Modern Physics MCQ Quiz - Objective Question with Answer for Modern Physics - Download Free PDF

Ans.(4)

Sol.

As the wavelength λ of the incident light falling on cathode is gradually increased, there will be a value of λ above which the photoelectric emission will stop. I.e., the photoelectric emission will stop. i.e, the photoelectric current will become zero. As λ is decreased, the energy of the incident photons increases (E = hc/λ). Due to it, a greater number of photons will be successful in ejecting the electrons from the surface of cathode. As a result of it, the photoelectric current increases, i.e., I ∝ 1/λ

Concept:

The photoelectric effect occurs when light of sufficient energy strikes a metal surface, causing electrons to be ejected. The energy of a photon is given by:

E = h × f, where E is the energy of the photon, h is Planck’s constant, and f is the frequency of the light.

The work function (Φ) is the minimum energy required to eject an electron. For the photoelectric effect to occur, the energy of the photon must be greater than or equal to the work function: h × f = Φ.

The frequency of light is related to its wavelength by the equation: f = c / λ, where c is the speed of light and λ is the wavelength.

Thus, the longest wavelength λ_max that can initiate the photoelectric effect can be found using the equation:

λ_max = h × c / Φ, where h = 6.626 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s, and Φ = 9 eV = 9 × 1.602 × 10⁻¹⁹ J.

Calculation:

Given: Φ = 9 eV = 9 × 1.602 × 10⁻¹⁹ J = 1.4418 × 10⁻¹⁸ J

Using the formula: λ_max = h × c / Φ

λ_max = (6.626 × 10⁻³⁴ × 3 × 10⁸) / (1.4418 × 10⁻¹⁸)

λ_max ≈ 1.37 × 10⁻⁷ m

∴ The longest wavelength of light that can initiate the photoelectric effect is 1.37 × 10⁻⁷ m. Option 1) is correct.

Calculation:

We are given:

• Thickness of the GaAs sample, d = 0.46 μm = 0.46 × 10^-4 cm
• Photon energy, hv = 2 eV
• Absorption coefficient, α = 5 × 10⁴ cm^-1
• Incident power, P = 10 mW = 10 × 10^-3 W

The absorption of light in the sample follows the exponential law:

I(x) = I₀ e^-αx

where I(x) is the intensity at depth x, and I₀ is the incident intensity.

To find the total energy absorbed per second, we need to calculate the absorbed power.

The absorbed power P_abs can be found using the relation:

P_abs = P_in (1 - e^-αd)

Substituting the given values:

αd = 5 × 10⁴ cm^-1 × 0.46 × 10^-4 cm = 2.3

Given that e^-2.3 = 0.1, we have:

P_abs = 10 × 10^-3 W × (1 - 0.1) = 10 × 10^-3 W × 0.9 = 9 × 10^-3 W

Therefore, the total energy absorbed by the sample per second is:

Final Answer: 9 × 10^-3 J/s

The correct answer is Transistor-Transistor Logic.

Key Points

• TTL stands for Transistor-Transistor Logic, a class of digital circuits built from bipolar junction transistors (BJTs) and resistors.
• It was invented in 1963 by James L. Buie of TRW, which was later marketed by Texas Instruments.
• TTL is known for its high speed and low power consumption compared to earlier logic families like DTL (Diode-Transistor Logic).
• TTL circuits are widely used in computers, industrial controls, consumer electronics, and instrumentation.

Additional Information

• Logic Families
  • Logic families are groups of electronic logic gates that use similar circuit configurations and are compatible in terms of voltage and current levels.
  • Examples include TTL, CMOS (Complementary Metal-Oxide-Semiconductor), and ECL (Emitter-Coupled Logic).
• CMOS vs. TTL
  • CMOS technology offers higher noise immunity and lower power consumption compared to TTL.
  • However, TTL circuits are faster and were more widely used in the 1970s and 1980s.
• Bipolar Junction Transistors (BJTs)
  • BJTs are a type of transistor that uses both electron and hole charge carriers.
  • They are known for their high current density and fast switching speed.
• Applications of TTL
  • TTL circuits have been used in early microprocessors, memory chips, and various digital logic devices.
  • They are also found in some signal processing and data communication equipment.

