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101. Which scientist first demonstrated the wave nature of light through the double-slit experiment?
ⓐ. James Clerk Maxwell
ⓑ. Isaac Newton
ⓒ. Albert Einstein
ⓓ. Thomas Young
Correct Answer: Thomas Young
Explanation: In 1801, Thomas Young performed the double-slit experiment, showing that light produces an interference pattern characteristic of waves. Newton supported the particle theory, Maxwell developed electromagnetic theory, and Einstein later explained the photoelectric effect.
102. Which property of light is explained by the phenomenon of interference?
ⓐ. Wave nature of light
ⓑ. Particle nature of light
ⓒ. Quantization of energy
ⓓ. Emission of photons from metals
Correct Answer: Wave nature of light
Explanation: Interference patterns arise when two coherent light waves overlap, creating regions of constructive and destructive superposition. This is possible only if light behaves as a wave.
103. Which equation represents the condition for constructive interference in a double-slit experiment?
ⓐ. $d \sin \theta = (2n+1)\dfrac{\lambda}{2}$
ⓑ. $d \sin \theta = n\lambda$
ⓒ. $d \cos \theta = n\lambda$
ⓓ. $d \tan \theta = (n+1)\lambda$
Correct Answer: $d \sin \theta = n\lambda$
Explanation: For constructive interference, the path difference between two waves must be an integral multiple of wavelength, i.e., $n\lambda$. Odd multiples of $\lambda/2$ give destructive interference.
104. Which experiment confirms the diffraction of light?
ⓐ. Rutherford’s gold foil experiment
ⓑ. Young’s double-slit experiment
ⓒ. Fresnel’s single-slit experiment
ⓓ. Millikan’s oil drop experiment
Correct Answer: Fresnel’s single-slit experiment
Explanation: Diffraction is bending of light around edges. Fresnel’s single-slit experiment produced characteristic diffraction fringes, proving light behaves as a wave.
105. The polarization of light demonstrates that:
ⓐ. Light is a longitudinal wave
ⓑ. Light has only electric field
ⓒ. Light is a stream of particles
ⓓ. Light is a transverse wave
Correct Answer: Light is a transverse wave
Explanation: Only transverse waves can be polarized. Since light exhibits polarization, it proves that light waves are transverse electromagnetic waves, with electric and magnetic fields oscillating perpendicular to propagation.
106. Which electromagnetic theory of light was given by Maxwell?
ⓐ. Light is made of photons only
ⓑ. Light is an electromagnetic wave traveling in vacuum
ⓒ. Light is corpuscular in nature
ⓓ. Light is always longitudinal
Correct Answer: Light is an electromagnetic wave traveling in vacuum
Explanation: Maxwell showed that oscillating electric and magnetic fields propagate through space as electromagnetic waves at the speed of light, unifying optics and electromagnetism.
107. Which phenomenon cannot be explained by the wave theory of light?
ⓐ. Diffraction
ⓑ. Interference
ⓒ. Photoelectric effect
ⓓ. Polarization
Correct Answer: Photoelectric effect
Explanation: Wave theory cannot explain the immediate ejection of electrons in the photoelectric effect. Einstein explained it using photons (particle nature). Wave theory explains A, B, and D successfully.
108. In Young’s double-slit experiment, the distance between slits is $d$, wavelength is $\lambda$, and distance to screen is $D$. The fringe width is given by:
ⓐ. $\dfrac{d}{D\lambda}$
ⓑ. $\dfrac{D\lambda}{d}$
ⓒ. $\dfrac{\lambda}{dD}$
ⓓ. $\dfrac{\lambda}{d}$
Correct Answer: $\dfrac{D\lambda}{d}$
Explanation: Fringe width is given by $\beta = \dfrac{D\lambda}{d}$. Larger screen distance and wavelength increase fringe spacing, while smaller slit separation increases fringe separation.
109. Which color of visible light diffracts the most?
ⓐ. Red
ⓑ. Violet
ⓒ. Blue
ⓓ. Green
Correct Answer: Red
Explanation: Amount of diffraction depends on wavelength. Red light has the longest wavelength in the visible spectrum, so it diffracts the most. Violet has the shortest wavelength and diffracts the least.
110. Which principle explains the interference of light waves?
ⓐ. Uncertainty principle
ⓑ. Pauli’s exclusion principle
ⓒ. Principle of superposition
ⓓ. Law of conservation of energy
Correct Answer: Principle of superposition
Explanation: Interference occurs when two or more light waves overlap, and their resultant displacement is given by the algebraic sum of individual displacements (superposition principle).
111. Which relation connects wavelength $\lambda$, frequency $\nu$, and velocity $c$ of electromagnetic waves?
ⓐ. $c = \dfrac{\nu}{\lambda}$
ⓑ. $c = \dfrac{\lambda}{\nu}$
ⓒ. $c = \lambda + \nu$
ⓓ. $c = \lambda \nu$
Correct Answer: $c = \lambda \nu$
Explanation: The velocity of an electromagnetic wave equals the product of its wavelength and frequency. In vacuum, $c = 3 \times 10^8$ m/s. The other options are incorrect rearrangements.
112. What is the SI unit of wave number?
ⓐ. m
ⓑ. Hz
ⓒ. m$^{-1}$
ⓓ. s
Correct Answer: m$^{-1}$
Explanation: Wave number is defined as reciprocal of wavelength: $$\tilde{\nu} = \frac{1}{\lambda}$$ Its SI unit is per meter (m$^{-1}$).
113. If the wavelength of a wave is doubled, its frequency (in vacuum):
ⓐ. Remains the same
ⓑ. Becomes double
ⓒ. Becomes half
ⓓ. Becomes zero
Correct Answer: Becomes half
Explanation: Since $c = \lambda \nu$, if wavelength doubles, frequency must become half to keep velocity constant in vacuum.
