250+ TOP MCQs on Particle in a Box and Answers

Engineering Physics Multiple Choice Questions on “Particle in a 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
Clarification: The simplest quantum-mechanical 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
c) 0 < X < L
d) x > L
Answer: c
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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 ___________
a) L/2
b) 2/L
c) (sqrt{L/2})
d) (sqrt{2/L})
Answer: d
Clarification: The wave function for the particle in a box is normalizable, when the value of the coefficient of sin is equal to (sqrt{2/L})
. 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
Clarification: 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) (frac{pi^2hbar^2}{2mL^2})
c) (frac{pi^2hbar^2}{2mL})
d) (frac{pi^2hbar}{2mL})
Answer: b
Clarification: The minimum energy possessed by a particle inside a box with infinitely hard walls is equal to (frac{pi^2hbar^2}{2mL^2}). 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) (sqrt{frac{2}{L}}sinfrac{nx}{L})
b) (sqrt{frac{2}{L}}sinfrac{npi x}{L})
c) (sqrt{frac{2}{L}}sinfrac{x}{L})
d) (sqrt{frac{2}{L}}sinfrac{pi x}{L})
Answer: b
Clarification: The wave function for the particle in a box is given by: (sqrt{frac{2}{L}}sinfrac{npi x}{L}). The Energy possessed by the particle is given by: (frac{n^2pi^2hbar^2}{2mL^2}).

10. The wave function for which quantum state is shown in the figure?
engineering-physics-questions-answers-particle-box-q10
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: 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
Clarification: Here, m = 9.1 X 10-31 kg, L = 10-9m.
Therefore, E = (frac{pi^2hbar^2}{2mL^2})
= 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.

250+ TOP MCQs on Lenses and Answers Pdf

Engineering Physics Questions and Answers focuses on “Lenses ”.

1. What is the SI unit for the power of the lens?
a) Watt
b) Dioptre
c) Meter
d) Centimeter

Answer: b
Clarification: SI unit for the power of lens is Dioptre. SI unit for focal length is metre. SI unit for quantifying the rate of energy transfer is watt.

2. The lens power is 1D, what will be its focal length in centimeter?
a) 100
b) 0.91
c) 0.1
d) 10

Answer: b
Clarification: Since, Lens power is reciprocal of focal length in meter, i.e.,
P=1/f
=> f=1/P
=> f=1/1
therefore, f=1 meter which is, 100 cm

3. Name the lens used in a simple microscope.
a) Biconvex
b) Biconcave
c) Plano-convex
d) Cylindrical

Answer: a
Clarification: Biconvex Lens is used in a simple microscope. Whereas, plano-convex Lens is used in a Kellner’s eyepiece, Ransden eyepiece and Gauss eyepiece.

4. If the magnification is positive and greater than unity, what does it indicate?
a) Virtual image
b) Real image
c) Distorted image
d) Erect image

Answer: a
Clarification: For virtual image, the magnification is positive and greater than unity.
Since, m = h2/h1 = length of image / length of object = distance of image / distance of object
For real image, the magnification is negative.

5. What will be the power of convex lens having focal length 50 cm?
a) +2 D
b) -2 D
c) +50 D
d) -6 D

Answer: a
Clarification: The reciprocal of focal length is power.
.i.e., P=1/f (f in meter)
=> P = 1/50 * 10-2 = 1/50 * 0.001
=> f = 2 D.
Foe convex lens P = positive. Therefore, P = +2 D.

6. What will be the focal length of a convex lens having power -1.5 D?
a) -66.66 cm
b) +66.66 cm
c) +1.5 cm
d) -1.5 cm

Answer: a
Clarification: The reciprocal of focal length is power.
.i.e., P=1/f (f in meter)
=> f = 1 / P * 10-2 = – {1/ 1.5* 10-2} = -{ 1/1.5 * 0.001}
=> f = – 1000 / 15
Therefore, f = -66.66 cm.

7. How will be the image formed, when an object is placed 12 cm from a convex lens, whose focal length is 10 cm?
a) Virtual and enlarged
b) Virtual and reduced in size
c) Real and enlarged
d) Real and reduced in size

Answer: c
Clarification: here, the object is beyond F and before 2F .i.e., in between F and 2F so, the image formed by convex lens will be real, inverted and magnified (or enlarged).

