Exams › NEET › Physics › Dual Nature of Radiation and Matter
108 questions with worked solutions.
Answer: 0.1 Å
The radius of the circular path and the magnetic field allow us to calculate the momentum of the α-particle using the relation r = mv/qB. Using the de Broglie wavelength formula λ = h/p, we find the wavelength to be approximately 0.1 Å.
Answer: √hν/m
The maximum kinetic energy of the emitted electron is given by Einstein's photoelectric equation: K.E. = h(f - f₀), where f is the incident frequency and f₀ is the cut-off frequency. Substituting f = 2ν and f₀ = ν, we get K.E. = hν. The maximum velocity is then v = √(2K.E./m) = √(2hν/m).
Answer: > 2.8 × 10⁻⁹ m
The energy of the incident photon is calculated using E = hc/λ. Subtracting the work function gives the kinetic energy of the emitted electron. Using de Broglie's equation, λ = h/p, the wavelength of the electron is found to be greater than 2.8 × 10⁻⁹ m.
Answer: V₀
The stopping potential depends only on the frequency of the incident light and not on its intensity or distance from the source. Since the frequency remains unchanged, the stopping potential will remain V₀.
Answer: hc/3λ
Using the photoelectric equation, K.E. = hc/λ - ϕ, where ϕ is the work function. For λ and λ/2, the kinetic energies are related as K.E.2 = 3 × K.E.1. Solving these equations gives ϕ = hc/3λ.
Answer: 4λ
Using the photoelectric equation, we know that stopping potential is related to the energy of the incident light and the work function. For wavelength λ, the stopping potential is 3V₀, and for 2λ, it is V₀. Solving these equations gives the threshold wavelength as 4λ.
Answer: 0.65 eV
The photoelectric equation is given by: K.E. = hν - φ, where φ is the work function. Initially, hν = K.E. + φ = 0.5 + φ. When hν increases by 20%, the new energy is 1.2hν, and the new K.E. is 0.8 eV. Substituting, 0.8 = 1.2hν - φ. Solving these equations gives φ = 0.65 eV.
Q8. Calculate the kinetic energy of an electron accelerated through a potential difference of 100 V.
Answer: K.E. = eV = 1.6 × 10⁻¹⁹ × 100 = 1.6 × 10⁻¹⁷ J
The kinetic energy of a charged particle accelerated through a potential difference is given by K.E. = eV. Substituting e = 1.6 × 10⁻¹⁹ C and V = 100 V, we get K.E. = 1.6 × 10⁻¹⁷ J.
Answer: All of the above.
All the given statements are correct. (A) relates intensity to photocurrent, (B) highlights that photocurrent depends on light intensity and not applied voltage, and (C) correctly states the stopping potential's dependence on frequency and independence from intensity.
Answer: K.E. = hν − hν₀ = hc/λ − hν₀ = 12375/5000 − 1.9 = 0.58 eV
The kinetic energy of the electron is calculated using the photoelectric equation: K.E. = Photon Energy - Work Function. Substituting the given values, K.E. = 2.48 eV - 1.9 eV = 0.58 eV. Thus, option A is correct.
Q11. Calculate the threshold wavelength (λ) for a material with a work function W = 6.2 eV.
Answer: Threshold wavelength (λ) = 3000 × 10⁻¹⁰ m
The threshold wavelength is calculated using the formula λ = hc/W, where h = 6.63 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s, and W = 6.2 eV = 6.2 × 1.6 × 10⁻¹⁹ J. Substituting these values gives λ ≈ 3000 × 10⁻¹⁰ m.
Q12. Calculate the energy of a photon with wavelength λ = 0.21 m.
Answer: E = hν = hc/λ = 6.6 × 10⁻³⁴ × 3 × 10⁸ / 0.21 = 0.94 × 10⁻²³ J
The energy of a photon is calculated using the formula E = hc/λ. Substituting h = 6.6 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s, and λ = 0.21 m, we get E = (6.6 × 10⁻³⁴ × 3 × 10⁸) / 0.21 = 0.94 × 10⁻²³ J.
Answer: E = KEₘₐₓ + ϕ
The total incident energy E is the sum of the maximum kinetic energy KEₘₐₓ of the electron and the work function ϕ, as per the photoelectric equation: E = KEₘₐₓ + ϕ.
Answer: 654 nm
The wavelength of light emitted can be calculated using the formula λ = hc/E, where h is Planck's constant, c is the speed of light, and E is the energy gap. Substituting the values (h = 6.63 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s, E = 1.9 eV = 1.9 × 1.6 × 10⁻¹⁹ J), we get λ ≈ 654 nm.
Answer: \( \frac{2E}{C} \)
When radiation falls on a perfectly reflecting surface, the momentum transferred is twice the momentum of the incident radiation because the radiation reverses direction. The momentum of the radiation is given by \( p = \frac{E}{c} \), so the total momentum transferred is \( 2p = \frac{2E}{c} \).
Answer: 103 V
The de Broglie wavelength is given by λ = h / √(2meV), where h is Planck's constant, m is the mass of the electron, e is the charge of the electron, and V is the potential difference. Substituting the given wavelength and solving for V, we find V ≈ 103 V.
Answer: 12.2 × 10⁻¹² m
The de Broglie wavelength is calculated using λ = h / √(2mₑeV). Substituting h = 6.63 × 10⁻³⁴ J·s, mₑ = 9.31 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C, and V = 10,000 V, we get λ ≈ 12.2 × 10⁻¹² m.
Answer: λ₀
The de Broglie wavelength is inversely proportional to the momentum of the particle. Since the electron is decelerated by the electric field, its velocity decreases, and hence its momentum changes. However, the problem does not provide sufficient information to calculate the exact change in wavelength, so the initial wavelength λ₀ remains the best approximation.
Answer: 2 : 1
The de Broglie wavelength is inversely proportional to the square root of the mass for particles with the same kinetic energy. Since the α-particle has four times the mass of the proton, its de Broglie wavelength will be half that of the proton. Thus, the ratio λₚ : λₐ is 2:1.
Answer: h / √3mₖT
The de-Broglie wavelength is given by λ = h / p, where p is the momentum. For a neutron in thermal equilibrium, its kinetic energy is (3/2)kT, so p = √(2mₙ(3/2)kT). Substituting this into the de-Broglie formula gives λ = h / √(3mₙkT).