114. The frequency of visible light of wavelength 600 nm is approximately:
ⓐ. $3.0 \times 10^8 \,\text{Hz}$
ⓑ. $5.0 \times 10^{14} \,\text{Hz}$
ⓒ. $6.0 \times 10^6 \,\text{Hz}$
ⓓ. $1.0 \times 10^{10} \,\text{Hz}$
Correct Answer: $5.0 \times 10^{14} \,\text{Hz}$
Explanation: Frequency is given by $\nu = \dfrac{c}{\lambda}$. $$\nu = \frac{3.0 \times 10^8}{600 \times 10^{-9}} = 5.0 \times 10^{14} \,\text{Hz}$$
115. Which parameter of a wave is directly proportional to photon energy?
ⓐ. Wavelength
ⓑ. Velocity
ⓒ. Frequency
ⓓ. Amplitude
Correct Answer: Frequency
Explanation: Photon energy is given by $E = h\nu$. Higher frequency means higher photon energy. Wavelength is inversely related ($E \propto \dfrac{1}{\lambda}$).
116. If the velocity of electromagnetic radiation in vacuum is constant, what happens when frequency increases?
ⓐ. Wavelength increases
ⓑ. Wavelength decreases
ⓒ. Wavelength remains constant
ⓓ. Wave number decreases
Correct Answer: Wavelength decreases
Explanation: Since $c = \lambda \nu$, for constant $c$, wavelength is inversely proportional to frequency. Higher frequency corresponds to shorter wavelength.
117. Which of the following correctly expresses wave number?
ⓐ. $\tilde{\nu} = \lambda \nu$
ⓑ. $\tilde{\nu} = \dfrac{\nu}{c}$
ⓒ. $\tilde{\nu} = \dfrac{1}{\nu}$
ⓓ. $\tilde{\nu} = \dfrac{1}{\lambda}$
Correct Answer: $\tilde{\nu} = \dfrac{1}{\lambda}$
Explanation: Wave number is the reciprocal of wavelength. It represents the number of waves per unit distance.
118. In electromagnetic waves, the electric field vector is always:
ⓐ. Parallel to the direction of wave propagation
ⓑ. Perpendicular to magnetic field and wave propagation
ⓒ. Along the magnetic field direction
ⓓ. Random in orientation
Correct Answer: Perpendicular to magnetic field and wave propagation
Explanation: EM waves are transverse. Electric field, magnetic field, and propagation direction are mutually perpendicular (right-handed system).
119. If the wave number of light is $2 \times 10^6 \,\text{m}^{-1}$, its wavelength is:
ⓐ. $2 \times 10^6 \,\text{m}$
ⓑ. $5 \times 10^{-7} \,\text{m}$
ⓒ. $2 \times 10^{-6} \,\text{m}$
ⓓ. $5 \times 10^{6} \,\text{m}$
Correct Answer: $5 \times 10^{-7} \,\text{m}$
Explanation: Wavelength is reciprocal of wave number: $$\lambda = \frac{1}{\tilde{\nu}} = \frac{1}{2 \times 10^6} = 5 \times 10^{-7} \,\text{m}$$
120. Which statement about electromagnetic waves is correct?
ⓐ. They require a medium to propagate.
ⓑ. Their velocity depends on frequency in vacuum.
ⓒ. They consist of oscillating electric and magnetic fields perpendicular to each other.
ⓓ. They are longitudinal in nature.
Correct Answer: They consist of oscillating electric and magnetic fields perpendicular to each other.
Explanation: EM waves are transverse in nature. They do not require a medium, travel at constant speed $c$ in vacuum, and involve mutually perpendicular electric and magnetic fields.
121. Which relation is correct between frequency $\nu$ and time period $T$?
ⓐ. $\nu = T$
ⓑ. $\nu = \dfrac{1}{T}$
ⓒ. $\nu = T^2$
ⓓ. $\nu = \dfrac{1}{T^2}$
Correct Answer: $\nu = \dfrac{1}{T}$
Explanation: Frequency is the reciprocal of time period. If a wave completes one oscillation in time $T$, then in one second it completes $\dfrac{1}{T}$ oscillations.
122. What is the velocity of a wave with frequency $5 \times 10^{14}\,\text{Hz}$ and wavelength $6 \times 10^{-7}\,\text{m}$?
ⓐ. $3 \times 10^8 \,\text{m/s}$
ⓑ. $5 \times 10^7 \,\text{m/s}$
ⓒ. $8 \times 10^8 \,\text{m/s}$
ⓓ. $1 \times 10^{14} \,\text{m/s}$
Correct Answer: $3 \times 10^8 \,\text{m/s}$
Explanation: Velocity is given by $c = \lambda \nu$. Substituting: $$c = (6 \times 10^{-7})(5 \times 10^{14}) = 3 \times 10^8 \,\text{m/s}$$
123. If the velocity of a wave is constant, what happens when the wavelength decreases?
ⓐ. Frequency decreases
ⓑ. Frequency increases
ⓒ. Frequency remains same
ⓓ. Wave number decreases
Correct Answer: Frequency increases
Explanation: Since $c = \lambda \nu$, when $\lambda$ decreases, $\nu$ must increase to keep velocity constant.
124. Which of the following is dimensionally equivalent to wave number?
ⓐ. $[M^0L^{-1}T^0]$
ⓑ. $[M^0L^0T^{-1}]$
ⓒ. $[M^0L^1T^0]$
ⓓ. $[M^1L^{-1}T^0]$
Correct Answer: $[M^0L^{-1}T^0]$
Explanation: Wave number is reciprocal of wavelength, so its dimension is length$^{-1}$. Hence the dimensional formula is $[M^0L^{-1}T^0]$.
125. What is the wavelength of a radio wave with frequency $100 \,\text{MHz}$?
ⓐ. $3.0 \,\text{m}$
ⓑ. $0.3 \,\text{m}$
ⓒ. $30 \,\text{m}$
ⓓ. $300 \,\text{m}$
Correct Answer: $3.0 \,\text{m}$
Explanation: Using $c = \lambda \nu$: $$\lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{100 \times 10^6} = 3 \,\text{m}$$ Oops, but wait — 100 MHz = $1 \times 10^8$. So, $$\lambda = \frac{3 \times 10^8}{1 \times 10^8} = 3 \,\text{m} $$
126. Which has the highest frequency among electromagnetic waves?
ⓐ. Infrared
ⓑ. X-rays
ⓒ. Gamma rays
ⓓ. Radio waves
Correct Answer: Gamma rays
Explanation: Gamma rays occupy the extreme high-frequency end of the EM spectrum. They have the shortest wavelengths and highest photon energies.
