Explain the following terms with suitable examples:
Question: Explain the following terms with suitable examples: (i)Schottky defect (ii)Frenkel defect (iii)Interstitials and (iv)F-centres Solution: (i)Schottky defect:Schottky defect is basically a vacancy defect shown by ionic solids. In this defect, an equal number of cations and anions are missing to maintain electrical neutrality. It decreases the density of a substance. Significant number of Schottky defects is present in ionic solids. For example, in NaCl, there are approximately 106Schottk...
Read More →An arc of length 20π cm subtends an angle of 144°
Question: An arc of length 20 cm subtends an angle of 144 at the centre of a circle. Find the radius of the circle. Solution: We know that the arc lengthlof a sector of an anglein a circle of radiusris $l=\frac{\theta}{360^{\circ}} \times 2 \pi r$ It is given $l=20 \pi \mathrm{cm}$ and angle $\theta=144^{\circ}$. Now we substitute the value oflandin above formula to find the value of radiusrof circle. $20 \pi \mathrm{cm}=\frac{144^{\circ}}{360^{\circ}} \times 2 \pi r$ $r=\frac{20 \pi \times 360^...
Read More →In terms of band theory, what is the difference
Question: In terms of band theory, what is the difference (i)Between a conductor and an insulator (ii)Between a conductor and a semiconductor Solution: (i)The valence band of a conductor is partially-filled or it overlaps with a higher energy, unoccupied conduction band. On the other hand, in the case of an insulator, the valence band is fully- filled and there is a large gap between the valence band and the conduction band. (ii)In the case of a conductor, the valence band is partially-filled or...
Read More →Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell.
Question: Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell? Solution: For a face-centredunit cell: $a=2 \sqrt{2} r$ It is given that the atomic radius,r= 0.144 nm So, $a=2 \sqrt{2} \times 0.144 \mathrm{~nm}$ = 0.407 nm Hence, length of a side of the cell = 0.407 nm...
Read More →Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell.
Question: Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell? Solution: For a face-centredunit cell: $a=2 \sqrt{2} r$ It is given that the atomic radius,r= 0.144 nm So, $a=2 \sqrt{2} \times 0.144 \mathrm{~nm}$ = 0.407 nm Hence, length of a side of the cell = 0.407 nm...
Read More →A uniform rod of length
Question: A uniform rod of length ' $l$ ' is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed $\omega$ the rod makes an angle $\theta$ with it (see figure). To find $\theta$ equate the rate of change of angular momentum (direction going into the paper) $\frac{m l^{2}}{12} \omega^{2} \sin \theta \cos \theta$ about the centre of mass (CM) to the\ torque provided by the horizontal and vertical forces $F_{H}$ and $F_{V}$ about the CM. The v...
Read More →Find the angle subtended at the centre of a circle of radius
Question: Find the angle subtended at the centre of a circle of radius 5 cm by an arc of length (5/3) cm. Solution: We know that the arc lengthlof a sector of an anglein a circle of radiusris $l=\frac{\theta}{360^{\circ}} \times 2 \pi r$ It is given that $r=5 \mathrm{~cm}$ and length $l=\frac{5 \pi}{3} \mathrm{~cm}$. Substituting these value in above equation, $\frac{5 \pi}{3}=\frac{\theta}{360^{\circ}} \times 2 \pi \times 5$ $5 \pi \times 360^{\circ}=\theta \times 2 \pi \times 5 \times 3$' $\th...
Read More →Classify each of the following as being either a p-type or an n-type semiconductor:
Question: Classify each of the following as being either ap-typeor ann-typesemiconductor: (i)Ge doped with In(ii)B doped with Si. Solution: (i)Ge (a group 14 element) is doped with In (a group 13 element). Therefore, a hole will be created and the semiconductor generated will be ap-typesemiconductor. (ii)B (a group 13 element) is doped with Si (a group 14 element). Thus, a hole will be created and the semiconductor generated will be ap-type semiconductor....
Read More →Ferric oxide crystallises in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions.
Question: Ferric oxide crystallises in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide. Solution: Let the number of oxide (O2) ions bex. So, number of octahedral voids =x It is given that two out of every three octahedral holes are occupied by ferric ions. So, number of ferric $\left(\mathrm{Fe}^{3+}\right)$ ions $=\frac{2}{3} x$ Therefore, ratio of the number of Fe3+ions to the number of O2...
Read More →Non-stoichiometric cuprous oxide,
Question: Non-stoichiometric cuprous oxide, Cu2O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is ap-type semiconductor? Solution: In the cuprous oxide (Cu2O) prepared in the laboratory, copper to oxygen ratio is slightly less than 2:1. This means that the number of Cu+ions is slightly less than twice the number of O2ions. This is because some Cu+ions have been replaced by Cu2+ions. Every Cu2+ion r...
