PH8252 Physics for Information Science MCQ PDF

Preview

Summary

SEMESTER Exam 2021...

Description

PH8252 Physics for Information Science

Regulations 2017

Answer: b Explanation: The conducting property of a solid is not a function of a total number of electrons in the metal, but it is due to the number of valance electrons called free electrons. 2. The free electrons collide with the lattice elastically. a) True b) False

PH8252 Physics for Information Science CSE & IT 2nd Semester Anna University Regulations 2017 UNIT I ELECTRICAL PROPERTIES OF MATERIALS TOPIC 1.1 CLASSICAL FREE ELECTRON THEORY 1. What does the conductivity of metals depend upon? a) The nature of the material b) Number of free electrons c) Resistance of the metal d) Number of electrons

Answer: a Explanation: The free electrons move randomly in all directions. The free electrons collide with each other and also with the lattice Elastically, without loss in energy. 3. What happens to the free electrons when an electric field is applied? a) They move randomly and collide with each other b) They move in the direction of the field c) They remain stable d) They move in the direction opposite to that of the field Answer: d Explanation: The free electrons move in the direction opposite to that of field direction. Since they are assumed to be a perfect gas as they obey classical kinetic theory of gases and the electron velocities in the metal obey the Maxwell-Boltzmann statistics. 4. Thermal conductivity is due to photons. a) True b) False Answer: a Explanation: Thermal conductivity is due to both photons and free electrons and not just photons. 5. Which of the following theories cannot be explained by classical theory? a) Electron theory b) Lorentz theory

Downloaded From: https://cse-r17.blogspot.com

1

PH8252 Physics for Information Science

Regulations 2017

c) Photo-electric effect d) Classical free electron theory

c) 6.9973×10-3 m2/Vs d) 0.69973m/s

Answer: c Explanation: Classical theory states that all free electrons will absorb energy. This theory cannot explain the photo electric effect.

Answer: c Explanation: Mobility of the electrons = 1/ ƿne Mobility = 6.9973×10-3 m2/Vs.

6. Which of the following theories can be adopted to rectify the drawbacks of classical theory? a) Compton theory b) Quantum theory c) Band theory d) Electron theory

9. Calculate the drift velocity of the free electrons with mobility of 3.5×10-3 m2/Vs in copper for an electric field strength of 0.5 V/m. a) 3.5 m/s b) 1.75×103 m/s c) 11.5 m/s d) 1.75×10-3 m/s

Answer: b Explanation: In classical theory, the properties of metals, such as electrical and thermal conductivities are well explained on the assumption that the electrons in the metal freely moves like the particles of a gas. Hence it can be used to rectify the drawbacks of classical theory. 7. What is the level that acts as a reference which separated the vacant and filled states at 0K? a) Excited level b) Ground level c) Valance orbit d) Fermi energy level Answer: d Explanation: Fermi energy level is the maximum energy level up to which the electrons can be filled at 0K. Thus it acts as reference level which separated the vacant and filled states at 0K. 8. A uniform silver wire has a resistivity of 1.54×10-18 ohm/m at room temperature. For an electric field along the wire of 1 volt/cm. Compute the mobility, assuming that there are 5.8×1028 conduction electrons/m3. a) 1.54 m2/Vs b) 6.9973m2/Vs

Answer: d Explanation: Drift velocity = μE Drift velocity = 3.5×10-3×0.5 = 1.75×103m/s. 10. The Fermi temperature of a metal is 24600K. Calculate the Fermi velocity. a) 0.5m/s b) 1.38m/s c) 0.8633×106m/s d) 9.11×10-3m/s Answer: c Explanation: EF = KB TF = 1⁄2 vF = vF = 0.8633×106m/s.