127. If a wave has wave number $1 \times 10^7 \,\text{m}^{-1}$, what is its wavelength?
ⓐ. $1 \times 10^7 \,\text{m}$
ⓑ. $1 \times 10^{-7} \,\text{m}$
ⓒ. $1 \times 10^{14} \,\text{m}$
ⓓ. $1 \times 10^{-14} \,\text{m}$
Correct Answer: $1 \times 10^{-7} \,\text{m}$
Explanation: Wavelength is reciprocal of wave number. $\lambda = \dfrac{1}{1 \times 10^7} = 1 \times 10^{-7}\,\text{m}$.
128. Which unit combination correctly represents frequency?
ⓐ. $\text{m/s}$
ⓑ. $\text{s}^{-1}$
ⓒ. $\text{Hz m}$
ⓓ. $\text{m}^{-1}$
Correct Answer: $\text{s}^{-1}$
Explanation: Frequency is number of oscillations per second. Its SI unit is hertz (Hz), which equals $\text{s}^{-1}$.
129. The speed of electromagnetic waves in a medium is less than in vacuum because:
ⓐ. Medium destroys wave energy
ⓑ. Photons lose mass in medium
ⓒ. Electric and magnetic fields slow down due to polarization
ⓓ. Medium has refractive index greater than 1
Correct Answer: Medium has refractive index greater than 1
Explanation: The velocity of light in a medium is given by $v = \dfrac{c}{n}$. Since $n>1$, velocity decreases in a medium compared to vacuum.
130. If the time period of an EM wave is $2 \times 10^{-15}\,\text{s}$, its frequency is:
ⓐ. $5 \times 10^{14}\,\text{Hz}$
ⓑ. $2 \times 10^{-15}\,\text{Hz}$
ⓒ. $5 \times 10^{-16}\,\text{Hz}$
ⓓ. $2 \times 10^{15}\,\text{Hz}$
Correct Answer: $5 \times 10^{14}\,\text{Hz}$
Explanation: Frequency is inverse of time period. $$\nu = \frac{1}{T} = \frac{1}{2 \times 10^{-15}} = 5 \times 10^{14}\,\text{Hz}$$
131. Which region of the electromagnetic spectrum has the longest wavelength?
ⓐ. Ultraviolet
ⓑ. X-rays
ⓒ. Radio waves
ⓓ. Infrared
Correct Answer: Radio waves
Explanation: Radio waves have wavelengths ranging from meters to kilometers, making them the longest in the EM spectrum. Ultraviolet and X-rays have much shorter wavelengths, and infrared lies between microwaves and visible light.
132. Microwaves are commonly used in:
ⓐ. Medical imaging of bones
ⓑ. Communication and radar systems
ⓒ. Photosynthesis in plants
ⓓ. Sterilization of surgical instruments
Correct Answer: Communication and radar systems
Explanation: Microwaves (wavelength 1 mm to 1 m) are widely used in satellite communication, radar, and microwave ovens due to their ability to penetrate atmosphere and heat water molecules.
133. Which region of the electromagnetic spectrum is responsible for heating effect in sunlight?
ⓐ. Radio waves
ⓑ. Infrared rays
ⓒ. Ultraviolet rays
ⓓ. X-rays
Correct Answer: Infrared rays
Explanation: Infrared rays are strongly absorbed by water molecules in skin and materials, producing heat. This is why they are called “heat rays.”
134. The visible region of the EM spectrum lies approximately between:
ⓐ. 100–400 nm
ⓑ. 400–700 nm
ⓒ. 700–1000 nm
ⓓ. 1–10 µm
Correct Answer: 400–700 nm
Explanation: The human eye detects light in the wavelength range of 400 nm (violet) to 700 nm (red). Below 400 nm lies UV, and above 700 nm lies infrared.
135. Which type of electromagnetic wave is mainly responsible for sunburn?
ⓐ. Ultraviolet rays
ⓑ. Infrared rays
ⓒ. Visible light
ⓓ. Radio waves
Correct Answer: Ultraviolet rays
Explanation: UV rays (100–400 nm) can damage skin cells and cause sunburn. They have higher energy photons than visible and infrared light.
136. X-rays are mainly used in:
ⓐ. Cooking food
ⓑ. Sterilization of medical instruments
ⓒ. Medical imaging and crystallography
ⓓ. Wireless communication
Correct Answer: Medical imaging and crystallography
Explanation: X-rays penetrate soft tissues but are absorbed by bones, making them useful in medical imaging. They also help in determining crystal structures in X-ray diffraction studies.
137. Which region of the EM spectrum has the highest frequency?
ⓐ. Ultraviolet
ⓑ. Visible
ⓒ. Gamma rays
ⓓ. Infrared
Correct Answer: Gamma rays
Explanation: Gamma rays lie at the extreme high-frequency end of the spectrum. They have very short wavelengths ($<10^{-12}\,\text{m}$) and the highest photon energies.
138. Which radiation is often called “black light” and used in forensic investigations?
ⓐ. Infrared rays
ⓑ. Ultraviolet rays
ⓒ. Gamma rays
ⓓ. Microwaves
Correct Answer: Ultraviolet rays
Explanation: UV light is invisible to the naked eye but causes certain substances to fluoresce, making it useful in forensic analysis and currency verification.
139. Which radiation is most dangerous due to its high penetration power and ionizing ability?
ⓐ. Infrared rays
ⓑ. Visible light
ⓒ. Gamma rays
ⓓ. Microwaves
Correct Answer: Gamma rays
Explanation: Gamma rays are highly energetic, penetrate deeply, and cause ionization. They are dangerous to living tissue and are used in cancer treatment (radiotherapy).
