GP Quiz Test

γ-rays have very high photon energy. Pair production needs a high-energy photon: γ → e⁻ + e⁺ Minimum condition: h f ≥ 2m₀c² This condition is typically met by γ-ray photons, so γ-rays can produce electron–positron pairs.

In the depletion region of an unbiased P-N junction diode there are only fixed or immobile positive and negative atoms. Note: As the depletion layer contains no free or mobile charge carriers but only fixed and immobile ions, this layer (or region) behaves like an insulator.

The minimum energy needed to just eject an electron from a metal surface is called the work function. Work function is written as φ. At threshold condition Photon energy = work function h f₀ = φ Stopping potential is the voltage needed to stop the fastest electrons, not the minimum removal energy.

Electrons in a metal are bound to the surface. A minimum energy is needed to just free an electron from the metal. This minimum required energy is called the work function (φ). If photon energy is E = h f, then emission occurs only when: h f ≥ φ So the energy required to escape the electron from the metal surface is the work function

In this condition P−N junction is reverse biased. and in reversed biased it worked as insulator because no current pass

Phase wave rectifier is not a type of rectifier.

Photoelectric emission rate depends on how many photons hit the metal per second. Higher light intensity means more photons per second. More photons per second means more electrons ejected per second (if frequency is above threshold). So number of ejected photoelectrons increases with intensity. Correct option: A) in intensity of light

In unbiased condition of PN-junction, depletion region is generated which stops the movement of charge carriers.

Nuclear force is the strong attractive force that holds nucleons together in the nucleus. It acts between protons and neutrons and keeps the nucleus stable. Without this force, protons would repel each other and the nucleus would break apart.

Photoelectric effect shows that light transfers energy in discrete packets (photons), not continuously like a wave. Key facts: Energy of one photon E = h f One photon gives its energy to one electron. If h f < φ, no emission occurs even with high intensity. This matches the photon (quantum) model of light. Therefore, photoelectric effect supports Einstein’s quantum theory of light.

When an electron absorbs energy, it gains energy. So it moves from a lower energy level to a higher energy level. This higher level is called the excited state.

Full wave rectifier, V rms = V max /√2 = 20/√2 = 14.14

In the Lyman series, the electron always falls to the first energy level. So the final level is n = 1. This series lies in the ultraviolet region.

PN junction Diode works as an insulator if connected in reverse bias with no current flows.

The spectral series depends on the final orbit. If the electron jumps to n = 1, the series is called Lyman series. So when electron falls into n₁ orbit, it gives the Lyman series. Short trick: Final orbit decides the series n = 1 → Lyman n = 2 → Balmer n = 3 → Paschen n = 4 → Brackett n = 5 → Pfund

After the wave has been half-rectified; the frequency of the wave remains same. When the wave has been fully rectified, then the frequency doubles.

Particle nature of light means light transfers energy in discrete packets (photons). Photoelectric effect shows this clearly: Electrons are ejected only when photon energy is enough. Photon energy: E = h f If f < f₀, no emission even with high intensity. This is a particle (photon) behavior. Refraction, interference, and polarization are wave phenomena.

There is no net charge on N-type or P-type substances because are formed by neutral atoms.

Reverse biased diode ⇒ current ≈ 0 Voltage drop across resistor ≈ 0 So Vₐ − Vᵦ ≈ battery voltage = 6 V

In the Balmer series, all transitions end at n = 2. The lines are: 1st line: 3 → 2 2nd line: 4 → 2 3rd line: 5 → 2 4th line: 6 → 2 So the fourth line of the Balmer series is the transition from 6 to 2.

Einstein photoelectric equation: Kₘₐₓ = E − φ Given: E = 6.2 eV φ = 4.2 eV Calculation: Kₘₐₓ = 6.2 eV − 4.2 eV Kₘₐₓ = 2.0 eV Convert eV to joule: 1 eV = 1.6 × 10⁻¹⁹ J Kₘₐₓ = 2.0 × (1.6 × 10⁻¹⁹) J Kₘₐₓ = 3.2 × 10⁻¹⁹ J Kₘₐₓ ≈ 3 × 10⁻¹⁹ J

Short trick: Atom → line spectrum Molecule → band spectrum Hot solid → continuous spectrum Explanation: Atomic spectra consist of separate bright or dark lines at specific wavelengths. This happens because electrons in atoms can change energy only by fixed amounts. So atoms produce line spectra, not continuous or band spectra. Continuous spectrum is given by hot solids or liquids. Band spectrum is usually given by molecules. So atomic spectra is an example of line spectra.

It is used to convert AC into DC (rectifier).

In photoelectric effect, the maximum kinetic energy of emitted electrons is given by Einstein’s equation: Kₘₐₓ = h f − ϕ Where: Kₘₐₓ = maximum kinetic energy of photoelectrons h = Planck constant f = frequency of incident light ϕ = work function of the metal So: If frequency f increases → Kₘₐₓ increases Intensity changes the number of emitted electrons, not their kinetic energy (for fixed f above threshold).

