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1. The central idea of dual nature is best described as
ⓐ. radiation is always a wave and matter is always a particle
ⓑ. radiation is always a particle and matter is always a wave
ⓒ. radiation and matter both lose their ordinary physical properties
ⓓ. both radiation and matter can show wave-particle duality
Correct Answer: both radiation and matter can show wave-particle duality
Explanation: Dual nature means that the same physical entity may need different descriptions in different experiments. Light behaves like a wave in interference and diffraction, but it behaves like particles in the photoelectric effect. Matter usually appears as particles in daily life, but moving microscopic particles such as electrons can show diffraction. The word dual does not mean that the two descriptions appear equally strongly in every situation. The observed behaviour depends on the scale and the experiment being performed.
2. Photoelectric effect gives evidence for the particle nature of radiation because it shows that
ⓐ. light always bends around obstacles like a wave
ⓑ. light transfers energy in discrete packets to electrons
ⓒ. electrons inside the metal become electromagnetic waves
ⓓ. the speed of light changes after striking every metal
Correct Answer: light transfers energy in discrete packets to electrons
Explanation: In the photoelectric effect, suitable radiation can eject electrons from a metal surface. The effect is explained by assuming that light energy comes in packets called photons. A photon can give its energy to one electron, and the electron may escape if the supplied energy is enough. This packet-like transfer is not the same as spreading energy continuously over a wavefront. The phenomenon became strong evidence that radiation has a particle aspect.
3. Which observation is the clearest evidence for the wave nature of matter?
ⓐ. Heating a metal wire until electrons are emitted
ⓑ. A metal surface becoming positively charged after losing electrons
ⓒ. Electron diffraction by a crystal
ⓓ. A photon carrying energy proportional to its frequency
Correct Answer: Electron diffraction by a crystal
Explanation: Diffraction is a wave behaviour because it involves spreading and interference. If electrons produce diffraction maxima after interacting with a crystal, they cannot be treated only as tiny classical particles for that experiment. Their motion must be associated with a wavelength, called the de Broglie wavelength. Heating a metal or charging a surface may involve electrons, but those observations do not by themselves show wave behaviour. Electron diffraction is significant because a particle beam displays a pattern normally associated with waves.
4. Which pairing gives the most suitable evidence-aspect connection?
ⓐ. Photoelectric effect — particle nature of radiation
ⓑ. Photoelectric effect — wave nature of matter
ⓒ. Electron diffraction — particle nature of radiation
ⓓ. Interference of light — particle nature of matter
Correct Answer: Photoelectric effect — particle nature of radiation
Explanation: The photoelectric effect is explained by light delivering energy in photon packets. This makes it evidence for the particle nature of radiation. Electron diffraction, on the other hand, supports the wave nature of matter. Interference of light supports the wave nature of radiation, not the particle nature of matter. Dual nature links two complementary ideas: light can act like particles, and particles such as electrons can act like waves.
5. In the phrase “dual nature of radiation and matter,” the word radiation mainly refers to
ⓐ. only charged particles moving through a metal
ⓑ. electromagnetic radiation such as light
ⓒ. only sound waves travelling through air
ⓓ. atoms permanently fixed inside a crystal lattice
Correct Answer: electromagnetic radiation such as light
Explanation: In dual-nature physics, radiation mainly refers to electromagnetic radiation, especially light used in photoelectric experiments. Light has familiar wave properties such as wavelength \( \lambda \) and frequency \( \nu \). It also has particle-like behaviour when described in terms of photons. Sound waves are mechanical waves, not electromagnetic radiation. This use of radiation is closely connected with photon energy and photoelectric emission.
6. Matter-wave behaviour is usually noticed for microscopic particles rather than ordinary large bodies because
ⓐ. large bodies do not have momentum
ⓑ. microscopic particles do not have mass
ⓒ. their matter wavelength is extremely small
ⓓ. matter waves need air as a material medium
Correct Answer: their matter wavelength is extremely small
Explanation: de Broglie’s idea associates a wavelength with moving matter. For ordinary large bodies, the momentum is so large that the associated wavelength becomes extremely tiny. Such a small wavelength cannot usually produce observable diffraction in everyday situations. Microscopic particles such as electrons can have wavelengths comparable to atomic spacings under suitable conditions. This is why electron diffraction is observable, while wave behaviour of a moving ball is not noticed.