140. Which electromagnetic radiation has wavelengths slightly longer than visible red light?
ⓐ. Ultraviolet
ⓑ. Infrared
ⓒ. Microwaves
ⓓ. X-rays
Correct Answer: Infrared
Explanation: Infrared lies just beyond the red end of the visible spectrum, with wavelengths from 700 nm to 1 mm. It is associated with heat radiation.
141. According to Planck’s quantum theory, energy is emitted or absorbed in:
ⓐ. Continuous amounts
ⓑ. Discrete packets called quanta
ⓒ. Varying amounts depending on temperature
ⓓ. Infinite energy waves
Correct Answer: Discrete packets called quanta
Explanation: Planck proposed that electromagnetic energy is quantized. Each packet of energy is called a quantum (or photon), with energy proportional to frequency: $E = h\nu$.
142. What is the value of Planck’s constant $h$?
ⓐ. $3.0 \times 10^8 \,\text{m/s}$
ⓑ. $9.1 \times 10^{-31}\,\text{kg}$
ⓒ. $6.626 \times 10^{-34}\,\text{J·s}$
ⓓ. $1.67 \times 10^{-27}\,\text{kg}$
Correct Answer: $6.626 \times 10^{-34}\,\text{J·s}$
Explanation: Planck’s constant is a fundamental constant in quantum theory. It relates the energy of a photon to its frequency, $E = h\nu$.
143. The energy of a photon of frequency $\nu$ is given by:
ⓐ. $E = mc^2$
ⓑ. $E = h\nu$
ⓒ. $E = \dfrac{hc}{\nu}$
ⓓ. $E = \dfrac{1}{2}mv^2$
Correct Answer: $E = h\nu$
Explanation: According to Planck, photon energy is directly proportional to frequency. Higher frequency means higher photon energy.
144. If wavelength of radiation is $\lambda$, the photon energy is:
ⓐ. $E = h\nu = \dfrac{hc}{\lambda}$
ⓑ. $E = \dfrac{\lambda}{hc}$
ⓒ. $E = h\lambda\nu$
ⓓ. $E = \dfrac{h}{c\lambda}$
Correct Answer: $E = h\nu = \dfrac{hc}{\lambda}$
Explanation: Since $c = \lambda\nu$, substituting in $E = h\nu$ gives $E = \dfrac{hc}{\lambda}$. This connects photon energy directly with wavelength.
145. Planck introduced his quantum hypothesis to explain:
ⓐ. Photoelectric effect
ⓑ. Blackbody radiation
ⓒ. Atomic spectra of hydrogen
ⓓ. Rutherford’s scattering
Correct Answer: Blackbody radiation
Explanation: Classical physics failed to explain blackbody radiation (ultraviolet catastrophe). Planck proposed quantization of energy to successfully match experimental results.
146. Which of the following phenomena is explained by particle nature of light?
ⓐ. Diffraction
ⓑ. Interference
ⓒ. Photoelectric effect
ⓓ. Polarization
Correct Answer: Photoelectric effect
Explanation: The photoelectric effect requires photons to knock electrons out of metal. Wave theory cannot explain instantaneous ejection or threshold frequency.
147. If a photon has wavelength $400 \,\text{nm}$, its energy is approximately:
ⓐ. $3.1 \,\text{eV}$
ⓑ. $6.2 \,\text{eV}$
ⓒ. $1.5 \,\text{eV}$
ⓓ. $13.6 \,\text{eV}$
Correct Answer: $3.1 \,\text{eV}$
Explanation: Energy = $\dfrac{hc}{\lambda}$. Using $h=6.626 \times 10^{-34}, c=3 \times 10^8, \lambda=400 \times 10^{-9}$: $$E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} \approx 4.97 \times 10^{-19}\,\text{J} \approx 3.1\,\text{eV}$$
148. The term “quantum” of light refers to:
ⓐ. An electron
ⓑ. A photon
ⓒ. A proton
ⓓ. A wave crest
Correct Answer: A photon
Explanation: A photon is the quantum of electromagnetic radiation. It carries energy $E=h\nu$ and has no rest mass.
149. Which observation cannot be explained by classical wave theory but is explained by Planck’s quantum theory?
ⓐ. Continuous energy emission
ⓑ. Photoelectric threshold frequency
ⓒ. Superposition of waves
ⓓ. Reflection of light
Correct Answer: Photoelectric threshold frequency
Explanation: Classical theory predicted photoelectric effect at any frequency given sufficient intensity. But experimentally, electrons are emitted only above a threshold frequency, which quantum theory explains via photons.
150. Which relation is correct for photon momentum?
ⓐ. $p = \dfrac{h}{\nu}$
ⓑ. $p = \dfrac{\lambda}{h}$
ⓒ. $p = \dfrac{h}{\lambda}$
ⓓ. $p = \dfrac{hc}{\lambda}$
Correct Answer: $p = \dfrac{h}{\lambda}$
Explanation: Photon momentum is related to wavelength by $p = \dfrac{h}{\lambda}$. Using energy $E = h\nu = \dfrac{hc}{\lambda}$, and $E=pc$, we get $p = \dfrac{h}{\lambda}$.
151. Which scientist gave the photon explanation of the photoelectric effect?
ⓐ. J.J. Thomson
ⓑ. Albert Einstein
ⓒ. Max Planck
ⓓ. Niels Bohr
Correct Answer: Albert Einstein
Explanation: In 1905, Einstein proposed that light consists of photons, each carrying energy $E = h\nu$. This theory explained why electron emission depends on frequency, not intensity, and earned him the Nobel Prize in 1921.
152. In the photoelectric effect, electrons are emitted from a metal surface when:
ⓐ. Light of any frequency is incident with high intensity
ⓑ. Light of frequency greater than a threshold frequency strikes the surface
ⓒ. Very low-intensity light is incident for long duration
ⓓ. The metal is heated to a high temperature
Correct Answer: Light of frequency greater than a threshold frequency strikes the surface
Explanation: Photoemission occurs only if photon energy $h\nu$ exceeds the metal’s work function $\phi$. Intensity increases number of electrons, but frequency decides if emission happens.