Read More →Find the area of ∆ABC whose vertices are:
Question: Find the area of ∆ABCwhose vertices are:(i)A(1, 2),B(2, 3) andC(3, 4)(ii)A(5, 7),B(4, 5) andC(4, 5)(iii)A(3, 8),B(4, 2) andC(5, 1)(iv)A(10, 6),B(2, 5) andC(1, 3) Solution: (i)A(1, 2),B(2, 3) andC(3, 4) are the vertices of ∆ABC. Then,(x1= 1,y1= 2), (x2= 2,y2= 3) and (x3= 3,y3= -4) Area of triangle $A B C$ $=\frac{1}{2}\left\{x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right\}$ $=\frac{1}{2}\{1(3-(-4))+(-2)(-4-2)+(-3)(2-3)\}$ $=\frac{1}{2}\{1...
Read More →A person of 80 kg mass is standing on the rim
Question: A person of $80 \mathrm{~kg}$ mass is standing on the rim of a circular platform of mass $200 \mathrm{~kg}$ rotating about its axis at 5 revolutions per minute $(\mathrm{rpm})$. The person now starts moving towards the centre of the platform. What will be the rotational speed (in rpm) of the platform when the person reaches its centre___________ Solution: $(9.00)$ Here $M_{0}=200 \mathrm{~kg}, m=80 \mathrm{~kg}$ Using conservation of angular momentum, $L_{i}=L_{f}$ $I_{1} \omega_{1}=I_...
Read More →What is a semiconductor?
Question: What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism. Solution: Semiconductors are substances having conductance in the intermediate range of 10-6to 104ohm1m1. The two main types of semiconductors are: (i)n-type semiconductor (ii)p-type semiconductor n-type semiconductor:The semiconductor whose increased conductivity is a result of negatively-charged electrons is called ann-type semiconductor. When the crystal of a group 14 elem...
Read More →Analysis shows that nickel oxide has the formula
Question: Analysis shows that nickel oxide has the formula Ni0.98O1.00. What fractions of nickel exist as Ni2+and Ni3+ions? Solution: The formula of nickel oxide is Ni0.98O1.00. Therefore, the ratio of the number of Ni atoms to the number of O atoms, Ni : O = 0.98 : 1.00 = 98 : 100 Now, total charge on 100 O2ions = 100 (2) = 200 Let the number of Ni2+ions bex. So, the number of Ni3+ions is 98 x. Now, total charge on Ni2+ions =x(+2) = +2x And, total charge on Ni3+ions = (98 x)(+3) = 294 3x Since,...
Read More →Copper crystallises into a fcc lattice with edge length
Question: Copper crystallises into a fcc lattice with edge length 3.61 108cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm3. Solution: Edge length,a= 3.61 108cm As the lattice is fcc type, the number of atoms per unit cell,z= 4 Atomic mass,M = 63.5 g mol1 We also know that, NA= 6.022 1023mol1 Applying the relation: $d=\frac{z \mathrm{M}}{a^{3} \mathrm{~N}_{\mathrm{A}}}$ $=\frac{4 \times 63.5 \mathrm{~g} \mathrm{~mol}^{-1}}{\left(3.61 \times 10^{-8} \mathr...
Read More →Find the ratio in which y-axis divides the line segment joining the points A(5, –6) and B(–1, 4) Also,
Question: Find the ratio in whichy-axis divides the line segment joining the pointsA(5, 6) andB(1, 4) Also, find the coordinates of the point of division. Solution: Section formula: if the point $(x, y)$ divides the line segment joining the points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ internally in the ratio $k: 1$, then the coordinates $(x, y)=\left(\frac{k x_{2}+x_{1}}{k+1}, \frac{k y_{2}+y_{1}}{k+1}\right)$ Letthe pointP(0,y)divides the line segment joining the pointsA(5...
Read More →If the radius of the octachedral void is r and radius of the atoms in close packing is R,
Question: If the radius of the octachedral void is r and radius of the atoms in close packing is R, derive relation between r and R. Solution: A sphere with centre O, is fitted into the octahedral void as shown in the above figure. It can be observed from the figure thatΔPOQ is right-angled POQ = 900 Now, applying Pythagoras theorem, we can write: $\mathrm{PQ}^{2}=\mathrm{PO}^{2}+\mathrm{OQ}^{2}$ $\Rightarrow(2 R)^{2}=(R+r)^{2}+(R+r)^{2}$ $\Rightarrow(2 R)^{2}=2(R+r)^{2}$ $\Rightarrow 2 \mathrm{...
Read More →Niobium crystallises in body-centred cubic structure.
