Width of the slit is a The path difference between two seconadry wavelets is given by, Nλ = a sinθ Since, θ is very small, sinθ = θ So, for the first order diffraction n =1, the angle is λ/a Now, as we know that θ must be very small θ = 0 because of which the diffraction pattern is minimum. Now for the interference case, for two interfering waves of intensity l₁ and l₂ we must have two slits seperated by a distance. We have the resultant intensity l = l₁ + l -2 +2 √l₁ l₂ cos θ Since, θ = 0 corresponding to angle λ/a, so cos θ = 1 Hence, l = l₁ + l +2 √l₁ l₂ cos θ l = l₁ + l +2 √l₁ l₂ cos 0 l = l₁ + l +2 √l₁ l₂ We observe the resultant intensity is the sum of two inetensities, so there is a maxima corresponding to the angle λ/a. That' the reason that at the same angle λ/a we get a maximum fro two narrow slits seperated by a distance a.
The sum of currents entering a junction equals the sum of currents exiting the junction, according to the junction rule. Alternatively, Δi = 0 Justification: Conservation of charge
The algebraic sum of charges in the potential around a closed loop with resistors and cells in the loop is zero, according to the loop rule.
Alternatively, ΔV: 0, where ΔV represents the potential change.
A cyclotron works on the idea that by continuously passing charged particles through an electric field, their energy can be raised to a high level. The charged particles move in a circular pattern inside the dee due to the magnetic field. The alternating electric field acts on the particle every time it passes from one dee to the next, and this field accelerates the particle, increasing its energy. (i) It's used to bombard nuclei with high-energy particles accelerated by a cyclotron and examine the nuclear process that results. (ii) It's utilised to inject ions into solids, change their characteristics, and even create new materials.
Here, q = ±8 x 10⁻⁹, 2a = 4 cm = 0.04 m, τ = 4 √ 3 τ=43 Nm, and θ = 60° (i) Now, p = q(2a) = 8 x 10⁻⁹x 0.04 = 3.2 x 10⁻¹⁰cm Torque on the electric dipole is τ = pEsinθ or E = τ / psinθ = 4√ 3/ 3.2 x 10⁻¹⁰ x sin60° = 4√ 3 x 2 / 3.2 x 10⁻¹⁰ x √ 3 = 2.5 x 10⁻¹⁰ NC⁻¹ (ii) Potential energy of the electric dipole is U = pE cosθ = -3.2 x 10⁻¹⁰ 2.5 x 10⁻¹⁰ x cos60° = -3.2 x 2.5 x 0.5 = -4 J
(i) de Broglie wavelength is given by λ = h/√2mqV As mass of proton < mass of deuteron and qp = qd and V is same. => λp > λd for same accelerating potential (ii) Momentum = h/λ since, λp > λd therefore, the momentum of proton will be less than that of deuteron.
(i) Number of photons emitted per second = power / energy of one photon = 2.0 x 10⁻³/ 6.63 x 10⁻³⁴ x 6 x 10¹⁴ = 5.03 x 10¹⁵ (ii) The relationship between photoelectric current and incident radiation intensity on a photosensitive surface.
Energy required to excite hydrogen atom from ground state to the excited state = Efinal = Einitial = -1.51 - (-13.6) = 12 .09 eV In other words, hydrogen atoms would be excited to the third energy level (r: 3) / second excited state. Alternatively,