TOPIC 1.2 EXPRESSION FOR ELECTRICAL CONDUCTIVITY TOPIC 1.3 THERMAL CONDUCTIVITY, EXPRESSION 1. Thermal conductivity is defined as the heat flow per unit time

Downloaded From: https://cse-r17.blogspot.com

2

PH8252 Physics for Information Science

a) When the temperature gradient is unity b) Across the wall with no temperature c) Through a unit thickness of the wall d) Across unit area where the temperature gradient is unity Answer: d Explanation: Thermal conductivity of a material is because of migration of free electrons and lattice vibrational waves. 2. Mark the matter with least value of thermal conductivity a) Air b) Water c) Ash d) Window glass Answer: a Explanation: For air, it is .024 W/ m degree i.e. lowest. 3. Which one of the following forms of water have the highest value of thermal conductivity? a) Boiling water b) Steam c) Solid ice d) Melting ice Answer: c Explanation: For ice, it is 2.25 W/m degree i.e. maximum. 4. The average thermal conductivities of water and air conform to the ratio a) 50:1 b) 25:1 c) 5:1 d) 15:1 Answer: b Explanation: For water, it is 0.55-0.7 W/m degree and for air it is .024 W/m degree. 5. Identify the very good insulator a) Saw dust b) Cork

Regulations 2017

c) Asbestos sheet d) Glass wool Answer: d Explanation: Glass wool has a lowest thermal conductivity of 0.03 W/m degree amongst given option. 6. Most metals are good conductor of heat because of a) Transport of energy b) Free electrons and frequent collision of atoms c) Lattice defects d) Capacity to absorb energy Answer: b Explanation: For good conductors, there must be electrons that are free to move. 7. Heat conduction in gases is due to a) Elastic impact of molecules b) Movement of electrons c) EM Waves d) Mixing of gases Answer: a Explanation: If there is elastic collision then after sometime molecules regain its natural position. 8. The heat energy propagation due to conduction heat transfer will be minimum for a) Lead b) Water c) Air d) Copper Answer: c Explanation: It is because air has lowest value of thermal conductivity amongst given options. 9. Cork is a good insulator because a) It is flexible b) It can be powdered c) Low density d) It is porous

Downloaded From: https://cse-r17.blogspot.com

3

PH8252 Physics for Information Science

Answer: d Explanation: Cork has thermal conductivity in the range of 0.05-0.10 which is very low so it can be porous. 10. Choose the false statement a) For pure metal thermal conductivity is more b) Thermal conductivity decreases with increase in the density of the substance c) Thermal conductivity of dry material is lower than that of damp material d) Heat treatment causes variation in thermal conductivity Answer: b Explanation: Thermal conductivity increase with increase in the density of a substance.

TOPIC 1.4 WIEDEMANN-FRANZ LAW TOPIC 1.5 ELECTRONS IN METALS TOPIC 1.6 PARTICLE IN A THREE DIMENSIONAL BOX 1. The walls of a particle in a box are supposed to be ____________ a) Small but infinitely hard b) Infinitely large but soft c) Soft and Small d) Infinitely hard and infinitely large Answer: d Explanation: The simplest quantummechanical problem is that of a particle in a box with infinitely hard walls and are infinitely large. 2. The wave function of the particle lies in which region? a) x > 0 b) x < 0

Regulations 2017

c) 0 < X < L d) x > L Answer: c Explanation: The particle cannot exist outside the box, as it cannot have infinite amount of energy. Thus, it’s wave function is between 0 and L, where L is the length of the side of the box. 3. The particle loses energy when it collides with the wall. a) True b) False Answer: b Explanation: The total energy of the particle inside the box remains constant. It does not loses energy when it collides with the wall. 4. The Energy of the particle is proportional to __________ a) n b) n-1 c) n2 d) n-2 Answer: c Explanation: In a particle inside a box, the energy of the particle is directly proportional to the square of the quantum state in which the particle currently is. 5. For a particle inside a box, the potential is maximum at x = ___________ a) L b) 2L c) L/2 d) 3L Answer: a Explanation: In a box with infinitely high barriers with infinitely hard walls, the potential is infinite when x = 0 and when x = L. 6. The Eigen value of a particle in a box is ___________

Downloaded From: https://cse-r17.blogspot.com

4

PH8252 Physics for Information Science

a) L/2 b) 2/L c) d) Answer: d Explanation: The wave function for the particle in a box is normalizable, when the value of the coefficient of sin is equal to

Regulations 2017

Answer: b Explanation: The wave function for the particle in a box is given by: . The Energy possessed by the particle is given by: . 10. The wave function for which quantum state is shown in the figure?