8. At what distance the object should be placed, when convex lens is used as simple magnifying glass?
a) Less than one focal length
b) Less than twice the focal length
c) More than one focal length
d) More than twice the focal length

Answer: a
9. What type of image is produced by a concave lens?
a) Always virtual and enlarged
b) Always real
c) Always virtual and reduced in size
d) Sometimes real and sometimes virtual

Answer: c
Clarification: Concave lens produces an image that is always virtual and reduced in size. While convex lens produces virtual image only when the object is in between focus and curvature.

10. At what distance the image will be formed when an object is placed 25 cm from a convex lens whose focal length is 10 cm?
a) 50 cm
b) 5.66 cm
c) 6.66 cm
d) 16.66 cm

Answer: d
Clarification: According to Gauss’s formula for lens:
=>1/v – 1/u = 1/f
=>1/v – 1/(-25) = 1/10
=>1/v = 1/10 – 1/25
=>1/v = 6/100
=>v = 100/6
therefore, v = 16.66 cm.

11. What is the least distance for distinct vision?
a) 25 m
b) 25 cm
c) 25 mm
d) 2.5 cm

Answer: b
Clarification: The least distance for distinct vision is 25 cm. It is defined as the minimum comfortable distance between the object and the eye lens to form a clear image.

12. A ray of light travels from a medium of refractive index m1 to a medium of refractive index m2. If “i” and “r” are the angle of incidence and refraction, then what is sin(i)/sin(r)?
a) m1
b) m2
c) m12
d) m21

Answer: d
Clarification: By Snell’s law:
m1*sin(i) = m2*sin(r)
=> sin(i)/sin(r) = m2/m1 = m21
therefore, sin(i)/sin(r) = m21.

250+ TOP MCQs on Difference Between Prism and Grating Spectra and Answers

Advanced Engineering Physics Questions and Answers focuses on “Difference Between Prism and Grating Spectra”.

1. Light utilization efficiency of prism and grating respectively is _________
a) High and High
b) High and Low
c) Low and High
d) Low and Low
Answer: b
Clarification: In a grating, the light with the same wavelength is dispersed in several directions as higher-order light. The prism generally has higher light utilization efficiency despite other losses.

2. Wavelength dependency for dispersion is constant for a prism.
a) True
b) False
Answer: b
Clarification: For a prism, the wavelength dependency for a prism is variable. It is high for UV while low for visible light. For a grating, however, it is high and approximately constant.

3. The spectrum observed for a grating is much purer than a prism.
a) True
b) False
Answer: a
Clarification: It is true that the spectrum observed by a plane transmission grating is purer than that of a prism. This is so because for the grating, it is due to diffraction while for the prism, it is due to dispersion.

4. The relationship between the angle of diffraction of violet, θv, and the angle of diffraction for red, θr, for a grating, is ________
a) θv > θr
b) θv < θr
c) θv = θr
d) No relationship
Answer: b
Clarification: For a diffraction grating, the angle of diffraction for violet is less than the one for red. However, with a prism, the angle of deviation for the violet rays of light is more than for the red rays of light.

5. How many spectrums can be achieved by grating and a prism respectively?
a) 1 and 1
b) More than 1 and 1
c) 1 and More than 1
d) More than 1 and More than 1
Answer: b
Clarification: In a grating, a number of spectra can be achieved on the two side of the principal maxima while for a prism, only one spectrum can be obtained.

6. The wavelength range for a grating is _______
a) 200 nm – 400 nm
b) 400 nm – 800 nm
c) 800 nm – 1200 nm
d) 200 – 800 nm
Answer: b
Clarification: For a grating, the wavelength range is from 400 nm to 800 nm. For a prism, the wavelength range is from 365 mm to 920 mm.

7. The phenomenon used for obtaining grating spectra is diffraction. Which phenomenon is used for obtaining prism spectra?
a) Diffraction
b) Interference
c) Scattering
d) Dispersion
Answer: d
Clarification: The prism spectra is obtained by dispersion. It is the splitting of light into seven colors. Violet has the greatest deviation in this case, while for grating spectra violet has the least deviation.

8. Identify X and Y.
advanced-engineering-physics-questions-answers-q8
a) X: Grating Y: Prism
b) X: Prism Y: Grating
c) X: Grating Y: Slit
d) X: Slit Y: Grating
Answer: a
Clarification: X is a grating, as we can spectrums are observed on both sides of the principal maxima. While Y is grating, as only one spectrum is observed.

