A proton moving in the positive x direction at 4.3 Mm/s collides with a nucleus. The collision lasts 0.12 fs, and the average impulsive force is 42 i + 17 j micro - Newton. A) Find the velocity of the proton after the collision. B) Through what angle has the proton's motion been deflected?
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- If two nuclei are to fuse in a nuclear reaction, they must be moving fast enough so that the repulsive Coulomb force between them does not prevent them for getting within R1014mof one another. At this distance or nearer, the attractive nuclear force can overcome the Coulomb force, and the nuclei are able to fuse. (a) Find a simple formula that can be used to estimate the minimum kinetic energy the nuclei must have if they are to fuse. To keep the calculation simple, assume the two nuclei are identical and moving toward one another with the same speed v. (b) Use this minimum kinetic energy to estimate the minimum temperature a gas of the nuclei must have before a significant number of them will undergo fusion. Calculate this minimum temperature first for hydrogen and then for helium. (Hint: For fusion to occur, the minimum kinetic energy when the nuclei are far apart must be equal to the Coulomb potential energy when they are a distance R apart.)A solid copper sphere whose radius is 1.0 cm has a verythin surface coating of nickel. Some of the nickel atoms areradioactive, each atom emitting an electron as it decays. Halfof these electrons enter the copper sphere, each depositing 100 keVof energy there.The other half of the electrons escape, each carryingaway a charge e.The nickel coating has an activity of 3.70 *10^8 radioactivedecays per second. The sphere is hung from a long, nonconductingstring and isolated from its surroundings. (a) How longwill it take for the potential of the sphere to increase by 1000 V? (b)How long will it take for the temperature of the sphere to increaseby 5.0 K due to the energy deposited by the electrons? The heatcapacity of the sphere is 14 J/K.What is the magnitude of the repulsive electrostatic force between two protons in a nucleus? Consider the distance between the centers of the protons to be 3.5 x 10^-13 m.If these protons were released from rest, Calculate the magnitude of their initial acceleration?
- any speed up to 8.08 m/s (а) 8.33 (b) 9.15 N toward the nucleus m/s inward2. The nuclear radius of gold is approximately r = 7.0 fm (1.0 fm = 1.0 × 10-15 m). The radii of protons and a particles are 1.3 fm and 2.6 fm, respectively. (a) What energy a particles would be needed in head-on collision for the nuclear surfaces to just touch? (This is about where the nuclear force becomes effective.) (b) What energy protons would be needed?An alpha particle is whizzing by at a velocity determined by its KE of 5 Mev. a ) What is the velocity of the alpha particle? b) What is the magnetic field at the position of a nucleus 2nm away from the alpha particle at angle of 900 to the velocity of the alpha. Alpha particle has charge of 2e but mass of 4 protons.
- 10) Now you have a nucleus with 13 protons at x = 6.2 Angstroms on the x-axis. How much work would it take to bring in ANOTHER nucleus with 7 protons from 1 m away and place it at y = 8.0 Angstroms on the y-axis? 70.0 eV 116.7 eV -12.6 eV 129.3 eVAn alpha particle (charge +3.20 x 10^-19C, mass 6.64 x10^-27kg) is initially 5.2cm away from a fixed golden nucleus (charge +1.36 x10^-17C, mass 3.29x10^-25kg), and moving toward the nucleus with a speed of 8.1x10^5m/s. How close to the nucleus does te alpha particle get? Note: the nucleus diameter is approximately 10^-14m and the alpha particles's is 10^-15mA typical carbon nucleus contains 6 neutrons and 6 protons. The 6 protons are all positively charged and in very close proximity, with separations on the order of 10-15 meters, which should result in an enormous repulsive force. What prevents the nucleus from dismantling itself due to the repulsion of the electric force? a. The attractive nature of the strong nuclear force overpowers the electric force. b. The weak nuclear force barely offsets the electric force. c. Magnetic forces generated by the orbiting electrons create a stable minimum in which the nuclear charged particles reside. d. The attractive electric force of the surrounding electrons is equal in all directions and cancels out, leaving no net electric force.
- A beam of alpha particles is incident on a target of lead. A particular alpha particle comes in “head-on” to a particular lead nu- cleus and stops 6.50 * 10-14 m away from the center of the nucleus. (This point is well outside the nucleus.) Assume that the lead nucleus, which has 82 protons, remains at rest. The mass of the alpha particle is 6.64 * 10-27 kg. (a) Calculate the electrostatic potential energy at the instant that the alpha particle stops. Express your result in joules and in MeV. (b) What initial kinetic energy (in joules and in MeV) did the alpha particle have? (c) What was the initial speed of the alpha particle?c) The equation below describes the disintegration of a polonium nucleus into a lead nucleus and an alpha-particle. During the reaction energy Q is released. 210Po → He +²02Pb+Q 84 82 Calculate the loss of energy during the reaction. The masses in the atomic mass unit u are as follows: 210 206 Po= 209.98287 u, Pb = 205.97446 u and He = 4.002604 u. 84 82 You may assume that 1u is equivalent to 931 MeV. d) The lead nucleus recoils in the opposite direction to the emitted alpha particle conserving momentum. Hence calculate: i) The ratio of the recoil nucleus and alpha particle velocities ii) The kinetic energy distribution of these products.Consider an object of mass 56.6 kg. Assume that it s made up of equal numbers of protons, neutrons, and electrons. How many protons does this object contain? Question 1 options: 8.45E+27 1.69E+28 3.38E+28 6.76E+28