. It is the Eigen value of the wave function. 7. Particle in a box can never be at rest. a) True b) False Answer: a Explanation: If the particle in a box has zero energy, it will be at rest inside the well and it violates the Heisenberg’s Uncertainty Principle. Thus, the minimum energy possessed by a particle is not equal to zero. 8. What is the minimum Energy possessed by the particle in a box? a) Zero b) c) d) Answer: b Explanation: The minimum energy possessed by a particle inside a box with infinitely hard walls is equal to . The particle can never be at rest, as it will violate Heisenberg’s Uncertainty Principle. 9. The wave function of a particle in a box is given by ____________ a)

a) 1 b) 2 c) 3 d) 4 Answer: b Explanation: The shown wave function is for the 2nd principal quantum number, i.e., it is the wave function for the state when n = 2. 11. Calculate the Zero-point energy for a particle in an infinite potential well for an electron confined to a 1 nm atom. a) 3.9 X 10-29 J b) 4.9 X 10-29 J c) 5.9 X 10-29 J d) 6.9 X 10-29 J Answer: c Explanation: Here, m = 9.1 X 10-31 kg, L = 10-9m. Therefore, E = = 3.14 X 3.14 X 1.05 X 1.05 X 10-68/ 2 X 9.1 X 10-31 X 10-9 = 5.9 X 10-29 J.

b) c)

TOPIC 1.7 DEGENERATE STATES

d)

Downloaded From: https://cse-r17.blogspot.com

5

PH8252 Physics for Information Science

TOPIC 1.8 FERMI- DIRAC STATISTICS 1. Fermi-Dirac statistics is for the ________ a) Distinguishable particles b) Symmetrical Particles c) Particles with half integral spin d) Particles with integral spin

Regulations 2017

c) Symmetric d) Differentiable Answer: c Explanation: The particles which have antisymmetric wave function having half odd integral spin number and obey Pauli’s principle are called fermions.

Answer: c Explanation: The Maxwell-Boltzmann statistics is for the distinguishable particles, which are basically the classical particles like atoms and molecules.

5. Fermi-Dirac statistics cannot be applied to ________ a) Electrons b) Photons c) Fermions d) Protons

2. The difference between fermions and bosons is that bosons do not obey ______ a) Aufbau principle b) Pauli’s Exclusion Principle c) Hund’s Rule of Maximum Multiplicity d) Heisenberg’s Uncertainty Principle

Answer: b Explanation: Fermi-Dirac Statistics can be applied to particles having half odd integral spin number and obey Pauli’s principle which are electrons, fermions and protons. Photon has an integral spin number.

Answer: b Explanation: The particles that follow Pauli’s Exclusion Principle are defined as Fermions while that don’t are called bosons. Bosons have an integral spin number.

6. The distribution function is given by ________ a) b) c)

3. The Maxwell-Boltzmann law is given by the expression ni = ________ a) b) c) d) Answer: c Explanation: The correct expression for the Maxwell-Boltzmann law is ni = , where α depends on the volume and the temperature of the gas and β is equal to 1/kT. 4. The wave function of fermions is not _________ a) Continuous b) Single Valued

d) Answer: a Explanation: The distribution function is given by , where A = The energy EF is called the Fermi energy and is constant for a given system. 7. At T > 0K, the probability of a state with E > EF filled is zero. a) True b) False Answer: b Explanation: At T > 0 K, the probability that a state with E > EF is filled is 1⁄2. Hence, fermi energy is the energy at which the