7. For light travelling in vacuum, increasing the frequency \( \nu \) while keeping the speed \( c \) fixed makes the wavelength \( \lambda \)
ⓐ. increase in the same ratio
ⓑ. remain unchanged for all colours
ⓒ. become equal to the photon momentum
ⓓ. decrease because \( c=\nu\lambda \)
Correct Answer: decrease because \( c=\nu\lambda \)
Explanation: For electromagnetic radiation in vacuum, the relation between speed, frequency, and wavelength is \( c=\nu\lambda \). Since \( c \) is fixed in vacuum, \( \nu \) and \( \lambda \) vary inversely. A higher frequency therefore corresponds to a shorter wavelength. This relation is later used when photon energy is written as \( E=\frac{hc}{\lambda} \). Frequency and wavelength are not independent for light travelling in the same medium.
8. A beam of suitable light falls on a clean metal surface and electrons are emitted from the surface. This phenomenon is called
ⓐ. thermionic emission
ⓑ. electron diffraction
ⓒ. field emission
ⓓ. photoelectric effect
Correct Answer: photoelectric effect
Explanation: The photoelectric effect is the emission of electrons from a metal surface due to incident suitable radiation. The emitted electrons are called photoelectrons when they are produced by light. Thermionic emission is caused by heating, while field emission is caused by a very strong electric field. Electron diffraction is a wave-effect observation involving electron beams and crystals. The cause of emission is the key feature that separates these terms.
9. Which quantity represents the energy gained by a charge \( q \) when it moves through a potential difference \( V \)?
ⓐ. \( \frac{q}{V} \)
ⓑ. \( qV \)
ⓒ. \( \frac{V}{q} \)
ⓓ. \( q+V \)
Correct Answer: \( qV \)
Explanation: Potential difference is energy transferred per unit charge. Therefore, a charge \( q \) moving through a potential difference \( V \) gains energy \( qV \). For an electron accelerated through potential \( V \), the gained kinetic energy is written as \( eV \) when the charge magnitude is \( e \). This relation is a basic link between electrical energy and kinetic energy in dual-nature calculations. The symbol \( eV \) can also name the energy unit electron volt, so the physical context must be read carefully.
10. Match the observation or idea with the aspect it supports.
The proper matching is
| Column I | Column II |
|---|---|
| P. Photoelectric emission by light | 1. Wave nature of matter |
| Q. Diffraction of electrons by a crystal | 2. Particle nature of radiation |
| R. Interference of light | 3. Wave nature of radiation |
ⓐ. P-1, Q-2, R-3
ⓑ. P-3, Q-1, R-2
ⓒ. P-2, Q-1, R-3
ⓓ. P-2, Q-3, R-1
Correct Answer: P-2, Q-1, R-3
Explanation: Photoelectric emission by light is explained using photons, so it supports the particle nature of radiation. Diffraction of electrons by a crystal supports the wave nature of matter because diffraction is a wave phenomenon. Interference of light is a standard wave phenomenon of radiation. These three examples help separate the two halves of dual nature. The evidence must be matched to the behaviour observed, not merely to the object named in the situation.
11. A symbol-unit record contains one mismatched entry. Pick the mismatched entry.
| Entry | Quantity | Common unit |
|---|---|---|
| P | Frequency \( \nu \) | \(Hz\) |
| Q | Wavelength \( \lambda \) | \(m\) |
| R | Photon momentum \( p \) | \(kg\,m\,s^{-1}\) |
| S | Stopping potential \( V_0 \) | \(J\) |
ⓐ. Entry P
ⓑ. Entry Q
ⓒ. Entry R
ⓓ. Entry S
Correct Answer: Entry S
Explanation: Frequency \( \nu \) is measured in \(Hz\), and wavelength \( \lambda \) is measured in \(m\). Momentum has SI unit \(kg\,m\,s^{-1}\), so the unit listed for photon momentum \( p \) is suitable. Stopping potential \( V_0 \) is a potential difference, so its unit is \(V\), not \(J\). Joule \(J\) is a unit of energy, such as photon energy \(E\), work function \( \phi_0 \), or kinetic energy \(K\). The notation \(eV_0\) represents energy, but \(V_0\) alone represents potential.