153. Which equation represents Einstein’s photoelectric effect?
ⓐ. $h\nu = \phi + \dfrac{1}{2}mv^2$
ⓑ. $E = mc^2$
ⓒ. $h\nu = \phi – \dfrac{1}{2}mv^2$
ⓓ. $h\nu = \dfrac{1}{2}mv^2$
Correct Answer: $h\nu = \phi + \dfrac{1}{2}mv^2$
Explanation: The photon energy $h\nu$ is partly used to overcome the work function $\phi$, and the rest becomes the kinetic energy of the emitted electron.
154. The minimum energy needed to remove an electron from a metal surface is called:
ⓐ. Ionization energy
ⓑ. Threshold energy
ⓒ. Electron affinity
ⓓ. Work function
Correct Answer: Work function
Explanation: Work function is the minimum energy required to just remove an electron from the metal surface. Its value depends on the metal (e.g., sodium ≈ 2.3 eV).
155. Increasing light intensity above threshold frequency will:
ⓐ. Increase maximum kinetic energy of photoelectrons
ⓑ. Increase number of photoelectrons emitted per second
ⓒ. Decrease threshold frequency
ⓓ. Stop emission of electrons
Correct Answer: Increase number of photoelectrons emitted per second
Explanation: Intensity controls number of photons hitting the surface. More photons eject more electrons, but kinetic energy depends only on frequency, not intensity.
156. If frequency of incident light equals threshold frequency, then:
ⓐ. No electrons are emitted
ⓑ. Photoelectric effect does not occur
ⓒ. Electrons are emitted with maximum kinetic energy
ⓓ. Electrons are emitted with zero kinetic energy
Correct Answer: Electrons are emitted with zero kinetic energy
Explanation: At threshold frequency, photon energy equals the work function. Electrons are just released with negligible kinetic energy.
157. Which graph best represents Einstein’s photoelectric equation?
ⓐ. Kinetic energy vs. frequency is a straight line with slope $h$
ⓑ. Kinetic energy vs. frequency is a parabola
ⓒ. Kinetic energy vs. frequency is constant line
ⓓ. Kinetic energy vs. frequency decreases exponentially
Correct Answer: Kinetic energy vs. frequency is a straight line with slope $h$
Explanation: Einstein’s equation $\dfrac{1}{2}mv^2 = h\nu – \phi$ predicts a linear relation between kinetic energy and frequency, with slope equal to Planck’s constant $h$.
158. Which factor does not affect the kinetic energy of emitted photoelectrons?
ⓐ. Frequency of incident light
ⓑ. Intensity of incident light
ⓒ. Work function of metal
ⓓ. Planck’s constant
Correct Answer: Intensity of incident light
Explanation: Intensity increases the number of emitted electrons but not their kinetic energy. Kinetic energy depends on frequency and work function.
159. The stopping potential in a photoelectric experiment is directly related to:
ⓐ. Photon frequency
ⓑ. Light intensity
ⓒ. Threshold frequency
ⓓ. Refractive index of medium
Correct Answer: Photon frequency
Explanation: Stopping potential corresponds to maximum kinetic energy: $eV_0 = h\nu – \phi$. It increases with frequency above threshold but is independent of intensity.
160. Which key observation of photoelectric effect contradicted classical wave theory?
ⓐ. Dependence on frequency rather than intensity
ⓑ. Existence of diffraction of light
ⓒ. Continuous emission of radiation
ⓓ. Interference of light waves
Correct Answer: Dependence on frequency rather than intensity
Explanation: Classical wave theory predicted emission at any frequency given sufficient intensity. But experiments showed no electrons are emitted below threshold frequency, no matter the intensity, proving the photon nature of light.
161. What is meant by the term “spectrum” in physics?
ⓐ. Range of possible speeds of particles
ⓑ. Range of wavelengths or frequencies of radiation
ⓒ. Collection of electron orbitals in an atom
ⓓ. Set of quantized angular momenta
Correct Answer: Range of wavelengths or frequencies of radiation
Explanation: Spectrum refers to the ordered arrangement of electromagnetic waves according to wavelength or frequency. It can be continuous (like white light) or discrete (like atomic line spectra).
162. Which device is commonly used to obtain a spectrum of light?
ⓐ. Microscope
ⓑ. Ammeter
ⓒ. Electrometer
ⓓ. Spectrometer
Correct Answer: Spectrometer
Explanation: A spectrometer or a prism is used to separate light into its constituent wavelengths to form a spectrum. Other devices measure different physical quantities.
163. The splitting of white light into its component colors is called:
ⓐ. Reflection
ⓑ. Dispersion
ⓒ. Diffraction
ⓓ. Refraction
Correct Answer: Dispersion
Explanation: Dispersion occurs when light passes through a prism, separating into colors because different wavelengths refract at different angles. This produces a visible spectrum.
164. Which type of spectrum is obtained from an incandescent solid or liquid?
ⓐ. Continuous spectrum
ⓑ. Line spectrum
ⓒ. Band spectrum
ⓓ. Absorption spectrum
Correct Answer: Continuous spectrum
Explanation: Hot solids, liquids, and dense gases emit radiation of all wavelengths in a continuous range. This produces a continuous spectrum without gaps.
165. What kind of spectrum is produced when white light passes through a cold gas before entering a spectroscope?
ⓐ. Continuous spectrum
ⓑ. Emission spectrum
ⓒ. Absorption spectrum
ⓓ. Band spectrum
Correct Answer: Absorption spectrum
Explanation: Cold gas absorbs specific wavelengths corresponding to its atoms. These appear as dark lines superimposed on the continuous background, forming an absorption spectrum.
166. The hydrogen emission spectrum is an example of:
ⓐ. Continuous spectrum
ⓑ. Line spectrum
ⓒ. Band spectrum
ⓓ. Absorption spectrum
Correct Answer: Line spectrum
Explanation: Hydrogen atoms emit photons at specific wavelengths when electrons transition between quantized energy levels. This produces discrete bright lines in the spectrum.
