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Q11.3:
The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
Solution :
Step 1: Theoretical Foundation
According to the Einstein Photoelectric Equation, the relationship between the stopping potential ($V_0$) and the maximum kinetic energy ($K_{max}$) of the emitted photoelectrons is defined by the work required to stop the most energetic electrons. By the principle of conservation of energy [Per the work-energy theorem], the kinetic energy of the fastest electron is exactly equal to the potential energy barrier created by the retarding potential.
The governing equation is:
$K_{max} = e V_0$
Where:
- $K_{max}$ is the maximum kinetic energy (in Joules or eV).
- $e$ is the elementary charge of an electron ($1.602 \times 10^{-19}\text{ C}$).
- $V_0$ is the cut-off (stopping) voltage (in Volts).
Step 2: Given Variables
- Stopping Potential ($V_0$) = $1.5\text{ V}$
- Elementary Charge ($e$) = $1.602 \times 10^{-19}\text{ C}$
Step 3: Algebraic Calculation
Using the relationship $K_{max} = e V_0$, we substitute the known values:
$K_{max} = (1.602 \times 10^{-19}\text{ C}) \times (1.5\text{ V})$
$K_{max} = 2.403 \times 10^{-19}\text{ J}$
Step 4: Conversion to Electron-Volts (eV)
By definition, $1\text{ eV}$ is the energy gained by an electron accelerating through a potential difference of $1\text{ V}$. Therefore, since $K_{max} = e V_0$, the value in electron-volts is numerically equivalent to the value of the stopping potential in volts [By the definition of the electron-volt unit].
$K_{max} = 1.5\text{ eV}$
Visual Representation of Photoelectric Cut-off
| Unit System | Maximum Kinetic Energy ($K_{max}$) |
|---|---|
| SI Units (Joules) | $2.403 \times 10^{-19}\text{ J}$ |
| Atomic Units (eV) | $1.5\text{ eV}$ |
Final Solution: The maximum kinetic energy of the photoelectrons is $1.5\text{ eV}$ (or $2.403 \times 10^{-19}\text{ J}$).
More Questions from Class 12 Physics Dual Nature of Radiation and Matter EXERCISES
- Q11.1(a): Find the (a) maximum frequency of X-rays produced by $30$ kV electrons.
- Q11.1(a): Find the maximum frequency of X-rays produced by 30 kV electrons.
- Q11.1(b): Find the minimum wavelength of X-rays produced by 30 kV electrons.
- Q11.1(b): Find the (b) minimum wavelength of X-rays produced by $30$ kV electrons.
- Q11.10(a): What is the de Broglie wavelength of a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s?
- Q11.10(a): What is the de Broglie wavelength of (a) a bullet of mass $0.040$ kg travelling at the speed of $1.0$ km/s,
- Q11.10(b): What is the de Broglie wavelength of a ball of mass 0.060 kg moving at a speed of 1.0 m/s?
- Q11.10(b): What is the de Broglie wavelength of (b) a ball of mass $0.060$ kg moving at a speed of $1.0$ m/s, and
- Q11.10(c): What is the de Broglie wavelength of a dust particle of mass (1.0 \times 10^{-9}) kg drifting with a speed of 2.2 m/s ?
- Q11.10(c): What is the de Broglie wavelength of (c) a dust particle of mass $1.0 \times 10^{-9}$ kg drifting with a speed of $2.2$ m/s ?
- Q11.11: Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
- Q11.11: Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
- Q11.2(a): The work function of caesium metal is 2.14 eV. When light of frequency (6 \times 10^{14})Hz is incident on the metal surface, photoemission of electrons occurs. What is the maximum kinetic energy of the emitted electrons?
- Q11.2(a): The work function of caesium metal is $2.14$ eV. When light of frequency $6 \times 10^{14}$Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons,
- Q11.2(b): The work function of caesium metal is 2.14 eV. When light of frequency (6 \times 10^{14})Hz is incident on the metal surface, photoemission of electrons occurs. What is the Stopping potential?
- Q11.2(b): The work function of caesium metal is $2.14$ eV. When light of frequency $6 \times 10^{14}$Hz is incident on the metal surface, photoemission of electrons occurs. What is the (b) Stopping potential, and
- Q11.2(c): The work function of caesium metal is 2.14 eV. When light of frequency (6 \times 10^{14})Hz is incident on the metal surface, photoemission of electrons occurs. What is the maximum speed of the emitted photoelectrons?
- Q11.2(c): The work function of caesium metal is $2.14$ eV. When light of frequency $6 \times 10^{14}$Hz is incident on the metal surface, photoemission of electrons occurs. What is the (c) maximum speed of the emitted photoelectrons?
- Q11.3: The photoelectric cut-off voltage in a certain experiment is $1.5$ V. What is the maximum kinetic energy of photoelectrons emitted?
- Q11.4(a): Monochromatic light of wavelength $632.8$ nm is produced by a helium-neon laser. The power emitted is $9.42$ mW. (a) Find the energy and momentum of each photon in the light beam,
- Q11.4(a): Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW. Find the energy and momentum of each photon in the light beam,
- Q11.4(b): Monochromatic light of wavelength $632.8$ nm is produced by a helium-neon laser. The power emitted is $9.42$ mW. (b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and
- Q11.4(b): Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW. How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area ), and
- Q11.4(c): Monochromatic light of wavelength $632.8$ nm is produced by a helium-neon laser. The power emitted is $9.42$ mW. (c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
- Q11.4(c): Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW. How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
- Q11.5: In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be $4.12 \times 10^{-15}$ V s. Calculate the value of Planck’s constant.
- Q11.5: In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be (4.12 \times 10^{-15}) V s. Calculate the value of Planck’s constant.
- Q11.6: The threshold frequency for a certain metal is $3.3 \times 10^{14}$ Hz. If light of frequency $8.2 \times 10^{14}$ Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
- Q11.6: The threshold frequency for a certain metal is (3.3 \times 10^{14}) Hz. If light of frequency (8.2 \times 10^{14}) Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
- Q11.7: The work function for a certain metal is $4.2$ eV. Will this metal give hotoelectric emission for incident radiation of wavelength $330$ nm?
- Q11.7: The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?
- Q11.8: Light of frequency $7.21 \times 10^{14}$ Hz is incident on a metal surface. Electrons with a maximum speed of $6.0 \times 10^{5}$ m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
- Q11.8: Light of frequency (7.21 \times 10^{14}) Hz is incident on a metal surface. Electrons with a maximum speed of (6.0 \times 10^{5}) m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
- Q11.9: Light of wavelength $488$ nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-of f) potential of photoelectrons is $0.38$ V. Find the work function of the material from which the emitter is made.
- Q11.9: Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.
CBSE Solutions for Class 12 Physics Dual Nature of Radiation and Matter
Chapters in CBSE - Class 12 Physics
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- Ray Optics and Optical Instruments
- Wave Optics
- Atoms
- Nuclei
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