12. The photon energy relation written in terms of frequency is completed as \(E=\) ______.
ⓐ. \(h\lambda\)
ⓑ. \(\frac{h}{\nu}\)
ⓒ. \(h\nu\)
ⓓ. \(\frac{\nu}{h}\)
Correct Answer: \(h\nu\)
Explanation: Photon energy is directly proportional to the frequency of radiation. The proportionality constant is Planck’s constant \(h\), so the relation is \(E=h\nu\). A higher value of \( \nu \) means a more energetic photon. The expression \(h\lambda\) is not the energy of a photon, because wavelength enters through \(E=\frac{hc}{\lambda}\). Frequency and wavelength forms are connected through \(c=\nu\lambda\), not by replacing \( \nu \) with \( \lambda \) directly.
13. A photon has energy \(2.0\,eV\). Using \(1\,eV=1.6\times10^{-19}\,J\), its energy in joules is
ⓐ. \(3.2\times10^{-19}\,J\)
ⓑ. \(1.6\times10^{-19}\,J\)
ⓒ. \(2.0\times10^{-19}\,J\)
ⓓ. \(4.0\times10^{-19}\,J\)
Correct Answer: \(3.2\times10^{-19}\,J\)
Explanation: \( \textbf{Given:} \) Photon energy \(E=2.0\,eV\).
\( \textbf{Conversion value:} \) \(1\,eV=1.6\times10^{-19}\,J\).
\( \textbf{Required unit:} \) Energy in \(J\).
\( \textbf{Conversion rule:} \) Multiply the numerical value in \(eV\) by \(1.6\times10^{-19}\).
\( \textbf{Substitution:} \)
\[
E=2.0\times(1.6\times10^{-19})\,J
\]
\( \textbf{Calculation:} \)
\[
E=3.2\times10^{-19}\,J
\]
\( \textbf{Unit check:} \) The quantity remains energy; only the unit changes from \(eV\) to \(J\).
\( \textbf{Final answer:} \) The photon energy is \(3.2\times10^{-19}\,J\).
14. The relation \(p=\frac{h}{\lambda}\) tells us that photon momentum is measured in the unit
ⓐ. \(J\,s^{-1}\)
ⓑ. \(kg\,m^2\,s^{-2}\)
ⓒ. \(kg\,m\,s^{-1}\)
ⓓ. \(C\,V^{-1}\)
Correct Answer: \(kg\,m\,s^{-1}\)
Explanation: The symbol \(p\) represents momentum. Momentum has SI unit \(kg\,m\,s^{-1}\), whether it is used for a material particle or for photon momentum. In the photon relation \(p=\frac{h}{\lambda}\), Planck’s constant \(h\) and wavelength \( \lambda \) combine to give the same momentum unit. Joule \(J\) is an energy unit, \(Hz\) is a frequency unit, and volt \(V\) is a potential unit. The same symbol \(p\) should not be confused with photon energy \(E\).
15. Consider the following statements about basic quantities used in dual nature.
Statement I: \(h\) is Planck’s constant.
Statement II: \(c\) is the speed of light in vacuum.
Statement III: \(V_0\) is measured in \(kg\,m\,s^{-1}\).
ⓐ. I and II only
ⓑ. II and III only
ⓒ. I and III only
ⓓ. I, II and III
Correct Answer: I and II only
Explanation: Planck’s constant is denoted by \(h\), and it appears in relations such as \(E=h\nu\). The symbol \(c\) denotes the speed of light in vacuum, and it connects wavelength and frequency through \(c=\nu\lambda\). Stopping potential \(V_0\) is a potential difference, so its unit is \(V\). The unit \(kg\,m\,s^{-1}\) belongs to momentum, not potential. Keeping symbol-unit pairs clear is necessary before using photoelectric and photon formulas.
16. A radiation packet has energy \(4.8\times10^{-19}\,J\). Express this energy in \(eV\), using \(1\,eV=1.6\times10^{-19}\,J\).