167. Which type of spectrum consists of groups of closely spaced lines, often due to molecules?
ⓐ. Continuous spectrum
ⓑ. Absorption spectrum
ⓒ. Band spectrum
ⓓ. Line spectrum
Correct Answer: Band spectrum
Explanation: Band spectra are produced by molecules due to combined rotational and vibrational transitions. The lines are so close that they appear as bands.
168. Which branch of physics deals with the study of spectra?
ⓐ. Spectroscopy
ⓑ. Photometry
ⓒ. Crystallography
ⓓ. Electrodynamics
Correct Answer: Spectroscopy
Explanation: Spectroscopy is the study of interaction between matter and electromagnetic radiation, analyzing spectral lines to identify materials and atomic structure.
169. In a line spectrum, each line corresponds to:
ⓐ. Random vibrations of atoms
ⓑ. Heating of the sample
ⓒ. Continuous emission of photons
ⓓ. Specific electronic transition between quantized states
Correct Answer: Specific electronic transition between quantized states
Explanation: A spectral line arises when an electron jumps between two energy levels. The energy difference determines the frequency of the emitted or absorbed photon.
170. Which type of spectrum shows bright lines against a dark background?
ⓐ. Absorption spectrum
ⓑ. Band spectrum
ⓒ. Continuous spectrum
ⓓ. Emission spectrum
Correct Answer: Emission spectrum
Explanation: Emission spectra are formed when excited atoms or gases emit radiation at discrete wavelengths, producing bright lines on a dark background.
171. What is an emission spectrum?
ⓐ. Bright lines (or bands) observed against a dark background from a hot, low-pressure gas
ⓑ. Dark lines on a continuous rainbow from a cool gas
ⓒ. A continuous distribution from hot solids and liquids
ⓓ. A set of microwave absorption minima only
Correct Answer: Bright lines (or bands) observed against a dark background from a hot, low-pressure gas
Explanation: A rarefied gas excited by heat or electricity emits photons at discrete energies, giving bright lines. Each line corresponds to a specific downward electronic transition. The bright-line pattern is characteristic of the element (a “fingerprint”). In contrast, cool gases in front of a continuum source produce dark lines (absorption). Continuous spectra arise from hot dense matter.
172. Which arrangement produces an absorption spectrum?
ⓐ. Viewing a glowing discharge tube directly
ⓑ. Heating a solid until it glows and viewing it directly
ⓒ. Passing white light through a cool, low-pressure gas before analysis
ⓓ. Observing fluorescence from an excited vapor directly
Correct Answer: Passing white light through a cool, low-pressure gas before analysis
Explanation: A cool gas in front of a continuous source selectively absorbs photons matching upward transitions, creating dark lines on the continuum. This is Kirchhoff’s third law. Directly viewing the emitting gas gives emission lines, while hot dense bodies yield continuous spectra. Fluorescence again gives emission features.
173. For a given element and electronic transition, how do emission and absorption wavelengths compare?
ⓐ. Emission has shorter wavelengths than absorption
ⓑ. Absorption has shorter wavelengths than emission
ⓒ. They differ by a pressure-dependent shift only
ⓓ. They are exactly the same to first order (same line center)
Correct Answer: They are exactly the same to first order (same line center)
Explanation: The same energy gap $\Delta E$ is involved: absorption is $n_1 \!\to\! n_2$ (photon taken), emission is $n_2 \!\to\! n_1$ (photon given). Since $\Delta E = h\nu = hc/\lambda$, the line center wavelength is identical. Small environmental effects (Doppler, pressure) can broaden/shift both.
174. The dark Fraunhofer lines observed in sunlight are an example of:
ⓐ. Emission spectrum
ⓑ. Absorption spectrum
ⓒ. Continuous spectrum
ⓓ. Band spectrum
Correct Answer: Absorption spectrum
Explanation: The solar photosphere emits a continuum; cooler gases in the solar atmosphere absorb at specific wavelengths, imprinting dark lines. Each line corresponds to an atomic transition (e.g., Na D lines). Emission lines would appear bright on dark, not dark on bright.
175. What ultimately fixes the position of a spectral line in either emission or absorption?
ⓐ. Gas pressure alone
ⓑ. Source intensity
ⓒ. The energy difference between initial and final states
ⓓ. Slit width of the spectrometer
Correct Answer: The energy difference between initial and final states
Explanation: Line center satisfies $\Delta E = h\nu = hc/\lambda$. Quantum energy levels of the atom/molecule determine $\Delta E$. Pressure and motion affect line width/shape (broadening), and instrument slit affects resolution, not the fundamental line position.
176. A stellar spectrum shows strong dark Na “D” lines near $589.0$ and $589.6\,\text{nm}$. This most directly indicates:
ⓐ. Cool sodium atoms in the star’s outer layers absorbing the continuum
ⓑ. The star emits only sodium emission lines
ⓒ. Sodium exists only in the interstellar medium, not the star
ⓓ. The star lacks a continuous photospheric spectrum
Correct Answer: Cool sodium atoms in the star’s outer layers absorbing the continuum
Explanation: The continuum from the photosphere passes through cooler overlying gas; Na atoms absorb at their resonance wavelengths, producing dark lines. Emission would appear bright if we observed only the glowing Na gas. Interstellar Na can also contribute, but the stellar atmosphere explanation is primary.
177. Which situation yields a bright-line emission spectrum?
ⓐ. Incandescent solid viewed directly
ⓑ. White light through a cool gas into a spectroscope
ⓒ. Sunlight observed after passage through Earth’s atmosphere
ⓓ. Excited low-pressure neon gas observed directly in a discharge tube
Correct Answer: Excited low-pressure neon gas observed directly in a discharge tube
Explanation: A rarefied, excited gas emits at discrete wavelengths (neon signs). Hot dense solids/liquids give a continuous spectrum. White light through a cool gas gives absorption lines. Sunlight through the atmosphere adds additional absorption features.
