Dual Nature Of Radiation & Matter

The de-Broglie wavelength (λ) is inversely proportional to the momentum (p) of a body. This means that as the momentum of a body increases, its de-Broglie wavelength decreases. Conversely, as the momentum decreases, the de-Broglie wavelength increases. This relationship is consistent with the wave-particle duality of matter, as described by Louis de Broglie's hypothesis.

When an electron jumps across a potential difference of 1 V, it gains energy equal to the electric potential difference. This is because the electric potential difference is a measure of the work done per unit charge to move the charge between two points in an electric field. Therefore, the energy gained by the electron is equal to the potential difference, which in this case is 1 V.

The slope of the graph represents the ratio of Planck's constant (h) to the charge of an electron (e). This is because the stopping potential is directly proportional to the frequency of incident light, which is related to the energy of the photons. According to the photoelectric effect equation, the energy of a photon is given by E = hf, where h is Planck's constant and f is the frequency. The stopping potential is directly proportional to the energy of the photons, so the slope of the graph gives the ratio of h to e.

The photoelectric effect is the emission of electrons from a material when it absorbs electromagnetic radiation. It can be caused by X-rays, visible light, and gamma rays, as all of these forms of radiation have enough energy to dislodge electrons from their atoms. This phenomenon is not limited to a specific type of radiation but can occur with any form of electromagnetic radiation that has sufficient energy.

The question is asking for the color that has the minimum stopping potential. Stopping potential refers to the minimum potential difference required to stop the flow of electrons in a photoelectric effect experiment. According to the photoelectric effect, the stopping potential is directly proportional to the frequency of the incident light. The frequency of light is inversely proportional to its wavelength. Since red light has the longest wavelength among the given options, it has the lowest frequency and therefore requires the minimum stopping potential to stop the flow of electrons.

When the kinetic energy of an electron triples, its velocity also triples since kinetic energy is directly proportional to the square of velocity. The de-Broglie wavelength of an electron is inversely proportional to its velocity. Therefore, when the velocity triples, the de-Broglie wavelength changes by a factor of 1/3.

The maximum kinetic energy of a photoelectron is directly proportional to the frequency of the incident radiation. This means that as the frequency increases, the maximum kinetic energy of the photoelectron also increases.

The stopping potential in a photoelectric experiment is the minimum potential required to stop the photoelectrons from reaching the collector plate. It is given by the equation V = eV0, where V is the stopping potential and e is the charge of an electron. Since the stopping potential is given as 9V, we can rearrange the equation to solve for V0. The maximum velocity of the ejected electron can be determined using the equation V0 = √(2eV0/m), where m is the mass of the electron. Given that e/m is missing from the question, it is not possible to calculate the maximum velocity of the ejected electron.

The de-Broglie wavelength is inversely proportional to the momentum of a particle. Since protons and alpha particles have the same de-Broglie wavelength, it implies that they must have the same momentum. Therefore, momentum is the same for both protons and alpha particles.

The ratio of the maximum speeds of the emitted electrons is 1:2. This can be explained using the photoelectric effect. The work function of the metallic surface is the minimum energy required to remove an electron from the surface. When a photon with energy equal to or greater than the work function strikes the surface, an electron is emitted with a maximum kinetic energy equal to the difference between the photon energy and the work function. In this case, the first radiation with energy 1 eV is less than the work function, so no electrons are emitted. However, the second radiation with energy 2.5 eV is greater than the work function, so electrons are emitted with a maximum kinetic energy of 2.5 eV - 0.5 eV = 2 eV. Therefore, the ratio of the maximum speeds of the emitted electrons is 1:2.

The de-Broglie wavelength is inversely proportional to the momentum of a particle. Since all the particles mentioned (proton, neutron, electron, and α-particle) have the same energy, their momenta will be different due to their different masses. The momentum of a particle is given by its mass times its velocity, so the lighter the particle, the higher its velocity and momentum. Therefore, the electron, being the lightest particle, will have the highest momentum and the shortest de-Broglie wavelength, while the α-particle, being the heaviest, will have the lowest momentum and the longest de-Broglie wavelength.

The maximum energy of a photon in the radiation can be calculated using the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. Since the short wavelength end is given as 0.45 Ǻ, we can substitute this value into the equation to find the maximum energy of the photon.

The de Broglie wavelength of a particle is inversely proportional to its momentum. Therefore, if the de Broglie wavelength changes by 0.5% when the momentum changes by P, it means that the momentum has doubled. Since the initial momentum is P, the correct answer is 200 P.

The de-Broglie wavelength of a particle is given by the equation λ = h/mv, where λ is the wavelength, h is Planck's constant, m is the mass of the particle, and v is its velocity. In this case, the mass of the particle is 1 mg (0.001 g) and its velocity is 1 m/s. Plugging these values into the equation, we get λ = h/(0.001 g * 1 m/s) = h/(0.001 kg m/s) = 1000h/(kg m/s) = 1000h. Therefore, the correct answer is 1000h, which is equivalent to 106h.

The de-Broglie wavelength of a particle is given by the equation λ = h/p, where λ is the wavelength, h is the Planck's constant, and p is the momentum of the particle. In this case, the proton is accelerated through a potential difference, which implies that it gains kinetic energy. The momentum of the proton can be calculated using the equation p = √(2mE), where m is the mass of the proton and E is the kinetic energy gained. Since the proton is initially at rest, the kinetic energy gained is equal to the potential energy gained, which is given by E = qV, where q is the charge of the proton and V is the potential difference. Substituting these values into the equation for momentum and then into the equation for wavelength, we find that the de-Broglie wavelength of the proton is 1.23 Ǻ.