ⓐ. \(1.5\,eV\)
ⓑ. \(2.0\,eV\)
ⓒ. \(4.8\,eV\)
ⓓ. \(3.0\,eV\)
Correct Answer: \(3.0\,eV\)
Explanation: \( \textbf{Given energy:} \) \(E=4.8\times10^{-19}\,J\).
\( \textbf{Conversion value:} \) \(1\,eV=1.6\times10^{-19}\,J\).
\( \textbf{Required unit:} \) \(eV\).
\( \textbf{Conversion idea:} \) Divide the energy in \(J\) by the joule value of \(1\,eV\).
\( \textbf{Substitution:} \)
\[
E=\frac{4.8\times10^{-19}}{1.6\times10^{-19}}\,eV
\]
\( \textbf{Power of ten cancellation:} \)
\[
\frac{10^{-19}}{10^{-19}}=1
\]
\( \textbf{Numerical part:} \)
\[
\frac{4.8}{1.6}=3.0
\]
\( \textbf{Final answer:} \) The energy is \(3.0\,eV\), and using \(eV\) directly as \(J\) would miss the conversion factor.
17. The SI unit of Planck’s constant \(h\) can be written as
ⓐ. \(J\,s^{-1}\)
ⓑ. \(J\,s\)
ⓒ. \(kg\,m\,s^{-1}\)
ⓓ. \(V\,s^{-1}\)
Correct Answer: \(J\,s\)
Explanation: Planck’s constant \(h\) appears in the photon energy relation \(E=h\nu\). Since \(E\) is measured in \(J\) and frequency \( \nu \) is measured in \(s^{-1}\), the unit of \(h\) must be \(J\,s\). This also matches the dimensional meaning of action, because \(h\) connects energy with time or momentum with wavelength. The unit \(J\,s^{-1}\) is watt, which is power, not Planck’s constant. The unit check is useful because \(h\) appears in both photon and matter-wave formulas.
18. A photon relation is written as \(E=\frac{hc}{\lambda}\). If the wavelength \( \lambda \) is doubled, with \(h\) and \(c\) unchanged, the photon energy becomes
ⓐ. doubled
ⓑ. halved
ⓒ. unchanged
ⓓ. four times larger
Correct Answer: halved
Explanation: The relation \(E=\frac{hc}{\lambda}\) shows that photon energy is inversely proportional to wavelength. When \( \lambda \) becomes \(2\lambda\), the denominator doubles while \(h\) and \(c\) remain constant. Therefore, the energy becomes \( \frac{1}{2} \) of its original value. This is the same idea as saying that shorter wavelength radiation has higher frequency and hence higher photon energy. The mistake to avoid is treating wavelength and energy as directly proportional just because both appear in the same formula.
19. In the photoelectric effect, the symbol \( \phi_0 \) is used for
ⓐ. work function
ⓑ. photon momentum
ⓒ. stopping potential
ⓓ. de Broglie wavelength
Correct Answer: work function
Explanation: The symbol \( \phi_0 \) denotes the work function of a metal. It is the minimum energy required to liberate an electron from the metal surface. In photoelectric equations, it is compared with photon energy \(h\nu\). If \(h\nu\) is smaller than \( \phi_0 \), photoelectric emission does not occur. The subscript \(0\) helps connect it with the threshold condition, where \( \phi_0=h\nu_0 \).
20. A quantity has the unit \(kg\,m\,s^{-1}\). In dual-nature calculations, it is most naturally identified as
ⓐ. photon momentum \(p\)
ⓑ. photon energy \(E\)
ⓒ. stopping potential \(V_0\)
ⓓ. threshold frequency \( \nu_0 \)
Correct Answer: photon momentum \(p\)
Explanation: The unit \(kg\,m\,s^{-1}\) is the SI unit of momentum. A photon has momentum even though it has zero rest mass, and its momentum may be written as \(p=\frac{h}{\lambda}\). Photon energy is measured in \(J\), stopping potential is measured in \(V\), and threshold frequency is measured in \(Hz\). This distinction prevents mixing \(p=\frac{h}{\lambda}\) with \(E=\frac{hc}{\lambda}\). Both formulas contain wavelength, but they represent different physical quantities.