178. In an absorption line, the measured intensity at line center compared to the nearby continuum is:
ⓐ. Higher, because the gas emits more than it absorbs
ⓑ. Lower, because resonant photons are removed from the beam
ⓒ. Identical, because absorption and emission cancel exactly
ⓓ. Negative, because intensity reverses sign at line center
Correct Answer: Lower, because resonant photons are removed from the beam
Explanation: At line center, photons with $h\nu$ matching an upward transition are preferentially absorbed, creating a dip. The line depth depends on column density, temperature (population of lower state), and path length. Re-emission is generally isotropic, so fewer photons continue toward the observer.
179. How does increasing gas pressure typically affect spectral lines (emission or absorption)?
ⓐ. Increases line width via collisional (pressure) broadening
ⓑ. Decreases line width by reducing collisions
ⓒ. Leaves width unchanged, only shifts the line
ⓓ. Eliminates Doppler broadening completely
Correct Answer: Increases line width via collisional (pressure) broadening
Explanation: Collisions shorten the upper-state lifetime, giving uncertainty broadening $\Delta E\,\Delta t \!\gtrsim\! \hbar/2$ (Lorentzian wings). Higher pressure increases collision rates, broadening lines. Doppler broadening depends on temperature and motion, not pressure alone.
180. Which statement about emission vs absorption for the same transition is correct?
ⓐ. Emission occurs at a different frequency than absorption for the same atom
ⓑ. Absorption requires a continuous source, emission does not; therefore frequencies differ
ⓒ. Both occur at the same frequency; the sign of energy exchange differs
ⓓ. Neither corresponds to discrete quantum levels
Correct Answer: Both occur at the same frequency; the sign of energy exchange differs
Explanation: For a given transition, $\nu = \Delta E/h$ is fixed by the quantum levels; absorption uses $+h\nu$ to raise the atom, emission releases $h\nu$ as it falls. Observed profiles can broaden/shift slightly (Doppler/pressure), but the intrinsic line center is common to both.
181. The line spectrum of hydrogen is due to:
ⓐ. Vibration of protons inside the nucleus
ⓑ. Transitions of electrons between quantized energy levels
ⓒ. Continuous emission of photons from the atom
ⓓ. Absorption of neutrons by the atom
Correct Answer: Transitions of electrons between quantized energy levels
Explanation: Hydrogen’s spectral lines occur when an electron jumps between discrete energy levels. The emitted or absorbed photon has energy $\Delta E = h\nu = E_{n_2} – E_{n_1}$. This explains why hydrogen shows sharp line spectra instead of a continuum.
182. Which hydrogen series lies in the visible region of the spectrum?
ⓐ. Lyman series
ⓑ. Balmer series
ⓒ. Paschen series
ⓓ. Brackett series
Correct Answer: Balmer series
Explanation: The Balmer series corresponds to transitions from higher levels ($n \geq 3$) down to $n=2$. These lines (e.g., H$\alpha$ at 656.3 nm) fall in the visible range.
183. The first line of the Lyman series corresponds to transition:
ⓐ. $n=2 \to n=1$
ⓑ. $n=3 \to n=2$
ⓒ. $n=4 \to n=2$
ⓓ. $n=3 \to n=1$
Correct Answer: $n=2 \to n=1$
Explanation: Lyman series involves transitions ending at $n=1$. The first line is from $n=2 \to n=1$, producing ultraviolet radiation of wavelength 121.6 nm.
184. The Paschen series of hydrogen lies in which region of the electromagnetic spectrum?
ⓐ. Ultraviolet
ⓑ. Infrared
ⓒ. Visible
ⓓ. X-ray
Correct Answer: Infrared
Explanation: Paschen series corresponds to transitions from $n \geq 4$ to $n=3$. These wavelengths lie in the infrared region of the spectrum.
185. The Rydberg formula for hydrogen spectral lines is:
ⓐ. $\dfrac{1}{\lambda} = R_H\left(\dfrac{1}{n_2^2} – \dfrac{1}{n_1^2}\right)$
ⓑ. $\dfrac{1}{\lambda} = R_H\left(\dfrac{1}{n_1^2} – \dfrac{1}{n_2^2}\right)$
ⓒ. $\lambda = R_H\left(\dfrac{1}{n_1^2} – \dfrac{1}{n_2^2}\right)$
ⓓ. $\dfrac{1}{\nu} = R_H\left(\dfrac{1}{n_1^2} – \dfrac{1}{n_2^2}\right)$
Correct Answer: $\dfrac{1}{\lambda} = R_H\left(\dfrac{1}{n_2^2} – \dfrac{1}{n_1^2}\right)$
Explanation: Rydberg’s formula gives wavelengths of spectral lines, where $n_2 > n_1$. $R_H$ is the Rydberg constant ($1.097 \times 10^7 \, \text{m}^{-1}$).
186. The Brackett series in hydrogen corresponds to transitions ending at:
ⓐ. $n=1$
ⓑ. $n=2$
ⓒ. $n=3$
ⓓ. $n=4$
Correct Answer: $n=4$
Explanation: Brackett series is formed when electrons fall from higher levels ($n \geq 5$) to $n=4$. These lines occur in the infrared region.
187. Which hydrogen spectral line lies in the red region of the visible spectrum?
ⓐ. Lyman-$\alpha$
ⓑ. Balmer-$\alpha$ (H$\alpha$)
ⓒ. Balmer-$\beta$ (H$\beta$)
ⓓ. Paschen-$\alpha$
Correct Answer: Balmer-$\alpha$ (H$\alpha$)
Explanation: The Balmer-$\alpha$ line corresponds to transition $n=3 \to n=2$. Its wavelength (656.3 nm) lies in the red region of visible light, observed in hydrogen discharge tubes and stellar spectra.
188. Which transition produces the shortest wavelength line in a hydrogen series?
ⓐ. Transition from $n=3 \to n=2$
ⓑ. Transition from $n=\infty \to n=1$
ⓒ. Transition from $n=4 \to n=2$
ⓓ. Transition from $n=5 \to n=4$
Correct Answer: Transition from $n=\infty \to n=1$
Explanation: The shortest wavelength (highest energy) in a series occurs when an electron falls from infinity (ionization limit) to the series base level ($n=1$ for Lyman).