Downloaded From: https://cse-r17.blogspot.com

6

PH8252 Physics for Information Science

Regulations 2017

probability of occupation is 1⁄2 at any temperature above 0 K. 8. The expression for mean energy is given by ________ a) b) c) d) Answer: a Explanation: The expression for the mean energy of an electron at absolute zero is given by . The expression is called the zero point energy. 9. For all the quantum states with energy greater than Fermi energy to be empty in a Fermi-Dirac system, the temperature should be ______ a) 273 K b) 373 K c) 0 K d) 100 K Answer: c Explanation: We know that the Fermi-Dirac distribution is given by: FFD(E) = For all the quantum states with energy greater than Fermi energy to be empty, FFD(E) = 0, for E > EF and FFD(E) = 1, for E < EF Therefore, for E < EF As, E < EF, E- EF < 0. Therefore, to satisfy the given statement, T=0K Thus, we can define Fermi energy as the energy of the uppermost occupied level at 0 K. 10. What is the relationship between T1 and T2?

a) T1 > T2 b) T1 < T2 c) T1 = T2 d) Insufficient Information Answer: b Explanation: The given figure shows the variation of Fermi-Dirac distribution with energy E. In this case, T2 > T1, according to the expression: FFD(E) = . 11. The density of silver is 10.5 g/cm3 and its atomic weight is 108. If each atom contributes one electron for conduction, what is the fermi energy? a) 2.12 eV b) 3.31 eV c) 4.69 eV d) 5.51 eV Answer: d Explanation: Fermi Energy, Here, N /V = 10.5 X 6.02 X 1023/108 = 5.85 X 1028 m-3 Therefore, = 8.816 X 10-19J = 5.51 eV. 12. The Fermi energy of a material is 3.45 eV. What is the zero-point energy of the material? a) 1.02 eV b) 2.07 eV

Downloaded From: https://cse-r17.blogspot.com

7

PH8252 Physics for Information Science

Regulations 2017

c) 3.45 eV d) 4.16 eV Answer: b Explanation: As we know, the fermi energy of the material is 3.45 eV. Now, zero-point energy = = 3/5 X 3.45 eV = 2.07 eV. 13. The average energy of one electron silver is 3.306 eV. What is the fermi-energy of Silver at 0 K? a) 2.32 eV b) 3.78 eV c) 4.12 eV d) 5.51 eV Answer: d Explanation: We know, Average Energy, E = Here, E = 3.306 eV, N = 1 Therefore, we get: EF = 5E/3 = 5.51 eV. 14. In Fermi-Dirac Statistics, one energy state can be occupied by more than one particle. a) True b) False Answer: b Explanation: In Fermi-Dirac statistics, one energy state can be occupied by only one particle. Therefore, when one state is filled, the particle will go on to another state. 15. Which of the following is the curve for Fermi-Dirac statistics?

a) X b) Y c) Z d) None Answer: c Explanation: As seen the curve the order of the ration of f(K) with E/kT is in the order X > Y > Z. Here, Y is Maxwell-Boltzmann Statistics, X is Bose-Einstein Statistics and Z is Fermi-Dirac Statistics.

TOPIC 1.9 DENSITY OF ENERGY STATES TOPIC 1.10 ELECTRON IN PERIODIC POTENTIAL 1. The electron probability density is greatest at ____________ a) r = 0 b) r = n c) r = l d) r = me Answer: a Explanation: The probability of finding an electron is always the greatest at r = 0. As we move away from the nucleus, the probability decreases. 2. Which of the following is the correct expression for the total number of nodes? a) n – 1 b) l – 1

Downloaded From: https://cse-r17.blogspot.com

8

PH8252 Physics for Information Science

c) l + 1 d) n + 1 Answer: a Explanation: The principal quantum number, n, gives the shell to which the electron belongs. N – 1 gives us the total number of nodes in the orbital. 3. What orbital never has a zero probability of finding electrons? a) s b) px c) dxy d) dz2 Answer: d Explanation: The orbital dz2 has no nodal plane i.e., the probability of finding an electron is never zero for this orbital. For others, the nodal plane is given by n -1. 4. Nodes are the plane where the probability of finding an electron is 1. a) True b) False Answer: b Explanation: Nodal planes are described as the planes where the probability of finding an electron is not equal to zero. The total number of nodes in an orbital is n -1. 5. If Ψ is the wave function, the probability density function is given by _____________ a) |Ψ| b) |Ψ|2 c) |Ψ|3 d) |Ψ|4 Answer: b Explanation: The probability density function for a wave function, Ψ, is given by | Ψ|2. It is always greater than or equal to zero and less than or equal to one.