Question: Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm3, calculate atomic radius of niobium using its atomic mass 93 u. Solution: It is given thatthe density of niobium,d= 8.55 g cm3 Atomic mass, M = 93 gmol1 As the lattice is bcc type, the number of atoms per unit cell,z= 2 We also know that, NA= 6.022 1023mol1 Applying the relation: $d=\frac{z \mathrm{M}}{a^{3} \mathrm{~N}_{\mathrm{A}}}$ $\Rightarrow a^{3}=\frac{z \mathrm{M}}{d \mathrm{~N}_{\mathrm{A}}}$ $=\fra...
Read More →The line segment joining the point A(2, 1) and B(5, –8) is trisected at the points P and Q such that P is nearer to A.
Question: The line segment joining the pointA(2, 1) andB(5, 8) is trisected at the pointsPandQsuch thatPis nearer toA. IfPalso lies on the line given by 2x y+k= 0, find the value ofk. Solution: Let the pointsA(2, 1) andB(5, 8) is trisected at the pointsP(x,y)andQ(a,b).Thus,AP=PQ=QBTherefore,PdividesABinternally in the ratio 1 : 2.Section formula: if the point (x,y) divides the line segment joining the points (x1,y1) and (x2,y2) internally in the ratiom:n, then the coordinates (x,y) = $\left(\fra...
Read More →A cubic solid is made of two elements P and Q.
Question: A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-centre. What is the formula of the compound? What are the coordination numbers of P and Q? Solution: It is given that theatoms of Q are present at the corners of the cube. Therefore, number of atoms of $Q$ in one unit cell $=8 \times \frac{1}{8}=1$ It is also given that theatoms of P are present at the body-centre. Therefore, number of atoms of P in one unit cell = 1 This means th...
Read More →Silver crystallises in fcc lattice.
Question: Silver crystallises in fcc lattice. If edge length of the cell is 4.07 108cm and density is 10.5 g cm3, calculate the atomic mass of silver. Solution: It is given that the edge length,a= 4.077 108cm Density,d= 10.5 g cm3 As the lattice is fcctype, the number of atoms per unit cell,z= 4 We also know that, NA= 6.022 1023mol1 Using therelation: $d=\frac{z \mathrm{M}}{a^{3} \mathrm{~N}_{\mathrm{A}}}$ $\Rightarrow \mathrm{M}=\frac{d a^{3} \mathrm{~N}_{\mathrm{A}}}{z}$ $=\frac{10.5 \mathrm{g...
Read More →Silver crystallises in fcc lattice.
Question: Silver crystallises in fcc lattice. If edge length of the cell is 4.07 108cm and density is 10.5 g cm3, calculate the atomic mass of silver. Solution: It is given that the edge length,a= 4.077 108cm Density,d= 10.5 g cm3 As the lattice is fcctype, the number of atoms per unit cell,z= 4 We also know that, NA= 6.022 1023mol1 Using therelation: $d=\frac{z \mathrm{M}}{a^{3} \mathrm{~N}_{\mathrm{A}}}$ $\Rightarrow \mathrm{M}=\frac{d a^{3} \mathrm{~N}_{\mathrm{A}}}{z}$ $=\frac{10.5 \mathrm{g...
Read More →Find, in terms of π, the length of the arc that subtends an angle
Question: Find, in terms of $\pi$, the length of the arc that subtends an angle of $30^{\circ}$ at the centre of a circle of radius $4 \mathrm{~cm}$. Solution: The arc lengthlof a sector of an anglein a circle of radius r is given by $l=\frac{\theta}{360^{\circ}} \times 2 \pi r$ It is given that $r=4 \mathrm{~cm}$ and $\theta=30^{\circ}$. Substituting the value of $r$ and $\theta$ in above equation, $l=\frac{30^{\circ}}{360^{\circ}} \times 2 \pi \times 4 \mathrm{~cm}$ $=\frac{2 \pi}{3} \mathrm{~...
Read More →Calculate the efficiency of packing in case of a metal crystal for
Question: Calculate the efficiency of packing in case of a metal crystal for (i)simple cubic (ii)body-centred cubic (iii)face-centred cubic (with the assumptions that atoms are touching each other). Solution: (i)Simple cubic In a simple cubic lattice, the particles are located only at the corners of the cube and touch each other along the edge. Let the edge length of the cube be a and the radius of each particle ber. So, we can write: a= 2r Now, volume of the cubic unit cell =a3 = (2r)3 = 8r3 We...
Read More →Point A(3, 1), B(5, 1), C(a, b) and D(4, 3) are vertices of a parallelogram ABCD.
Question: PointA(3, 1),B(5, 1),C(a, b) andD(4, 3) are vertices of a parallelogramABCD. Find the values ofaandb. Solution: Given:PointA(3, 1),B(5, 1),C(a, b) andD(4, 3) are vertices of a parallelogramABCD.Diagonals of a parallelogram bisect each other. Mid point ofAC= Mid point ofBD Mid point of $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is $\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$. Mid point of $A C=\left(\frac{3+a}{2}, \frac{1+b}{2}\right)$ Mid point of $B D=\...
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