For an electron accelerated through potential V, its kinetic energy is K = eV Here V = 600 V de-Broglie wavelength (non-relativistic electron) λ = h / √(2 m e V) Useful direct form λ(Å) = 12.27 / √V Now substitute λ = 12.27 / √600 Calculate √600 √600 = √(6 × 100) √600 = 10√6 √6 ≈ 2.45 √600 ≈ 24.5 Now wavelength λ ≈ 12.27 / 24.5 λ ≈ 0.50 Å

Photoelectric emission needs one photon to give all its energy to one electron at one time. An electron cannot add energies from two separate photons to cross the work function in the basic photoelectric effect. Given Work function φ = 4.2 eV Energy of each photon E = 3.5 eV Check condition for emission E ≥ φ 3.5 eV < 4.2 eV So a single photon cannot eject an electron. Two photons of 3.5 eV do not combine their energies for one electron in ordinary photoelectric emission. Therefore emission is not possible.

In an energy level diagram, each allowed energy state Eₙ is shown by a horizontal line. These horizontal lines represent fixed energy levels of the atom. Vertical arrows are used only to show transitions between these levels. So, the energy levels Eₙ are represented by a series of horizontal lines.

The energy of nth quantized orbit is Eₙ = −13.6 / n² eV Given: n = 2 Put the value in formula E₂ = −13.6 / 2² eV E₂ = −13.6 / 4 eV E₂ = −3.4 eV

A hydrogen discharge tube contains excited hydrogen atoms. These atoms emit light at specific wavelengths only. So the spectrum appears as separate bright lines. That is why hydrogen filled discharge tube shows a line spectrum.

At low temperature, a hot body emits thermal radiation with a longer peak wavelength. Wien’s displacement law λₘₐₓ T = 2.9 × 10⁻³ m K If temperature T decreases λₘₐₓ increases Longer wavelengths lie in the infrared region. Infrared is not visible to human eyes. So both statements are correct. Correct option: D) Both A and C

Hertz discovered the effect, but could not explain it. Einstein explained it using the photon (quantum) idea of light. He said light energy comes in packets (photons) of energy: E = h f An electron uses energy φ to leave the metal, and the rest becomes kinetic energy: K.E(max) = h f − φ This explains: Why frequency matters (threshold frequency) Why intensity changes current (number of electrons), not K.E(max) So the successful explanation was given by Einstein.

In Compton scattering, the photon loses energy after collision with an electron. Photon energy and wavelength relation: E = h c / λ When E decreases, λ must increase. Therefore, the wavelength of the scattered photon increases.

Rectification is the process of converting alternating current (AC) into direct current (DC). This is achieved by using diodes to allow current to flow in one direction only.

Planck introduced the idea that energy is emitted or absorbed in discrete packets (quanta): E = h f Einstein extended this quantization idea to electromagnetic radiation itself. He said light consists of photons, each photon has energy E = h f. This photon concept explains the photoelectric effect. So Einstein extended Planck’s concept of quantization to electromagnetic waves.

Best evidence for wave nature of matter is diffraction or interference, because only waves show diffraction patterns. When electrons pass through a crystal, the regularly spaced atoms act like a diffraction grating. Electrons produce bright and dark diffraction maxima, just like X-rays do. This directly proves electrons behave as waves, consistent with de Broglie relation: λ = h/p Photoelectric effect and Compton effect are evidence for particle nature of light, not wave nature of matter.

Given λ = 5000 Å 1 Å = 10⁻¹⁰ m λ = 5000 × 10⁻¹⁰ m λ = 5.0 × 10⁻⁷ m Photon energy E = h c / λ Use the direct eV form E(eV) = 1240 / λ(nm) Convert wavelength to nm 5000 Å = 500 nm Now calculate photon energy E = 1240 / 500 E = 2.48 eV Photoelectric equation Kₘₐₓ = E − φ Given work function φ = 1.9 eV Now calculate maximum kinetic energy Kₘₐₓ = 2.48 − 1.9 Kₘₐₓ = 0.58 eV

de-Broglie wavelength λ = h / p Momentum p = m v All particles have the same velocity v, so p ∝ m λ = h / (m v) So λ ∝ 1/m Largest wavelength means smallest mass. Among the given particles, electron has the smallest mass. Therefore electron has the largest de-Broglie wavelength.

In Compton scattering, the photon gives some energy to the electron. So the scattered photon has less energy than the incident photon. Photon energy: E = h f If energy E decreases, frequency f also decreases. So both energy and frequency of the scattered photon are lower.

The Rydberg constant is a wave number constant. Its standard value is: R = 1.0974 × 10⁷ m⁻¹ Unit m⁻¹ means per meter, which is correct for wave number.

The first spectral series of hydrogen was discovered by J. J. Balmer. He studied the visible lines of hydrogen spectrum. This series is called the Balmer series. That is why the correct answer is J. J. Balmer.

Stopping potential relation: e V₀ = Kₘₐₓ Photoelectric maximum kinetic energy: Kₘₐₓ = h f − φ So, V₀ = (h f − φ) / e When distance changes from 10 cm to 20 cm, intensity decreases: I ∝ 1/r² I₂ / I₁ = (10/20)² = (1/2)² = 1/4 But stopping potential depends on frequency f, not on intensity. Changing intensity changes number of emitted electrons (photoelectric current), not Kₘₐₓ. Therefore V₀ remains the same

Kinetic energy gained by an electron accelerated through potential V is K = eV Given V = 100 V e = 1.6 × 10⁻¹⁹ C Now calculate K = (1.6 × 10⁻¹⁹) × 100 K = 1.6 × 10⁻¹⁹ × 10² K = 1.6 × 10⁻¹⁷ J