9. Which of the following lines represent the resolution of a prism?
engineering-physics-questions-answers-difference-between-prism-grating-spectra-q9
a) X
b) Y
c) Z
d) None
Answer: a
Clarification: In the following figure, X is the resolution of the prism. Y is the resolution of diffraction grating and Z is the QE of a camera.

10. Do the spectra observed by grating and prism depends on the nature of the material?
a) Yes for both
b) Yes for grating
c) Yes for prism
d) No for both
Answer: c
Clarification: The prism spectra observed depends on the nature of the material of the prism. While in the case of a diffraction grating, the spectra are independent on the nature of the material of the prism.

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250+ TOP MCQs on Helium Neon Laser and Answers

Engineering Physics Multiple Choice Questions on “Helium Neon Laser”.

1. Which of the following is a four-level laser?
a) ND: YAG
b) Ruby
c) He-Ne
d) Argon laser
Answer: c
Clarification: He-Ne laser is a four-level laser. Whenever an electric discharge is passed through the gas, the helium atoms gets to higher energy states, as the concentration of helium atoms is higher. It was one of the first successfully operated laser.

2. The difference between He-Ne Laser is __________
a) It gives pulsed output
b) It gives a non-continuous laser beam
c) It gives a continuous laser beam
d) No difference
Answer: c
Clarification: As we know, the ruby laser beam is discontinuous and pulsed. However, the He-Ne laser beam is a continuous laser beam. Also, it is a four-level laser while ruby laser is a three-level laser.

3. He-Ne laser is a type of ____________
a) Solid laser
b) Liquid laser
c) Gas laser
d) Diode laser
Answer: c
Clarification: He-Ne laser is the most widely used laser. It was the first continuous laser generated. Gas lasers are less prone to damage by overheating, as compared to solid laser.

4. Which pumping method is used in He-Ne laser?
a) Optical Pumping
b) Electrical Excitation
c) Chemical Pumping
d) Direct Conversion
Answer: b
Clarification: Generally, in He-Ne laser, an electric discharge is used to excite the atoms of the active medium. This process is known as Electrical Excitation and is normally used in gas lasers.

5. The He-Ne laser operates at a wavelength of ____________
a) 540 nm
b) 632 nm
c) 690 nm
d) 717 nm
Answer: b
Clarification: The helium-neon laser operates at a wavelength of 632.8 nanometers (nm), in the red portion of the visible spectrum. It is the most widely used gas laser.

6. The number of photons emitted for a 2.5 mW He-Ne laser is __________
a) 4.9 X 1015
b) 5.9 X 1015
c) 6.9 X 1015
d) 7.9 X 1015
Answer: b
Clarification: We know, for a He-Ne laser, λ = 6328 Å = 6.328 X 10-7 m
Power = 2.5 mW = 2.5 X 10-3 E
Energy = Power X Time = 0.0025 X 1
= 0.0025
Number of photons in each pulse = Energy X λ / hc
= 0.0025 X 6.6 X 10-7/6.6 X 10-34 X 3 X 108
= 7.9 X 1015.

7. The output of a He-Ne laser has pulse duration of 15 ms and average output power of 1 W per pulse. How much energy is released per pulse?
a) 5 mJ
b) 10 mJ
c) 15 mJ
d) 20 mJ
Answer: c
Clarification: As we know, Energy = Power X Time
= 1 W X 15 X 10-3 s
= 15 mJ.

8. When the transition takes place from En6 -> En5, what is the wavelength of produced beam?
a) 6328 Å
b) 33913 Å
c) 11523 Å
d) 7550 Å
Answer: b
Clarification: When the lasing transition is from En6 -> En5, it produces a beam of 33913 Å. The opper level is same in this case and in the 6328 Å transition.

9. He-Ne laser is used in Holography.
a) True
b) False
Answer: a
Clarification: He-Ne laser is highly coherent and monochromatic. Due to this, it is used in holography, spectrometers, prints, scanners, etc. It is widely used in Laboratories.

10. Which of the following shows the laser transition in Neon in He-Ne laser?
a) engineering-physics-questions-answers-helium-neon-laser-q10a
b) engineering-physics-questions-answers-helium-neon-laser-q10b
c) engineering-physics-questions-answers-helium-neon-laser-q10c
d) engineering-physics-questions-answers-helium-neon-laser-q10d
Answer: b
Clarification: The transition shown in the following figure is the correct transition that takes place in the Ne atoms. Between two or more layer, as soon as population inversion is achieved, the lasing action starts.
engineering-physics-questions-answers-helium-neon-laser-q10b

250+ TOP MCQs on Free Electron Theory and Answers

Engineering Physics Multiple Choice Questions on “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: b
Clarification: 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
Answer: a
Clarification: 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
Clarification: 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
Clarification: 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
c) Photo-electric effect
d) Classical free electron theory
Answer: c
Clarification: Classical theory states that all free electrons will absorb energy. This theory cannot explain the photo electric effect.