Each photon gives rise to one photoelectron. Also, intensity is proportional to incident photons and energy of each photon.

Hence, \( I \propto I_{int} \)

\( I_{int} \propto E \)

Energy of a photon, \( E = \dfrac{hc}{\lambda} \)

\( \implies I \propto \dfrac{1}{\lambda} \)

Option (c) is the closest approximation of the given equation.

Ans.(4)

Sol.

As the wavelength λ of the incident light falling on cathode is gradually increased, there will be a value of λ above which the photoelectric emission will stop. I.e., the photoelectric emission will stop. i.e, the photoelectric current will become zero. As λ is decreased, the energy of the incident photons increases (E = hc/λ). Due to it, a greater number of photons will be successful in ejecting the electrons from the surface of cathode. As a result of it, the photoelectric current increases, i.e., I ∝ 1/λ

Concept:

The photoelectric effect occurs when light of sufficient energy strikes a metal surface, causing electrons to be ejected. The energy of a photon is given by:

E = h × f, where E is the energy of the photon, h is Planck’s constant, and f is the frequency of the light.

The work function (Φ) is the minimum energy required to eject an electron. For the photoelectric effect to occur, the energy of the photon must be greater than or equal to the work function: h × f = Φ.

The frequency of light is related to its wavelength by the equation: f = c / λ, where c is the speed of light and λ is the wavelength.

Thus, the longest wavelength λ_max that can initiate the photoelectric effect can be found using the equation:

λ_max = h × c / Φ, where h = 6.626 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s, and Φ = 9 eV = 9 × 1.602 × 10⁻¹⁹ J.

Calculation:

Given: Φ = 9 eV = 9 × 1.602 × 10⁻¹⁹ J = 1.4418 × 10⁻¹⁸ J

Using the formula: λ_max = h × c / Φ

λ_max = (6.626 × 10⁻³⁴ × 3 × 10⁸) / (1.4418 × 10⁻¹⁸)

λ_max ≈ 1.37 × 10⁻⁷ m

∴ The longest wavelength of light that can initiate the photoelectric effect is 1.37 × 10⁻⁷ m. Option 1) is correct.

Calculation:

We are given:

• Thickness of the GaAs sample, d = 0.46 μm = 0.46 × 10^-4 cm
• Photon energy, hv = 2 eV
• Absorption coefficient, α = 5 × 10⁴ cm^-1
• Incident power, P = 10 mW = 10 × 10^-3 W

The absorption of light in the sample follows the exponential law:

I(x) = I₀ e^-αx

where I(x) is the intensity at depth x, and I₀ is the incident intensity.

To find the total energy absorbed per second, we need to calculate the absorbed power.

The absorbed power P_abs can be found using the relation:

P_abs = P_in (1 - e^-αd)

Substituting the given values:

αd = 5 × 10⁴ cm^-1 × 0.46 × 10^-4 cm = 2.3

Given that e^-2.3 = 0.1, we have:

P_abs = 10 × 10^-3 W × (1 - 0.1) = 10 × 10^-3 W × 0.9 = 9 × 10^-3 W

Therefore, the total energy absorbed by the sample per second is:

Final Answer: 9 × 10^-3 J/s

The correct answer is Transistor-Transistor Logic.

Key Points

• TTL stands for Transistor-Transistor Logic, a class of digital circuits built from bipolar junction transistors (BJTs) and resistors.
• It was invented in 1963 by James L. Buie of TRW, which was later marketed by Texas Instruments.
• TTL is known for its high speed and low power consumption compared to earlier logic families like DTL (Diode-Transistor Logic).
• TTL circuits are widely used in computers, industrial controls, consumer electronics, and instrumentation.

