to form a closely packed
triangular lattice. The number of discs per unit area is approximately equal to
(a) 0.86 m 2 (b) 0.43 m 2 (c) 0.29 m 2 (d) 0.14 m 2
Ans.: (c)
Solution: For closely packed triangular lattice,
1 1
a 2r , r 1 neff nC n f 1 nl
6 2
1 1
neff 3 0 1 0 neff 0.5
6 2
neff
Occupancy a 2 Closely packed hexagonal
A
0.5 0.5
0.29 m 2
3 3
2
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Q7. An ideal gas of non-relativistic fermions in 3-dimensions is at 0K. When both the number
density and mass of the particles are doubled, then the energy per particle is multiplied by
a factor
(a) 21 / 2 (b) 1 (c) 21 / 3 (d) 2 1 / 3
Ans.: (d)
2 2
Solution: EF
2m
3 2
n 3 at T 0 K
2 2
2/3 2/3
n 2n and m 2m EF 3 2 2n 3 2 n 21/ 3
4m 2m
Q8. When two different solids are brought in contact with each other, which one of the
following is true?
(a) Their Fermi energies become equal
(b) Their band gaps become equal
(c) Their chemical potentials become equal
(d) Their work functions become equal
Ans.: (c)
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JEST-2015
Q9. What is the area of the irreducible Brillouin zone of the crystal structure as given in the
figure?
2 2
(a)
3 A2
3 2
(b)
2 A2
2 2 o A
(c) 60
A2 B
2
A B A
(d)
3A 2
Ans.: (a)
Solution: Area of the Brillouin zone can be related to the area of normal cell as
2 2
Area of B.Z.
Area of cell A B B
600
A B A B sin A2 sin 600
2
3 2
A A
A B A
2 2
Area of Brillouin zone
3A2
Q10. For a 2 - dimensional honeycomb lattice as shown in the figure, the first Bragg spot
occurs for the grazing angle 1 , while sweeping the angle from 0 o . The next Bragg spot is
obtained at 2 given by A
3 a 120 o
(a) sin 1 3 sin 1 (b) sin 1 sin 1 120 o
2 B
120 o
3
(c) sin 1
sin 1
(d) sin 1 3 sin 1
2
Ans.: (c)
Solution: According to Bragg’s law, the condition for first Bragg spot and second spot is
2d1 sin 1 n and 2d 2 sin 2 n
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Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
d
2d1 sin 1 2d 2 sin 2 2 sin 1 1 sin 1
d2
For 2 - dimensional honeycomb lattice, the lattice constant ‘ a ’ and interplanar spacing
‘ d ’ is linked as
2
a a2 3 a
d a d1 a
1
2 2 2
a and d 2 a d1
2 4 2
0
60
3
2 sin 1 sin 1 a
2
ka
Q11. Given the tight binding dispersion relation E k E 0 A sin 2 , where E 0 and A are
2
constants and a is the lattice parameter. What is the group velocity of an electron at the
second Brillouin zone boundary?
a 2a a
(a) 0 (b) (c) (d)
h h 2h
Ans.: (a)
1 dE
Solution: Group velocity is defined as, vg
dk
ka dE ka ka aA
Since E E0 A sin 2 aA sin cos sin ka
2 dk 2 2 2
2
2a a 0
a a
K
In one dimension, the Brillouin zone boundary is
The 1st Brillouin zone boundaries lie at
a
2
The 2nd Brillouin zone boundaries lie at
a
Thus, the group velocity at the second Brillouin zone boundary is
aA 2 aA
vg 2 sin a sin 2 vg 0
a 2 a 2
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Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
Q12. The total number of Na and Cl ions per unit cell of NaCl is,
(a) 2 (b) 4 (c) 6 (d) 8
Ans.: (d)
Solution: Total number of Na and Cl ions per unit d is
1 1 1
N Cl nc n f , N Na
ne 1 ni
8 2 4
where nc number of ions at corner
n f number of ions at face
ne number of ions at edges
ni number of ions inside
1 1 1 Cl
N N Cl N Na
8 6 12 11 1 3 3 1 8 Na
8 2 4
Q13. For non-interacting Fermions in d dimensions, the density of states DE varies as
d
1
E 2
. The Fermi energy E F of an N particle system in 3, 2 and 1 dimensions
will scale respectively as,
(a) N 2 , N 2 / 3 , N (b) N , N 2 / 3 , N 2
(c) N , N 2 , N 2 / 3 (d) N 2 / 3 , N , N 2
Ans.: (d)
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JEST-2016
2
Q14. If k is the wavevector of incident light ( k , is the wavelength of light) and G is
a reciprocal lattice vector, then the Bragg’s law can be written as:
(a) k G 0 (b) 2k .G G 2 0
(c) 2k .G k 2 0 (d) k .G 0
Ans. : (b)
Solution: By means of Eward construction, we can write the Bragg’s law in B
vector form K
A G
G OB, K AO
K
O
For diffraction it is necessary that vector K G , that is vector AB
be equal in magnitude to the vector K or
K G K 2 2K G G 2 0
2
Q15. The number of different Bravais lattices possible in two dimensions is:
(a) 2 (b) 3 (c) 5 (d) 6
Ans. : (c)
Solution: Five Bravais lattices in 2D are:
(i) Square lattice
(ii) Rectangular P lattice
H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016
Phone: 011-26865455/+91-9871145498
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