Regulations 2017

6. For a 3p orbital, what are the total number of nodes? a) 3 b) 2 c) 1 d) 0 Answer: b Explanation: As we know, the total number of nodes in an orbital is given by n – 1. In 3p orbital, n = 3. Thus, total number of nodes is 2. 7. The probability of finding an electron is along two directions in which orbital? a) s b) p c) d d) p Answer: c Explanation: For a d-orbital, the probability of finding an electron is along two directions. For p orbitals, the probability of finding an electron is along one direction only and for an s-orbital it is uniform in every direction. 8. Arrange in the increasing number of radial nodes: 3p, 4s, 1s, 5d. a) 1s < 3p < 4s < 5d b) 1s < 4s < 3p < 5d c) 1s < 3p < 5d < 4s d) 5d < 1s < 3p < 4s Answer: c Explanation: The radial nodes in an orbital is given by n – l -1. For 3p it is 2, for 5d it is 3, for 4s it is 4 and for 1s it is 1. Thus, the correct order is: 1s < 3p < 5d < 4s. 9. Which of the following can be the quantum numbers for an orbital? a) n = 4, l = 4, m = 3 b) n = 2, l = 3, m = 1 c) n = 3, l = 2, m = -1 d) n = 3, l = 0, m = -3

Downloaded From: https://cse-r17.blogspot.com

9

PH8252 Physics for Information Science

Answer: c Explanation: In the given options, option n = 3, l = 2, m = -1 is the correct option because in this the value of l is between 0 – n-1 and value of m is between –l to +l. 10. For which quantum number, the probability of finding an electron is most? a) 1 b) 2 c) 3 d) 4 Answer: a Explanation: The probability of finding an electron in the first orbit is maximum. As we know, s orbitals have a uniform probability of finding an electron in every direction and 1s has no nodal plane. 11. Identify the orbital for the following probability density graph.

Regulations 2017

Answer: a Explanation: The s-orbitals have uniform probability density function in every direction. Its shape is spherical. Also, its azimuthal quantum number is zero. Therefore, all the nodes present in s-subshell are radial nodes. 13. Calculate the minimum uncertainty in the momentum of a 4He atom confined to 0.40 nm. a) 2.02 X 10-25 kg m/s b) 2.53 X 10-25 kg m/s c) 2.64 X 10-25 kg m/s d) 2.89 X 10-25 kg m/s Answer: c Explanation: We know that 4He atom is somewhere in the 0.40 nm region, therefore, Δx = 0.40 nm. Using, Δpx ≥ ℏ/Δ*x For minimum uncertainty, Δpx = 6.626 X 1034

Js/2π X 0.40 X 10-9 = 2.64 X 10-25 kg m/s.

a) 1s b) 2s c) 2p d) 3s Answer: b Explanation: The given figure is the probability density graph for the 2s orbital. As the graph touches the x-axis once, it means there is one nodal plane, i.e., there is one plane where the probability of finding an electron is zero. 12. S-orbitals have no angular node. a) True b) False

14. Which of the following is the correct expression for the magnetic moment of the electron? a) b) c) d) Answer: c Explanation: The magnetic moment of an electron is given by the expression: . Thus, when the principal quantum number, n, is zero the magnetic moment is zero as well. 15. The probability of finding an electron is zero in pxy along __________ a) x-axis b) y-axis

Downloaded From: https://cse-r17.blogspot.com

10

PH8252 Physics for Information Science

Regulations 2017

c) z-axis d) ne...