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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.
Solution :
Step 1: Identification of Given Physical Parameters
Based on the photoelectric effect experiments, we identify the following variables provided in the problem statement:
- Threshold frequency ($\nu_0$): $3.3 \times 10^{14} \text{ Hz}$
- Incident light frequency ($\nu$): $8.2 \times 10^{14} \text{ Hz}$
- Planck’s constant ($h$): $6.626 \times 10^{-34} \text{ J}\cdot\text{s}$
- Charge of an electron ($e$): $1.602 \times 10^{-19} \text{ C}$
Step 2: Theoretical Foundation
According to Einstein’s photoelectric equation, the maximum kinetic energy ($K_{max}$) of emitted photoelectrons is governed by the energy balance between the incident photon and the work function ($\phi_0$) of the metal surface [Per Einstein’s Photoelectric Equation]:
$K_{max} = h\nu - \phi_0$
Since the work function is defined as the minimum energy required for emission, $\phi_0 = h\nu_0$. Thus:
$K_{max} = h(\nu - \nu_0)$
The cut-off voltage ($V_0$), also known as the stopping potential, is defined as the potential required to halt the most energetic electrons, satisfying the relation $K_{max} = eV_0$ [By the principle of conservation of energy in an electric field].
Step 3: Derivation of the Cut-off Voltage Equation
Equating the two expressions for $K_{max}$:
$eV_0 = h(\nu - \nu_0)$
$V_0 = \frac{h}{e}(\nu - \nu_0)$
Step 4: Numerical Computation
Substitute the given values into the derived equation:
$V_0 = \frac{6.626 \times 10^{-34} \text{ J}\cdot\text{s}}{1.602 \times 10^{-19} \text{ C}} \times (8.2 \times 10^{14} - 3.3 \times 10^{14}) \text{ Hz}$
$V_0 = (4.136 \times 10^{-15} \text{ J}\cdot\text{s/C}) \times (4.9 \times 10^{14} \text{ Hz})$
$V_0 = 4.136 \times 4.9 \times 10^{-1} \text{ V}$
$V_0 = 20.2664 \times 0.1 \text{ V}$
$V_0 \approx 2.03 \text{ V}$
Step 5: Illustrative Representation of Photoelectric Emission
Final Solution: The cut-off voltage (stopping potential) for the photoelectric emission is $2.03 \text{ V}$.
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.
- Q11.7: 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
- Electric Charges and Fields
- Electrostatic Potential and Capacitance
- Current Electricity
- Moving Charges and Magnetism
- Magnetism and Matter
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- Alternating Current
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- Ray Optics and Optical Instruments
- Wave Optics
- Atoms
- Nuclei
- Semiconductor Electronics
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