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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?
Solution :
Step 1: Identifying Given Parameters and Constants
Based on the problem statement, we have the following physical quantities:
- Frequency of incident radiation ($ν$): $7.21 \times 10^{14} \, \text{Hz}$
- Maximum speed of emitted electrons ($v_{max}$): $6.0 \times 10^5 \, \text{m/s}$
- Planck’s constant ($h$): $6.626 \times 10^{-34} \, \text{J}\cdot\text{s}$
- Mass of an electron ($m$): $9.11 \times 10^{-31} \, \text{kg}$
Step 2: Theoretical Framework
According to Einstein’s Photoelectric Equation [Per the Law of Conservation of Energy applied to quantum interactions]:
$hν = \phi_0 + K_{max}$
Where $\phi_0$ is the work function defined as $\phi_0 = hν_0$ (where $ν_0$ is the threshold frequency) and $K_{max}$ is the maximum kinetic energy of the emitted photoelectrons.
Step 3: Calculating Maximum Kinetic Energy ($K_{max}$)
The maximum kinetic energy is calculated using the classical kinetic energy formula as the velocity is non-relativistic ($v \ll c$):
$K_{max} = \frac{1}{2}mv_{max}^2$
$K_{max} = 0.5 \times (9.11 \times 10^{-31} \, \text{kg}) \times (6.0 \times 10^5 \, \text{m/s})^2$
$K_{max} = 0.5 \times 9.11 \times 10^{-31} \times 36 \times 10^{10} \, \text{J}$
$K_{max} = 1.6398 \times 10^{-19} \, \text{J}$
Step 4: Deriving the Threshold Frequency ($ν_0$)
Rearranging the photoelectric equation to solve for the threshold frequency ($ν_0$):
$hν = hν_0 + K_{max}$
$hν_0 = hν - K_{max}$
$ν_0 = ν - \frac{K_{max}}{h}$
Substituting the numerical values:
$ν_0 = (7.21 \times 10^{14}) - \frac{1.6398 \times 10^{-19}}{6.626 \times 10^{-34}}$
Calculating the second term:
$\frac{1.6398 \times 10^{-19}}{6.626 \times 10^{-34}} \approx 0.24748 \times 10^{15} \, \text{Hz} = 2.475 \times 10^{14} \, \text{Hz}$
Final subtraction:
$ν_0 = 7.21 \times 10^{14} - 2.475 \times 10^{14}$
$ν_0 = 4.735 \times 10^{14} \, \text{Hz}$
Visual Representation of Photoelectric Effect
Final Solution: The threshold frequency for the metal surface is $4.74 \times 10^{14} \, \text{Hz}$.
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?
- Q11.3: 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?
- Q11.9: 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
- Electric Charges and Fields
- Electrostatic Potential and Capacitance
- Current Electricity
- Moving Charges and Magnetism
- Magnetism and Matter
- Electromagnetic Induction
- Alternating Current
- Electromagnetic Waves
- Ray Optics and Optical Instruments
- Wave Optics
- Atoms
- Nuclei
- Semiconductor Electronics
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