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
Answer: b
Clarification: 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
Clarification: 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
c) 6.9973×10-3 m2/Vs
d) 0.69973m/s
Answer: c
Clarification: Mobility of the electrons = 1/ƿne
Mobility = 6.9973×10-3 m2/Vs.

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: d
Clarification: Drift velocity = μE
Drift velocity = 3.5×10-3×0.5 = 1.75×10-3m/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
Clarification: EF = KB TF = 12 (mv_F^2)
vF = (sqrt{frac{2×K_B×T_F}{m}})
vF = 0.8633×106m/s.

250+ TOP MCQs on Nano Phase Materials and Answers

Engineering Physics Multiple Choice Questions on “Nano Phase Materials”.

1. In which of the following the atoms do not move from each other?
a) Shape memory alloys
b) Nano materials
c) Dielectrics
d) Static materials
Answer: b
Clarification: There exists a special type of material in which the atoms do not move away from each other and its size will be in the order of 1-100nanometers. These new materials are called nano-materials.

2. Which of the following uses radio frequency to produce nano-particles?
a) Plasma arching
b) Chemical vapour deposition
c) Sol-gel technique
d) Electro deposition
Answer: a
Clarification: Nano particles are produced by generating plasma using Radio frequency heating coils in plasma arching method. It consists of an evacuated chamber, wounded by high voltage RF coils.

3. Which of the following methods can be used to produce nano-powders of oxides?
a) Plasma arching
b) Sol-gel technique
c) Chemical vapour deposition
d) Mechanical crushing
Answer: c
Clarification: Chemical vapour deposition is used to prepare nano-powder. In this technique, initially, the material is heated to form a gas i=and is allowed to deposit on a solid surface under vacuum condition, which forms nano-powders on the surface of the solid.

4. Which of the following is used to make both nano-particles and nano-powders?
a) Chemical vapour deposition
b) Sol-gel technique
c) Plasma arching
d) Electro deposition
Answer: b
Clarification: Sol-gel technique is based on the hydrolysis of liquid precursors and the formation of colloidal solutions. Only this method can be used for the preparation of nano-particles and nano-powders.

5. Which method can be used to prepare iron nitriles nano-crystals using ammonia gas?
a) Pulsed laser deposition
b) Sol-gel technique
c) Electro-deposition
d) Mechanical crushing
Answer: d
Clarification: Mechanical crushing is the method sued in the preparation of metal oxide nano crystals. In this method, small balls are allowed to rotate inside a drum and are made to fall on a solid with a high gravitational force that crushes the solid into nano-crystals. Thus iron nitriles nano-crystal can be made by using mechanical crushing.

6. Nano-particles exhibit super plastic behaviour.
a) True
b) False
Answer: a
Clarification: The hardness of nano-phase materials varies from material to material. This may be due to the phase transformation, stress relief, density and grain boundaries. Therefore nano-particles exhibit super plastic behaviour.

7. Which of the following is used to modify the optical properties of a material system?
a) Electricity
b) Magnetic field
c) Pressure
d) Light
Answer: d
Clarification: In nonlinear optics, the modification of the optical properties of a material system is made by light. This is because according to electro optic effects when the light is passed through a material it changes the properties of the medium.

8. Find the odd one out.
a) Frequency mixing
b) Second-harmonic generation
c) Optical mixing
d) Raman and Rayleigh scattering
Answer: a
Clarification: The rest of the mentioned effects exhibit Bire fingence. When light passes through a material the incident ray splits into two rays, one of the same wavelength and the other of a different wavelength. This phenomenon is called double refraction. It is observed in NLO materials. Whereas frequency mixing is a passive material.

9. Which of the following is used in electro optic modulators?
a) Lithium tantalite
b) Barium sodium niobate
c) Lithium niobate
d) Lithium sodium niobate
Answer: c
Clarification: When an ordinary light is passed through the lithium niobate crystal, it is converted into polarised extraordinary light. Hence it is used in electro optic modulator.