Additional Information

• Logic Families
  • Logic families are groups of electronic logic gates that use similar circuit configurations and are compatible in terms of voltage and current levels.
  • Examples include TTL, CMOS (Complementary Metal-Oxide-Semiconductor), and ECL (Emitter-Coupled Logic).
• CMOS vs. TTL
  • CMOS technology offers higher noise immunity and lower power consumption compared to TTL.
  • However, TTL circuits are faster and were more widely used in the 1970s and 1980s.
• Bipolar Junction Transistors (BJTs)
  • BJTs are a type of transistor that uses both electron and hole charge carriers.
  • They are known for their high current density and fast switching speed.
• Applications of TTL
  • TTL circuits have been used in early microprocessors, memory chips, and various digital logic devices.
  • They are also found in some signal processing and data communication equipment.

Each photon gives rise to one photoelectron. Also, intensity is proportional to incident photons and energy of each photon.

Hence, \( I \propto I_{int} \)

\( I_{int} \propto E \)

Energy of a photon, \( E = \dfrac{hc}{\lambda} \)

\( \implies I \propto \dfrac{1}{\lambda} \)

Option (c) is the closest approximation of the given equation.

Concept:

• The energy of a photon (E) is related to its wavelength (λ) by the equation: \(E = \frac{hc}{λ},\) where h is Planck's constant, c is the speed of light, and λ is the wavelength of the light.
• The photoelectric work function (ϕ ) is the minimum energy required to release an electron from a metal surface. If the energy of the photon is less than the work function, photo-emission does not occur.
• Given the work function ϕ = 2.4 eV, the wavelength for which photo-emission does not take place can be calculated using the photon energy equation and the given work function.

Calculation:

Using the formula for the energy of a photon:

\(E = \frac{hc}{λ},\)

We know that the work function is given as 2.4 eV, and the energy of the photon should be greater than or equal to the work function for photo-emission to occur. Hence, solving for λ:

\(λ = \frac{hc}{E}\)

Substitute the known values:

\(λ = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{2.4 \times 1.6 \times 10^{-19}}\)

\(λ \approx 700 \, \text{nm}\)

Explanation of Options:

• Option 1: Incorrect. This wavelength corresponds to higher energy, which is enough to overcome the work function.
• Option 2: Incorrect. This wavelength also corresponds to a photon energy greater than the work function.
• Option 3: Correct. A wavelength of 700 nm gives photon energy just equal to the work function, so photo-emission will not occur beyond this wavelength.
• Option 4: Incorrect. This wavelength would have energy sufficient to release an electron from the surface.

∴ The correct answer is: Option 3 (700 nm).

Concept:

• In the photoelectric effect, the energy of a photon (E) is absorbed by the electrons in the metal. The energy required to release an electron is known as the work function (ϕ ), and the kinetic energy (K) of the emitted electron is given by: \(K = E - ϕ\).
• The work function of the metal is given as \(ϕ = \frac{E}{3}\), and the energy of the photon is E.
• Thus, the kinetic energy of the photoelectron will range between: \(0 \leq K \leq E - \frac{E}{3} = \frac{2E}{3}.\)

Explanation of Options:

• Option 1: Incorrect. The kinetic energy cannot be equal to \(\frac{2E}{3} \)directly as this formula does not take the work function into account properly.
• Option 2: Incorrect. This suggests that the kinetic energy is simply \(\frac{E}{3}\), which is incorrect based on the photoelectric equation.
• Option 3: Correct. The kinetic energy of the photoelectrons ranges from 0 to \(\frac{2E}{3}\) as explained above.
• Option 4: Incorrect. The kinetic energy cannot exceed \(\frac{2E}{3},\) since that is the maximum after subtracting the work function.

∴ The correct answer is: Option 3.