(c) Equal to the kinetic energy of the proton, (d) Greater than the kinetic energy of the proton. I got 1.672 xx 10^(-9) "eV". Hence, the photon’s energy is greater than the kinetic energy of the electron. ωt... Q: The position of a simple harmonic oscillator with mass m is given Hence, KE = (½) pv, the kinetic energy of the electron is greater than the kinetic energy of the proton. Calculate the de Broglie wavelength of a proton moving at 2.42 ✕ 108 m/s. The factors of the proton mass in the denominator simplify and we can write that the de Broglie wavelength from the proton is Planck’s constant divided by the square root of two times the proton’s mass times its kinetic energy. The nucleus is initially at rest. ____m. When we recall that momentum is equal to an object’s mass times its velocity, , that means we can apply the de Broglie wavelength equation to our scenario by writing that the wavelength of the proton equals Planck’s constant divided by the proton’s mass times its speed. where the photon energy was multiplied with the electronic charge to convert the energy in Joule rather than electron Volt. Hence, the de Broglie wavelengths of the α-particle and residual nucleus will be in the ratio = 1:1. Question 6: Which of the following is associated with an electron microscope? Click hereto get an answer to your question ️ If a proton and an electron have the same de Broglie wavelength, the ratio of the velocity of proton to the velocity of electron will be nearly 9: An electron and a proton have the same de Broglie wavelength. A: (a) Q: A small ball of mass m, initially at A, slides on the smooth circular surface ADB (see the figure be... A: since only gravitational force is acting mechanical energy can be conserved. a = (2.00i + 5.00j) m/s². This value is also known as 20 femtometers. (Assume the mass of the proton to be 1.673 x 10¯ 27 kg.) Learn more about our Privacy Policy. The wavelength of an electron moving 5.31 x 106 m/sec is 1.37 x 10-10 m or 1.37 Å. The angular speed is given by, Let’s call this kinetic energy of 2.0 megaelectron volts capital KE. Comparing with the proton above, the wavelength for the proton is about #10# times the wavelength for this #alpha# particle at this velocity, which again makes physical sense. Nagwa is an educational technology startup aiming to help teachers teach and students learn. What is the de Broglie wavelength of a proton whose kinetic energy is 2.0 megaelectron volts? k = C+273. By the conservation of momentum principle, we can assume that the momenta of the α-particle of mass m and the residual nucleus will be equal and opposite. Q: A 3.00-kg object undergoes an acceleration given by Question 7: Calculate the de-Broglie wavelength of an electron which has been accelerated from rest on application of a potential of 400volts. proton Compton wavelength †: Numerical value: 1.321 409 855 39 x 10-15 m : Standard uncertainty: 0.000 000 000 40 x 10-15 m : Relative standard uncertainty: 3.1 x 10-10: Concise form Electron microscope makes use of the matter waves associated with fast-moving electrons. The de Broglie wavelength is inversely proportional to momentum. Then the kinetic energy of the electron is? where k is the spring constant... Q: What is the direction of the magnetic force on a positive charge that moves as shown in each of the ... A: Direction of magnetic force can be determined using right hand rule. *Response times vary by subject and question complexity. Find (a) the resultant... A: The displacement vector A→can be written as, The planet’s mass is. (a) Zero (b) Infinity (c) Equal to the kinetic energy of the proton (d) Greater than the kinetic energy of the proton. One electron volt is equal to 1.602 times 10 to the negative 19th joules. In the end, we conclude that the lighter and slower the particle, the more of a wave characteristic it exhibits. c) Using P = √(2Km) . We want to solve for the de Broglie wavelength of this proton; we’ll call that . What is the wavelength (in 10-15 m) of a proton traveling at 15.6% of the speed of light? Using that value under our square root sign for the energy, when we compute , we find that it’s equal to 20 times 10 to the negative 15th meters.

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