189. The Pfund series of hydrogen corresponds to transitions to:
ⓐ. $n=2$
ⓑ. $n=3$
ⓒ. $n=4$
ⓓ. $n=5$
Correct Answer: $n=5$
Explanation: Pfund series is observed when electrons fall to $n=5$ from higher levels ($n \geq 6$). These lines lie in the far infrared region.
190. Which of the following best describes hydrogen line spectra?
ⓐ. Continuous, covering all wavelengths
ⓑ. Discrete lines corresponding to quantized electronic transitions
ⓒ. Broad bands due to molecular vibration
ⓓ. X-ray spectrum only
Correct Answer: Discrete lines corresponding to quantized electronic transitions
Explanation: Hydrogen emits radiation only at specific wavelengths determined by allowed electronic transitions. This produces a line spectrum, not continuous or band spectra.
191. The Lyman series of hydrogen spectral lines lies in which region of the electromagnetic spectrum?
ⓐ. Visible
ⓑ. Ultraviolet
ⓒ. Infrared
ⓓ. X-ray
Correct Answer: Ultraviolet
Explanation: The Lyman series arises from electron transitions ending at $n=1$. The large energy differences correspond to short wavelengths in the ultraviolet region.
192. The Balmer-$\alpha$ line (H$\alpha$) in the hydrogen spectrum is produced by the transition:
ⓐ. $n=3 \to n=2$
ⓑ. $n=4 \to n=3$
ⓒ. $n=2 \to n=1$
ⓓ. $n=5 \to n=2$
Correct Answer: $n=3 \to n=2$
Explanation: The Balmer-$\alpha$ line occurs when an electron falls from the third orbit to the second. Its wavelength is 656.3 nm in the red part of the visible spectrum.
193. The Paschen series corresponds to transitions of electrons ending at:
ⓐ. $n=1$
ⓑ. $n=2$
ⓒ. $n=3$
ⓓ. $n=4$
Correct Answer: $n=3$
Explanation: Paschen series involves transitions from higher levels to $n=3$. These spectral lines appear in the infrared region.
194. Which hydrogen spectral series lies entirely in the visible region?
ⓐ. Lyman
ⓑ. Balmer
ⓒ. Paschen
ⓓ. Brackett
Correct Answer: Balmer
Explanation: The Balmer series corresponds to transitions ending at $n=2$. These lines (400–700 nm) lie within the visible spectrum detectable by the human eye.
195. The Brackett series of hydrogen lines appears in which region?
ⓐ. Ultraviolet
ⓑ. Visible
ⓒ. Infrared
ⓓ. Microwave
Correct Answer: Infrared
Explanation: Brackett series arises when electrons fall to $n=4$ from higher levels. These lower-energy transitions give infrared wavelengths.
196. The Pfund series of hydrogen corresponds to electron transitions to:
ⓐ. $n=1$
ⓑ. $n=3$
ⓒ. $n=4$
ⓓ. $n=5$
Correct Answer: $n=5$
Explanation: The Pfund series involves transitions to the fifth orbit from higher levels ($n \geq 6$). These spectral lines fall in the far infrared region.
197. The general formula for hydrogen spectral series is:
ⓐ. $\dfrac{1}{\lambda} = R_H\left(\dfrac{1}{n_1^2} – \dfrac{1}{n_2^2}\right), \; n_2 > n_1$
ⓑ. $\dfrac{1}{\lambda} = R_H\left(\dfrac{1}{n_2^2} – \dfrac{1}{n_1^2}\right), \; n_1 > n_2$
ⓒ. $\dfrac{1}{\nu} = R_H\left(\dfrac{1}{n_2^2} – \dfrac{1}{n_1^2}\right)$
ⓓ. $\lambda = \dfrac{R_H}{n_1^2} – \dfrac{1}{n_2^2}$
Correct Answer: $\dfrac{1}{\lambda} = R_H\left(\dfrac{1}{n_1^2} – \dfrac{1}{n_2^2}\right), \; n_2 > n_1$
Explanation: The Rydberg formula calculates hydrogen wavelengths. Each spectral series has a fixed $n_1$ (Lyman = 1, Balmer = 2, Paschen = 3, Brackett = 4, Pfund = 5).
198. The first line of the Brackett series corresponds to which transition?
ⓐ. $n=5 \to n=4$
ⓑ. $n=6 \to n=4$
ⓒ. $n=4 \to n=3$
ⓓ. $n=7 \to n=5$
Correct Answer: $n=6 \to n=4$
Explanation: The Brackett series ends at $n=4$. Its first line comes from the transition $n=6 \to n=4$, with wavelength in the infrared.
199. Which of the following spectral series has the shortest wavelength limit?
ⓐ. Lyman
ⓑ. Balmer
ⓒ. Paschen
ⓓ. Brackett
Correct Answer: Lyman
Explanation: Wavelength decreases as the energy gap increases. The Lyman series ends at $n=1$, which has the largest $\Delta E$, thus the shortest wavelengths in the UV region.
200. The line spectrum of hydrogen atom provides evidence for:
ⓐ. Continuous nature of atomic energy levels
ⓑ. Quantization of electronic energy levels
ⓒ. Wave theory of matter only
ⓓ. Existence of neutrons in atoms
Correct Answer: Quantization of electronic energy levels
Explanation: Hydrogen emits discrete lines corresponding to specific transitions, showing electrons occupy quantized levels. This was crucial evidence for Bohr’s atomic theory.
The chapter Structure of Atom is one of the most significant topics in Class 11 Chemistry (NCERT/CBSE syllabus). It covers the evolution of atomic models starting from Dalton’s atomic theory to the revolutionary Bohr’s model and finally the quantum mechanical model. Students also learn about photoelectric effect, dual nature of matter, de Broglie’s hypothesis, and Heisenberg’s Uncertainty Principle. These concepts are directly applicable in solving numerical problems, predicting electron behavior, and understanding periodic trends. Across all 5 parts, this chapter contains 475 MCQs with solutions, designed for board exam preparation as well as competitive exams like JEE, NEET, and state entrance tests. This page brings you the second set of 100 solved MCQs with explanations. You can easily navigate between different parts and other chapters using the buttons and navigation options above.
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