Content Area - (118 X 556313855study. ? Y PHYSICS QUESTION × E2. Methods for the

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Content Area - (118 X 556313855study. ? Y PHYSICS QUESTION × E2. Methods for the × * 00384972.pdf × © Home! Chegg.com https://session.masteringphysics.com/myct/itemView?assignmentProblemID-97538875 Problem 23.85 11 of 12> Part A Rutherford backscattering spectrometry (RBS) is a technique used to determine the structure and composition of materials A beam of ions (typically helium ions) is accelerated to high energy and aimed at a sample. By analyzing the distribution and energy of the ions that are scattered from (that is deflected by collisions with) the atoms in the sample researchers can determine the sample's composition. To accelerate the ions to high energies, a tandem electrostatic accelerator may be used. In this device, negative ions (He) start at a pottial V 0 and are accelerated by a high positive voltage at the midpoint of the accelerator. The high voltage produces a constant electric field in the acceleration tube through which the ions move. When accelerated ions reach the midpoint, the electrons are stripped off, turning the negative ions into doubly positively charged ions (He. These positive ions are then repelled from the midpoint by the high positive voltage there and continue to accelerate to the far end of the accelerator, where again V 0 A helium ion (He+) that comes within about 10 fm of the center of the nucleus of an atom in the sample may induce a nuclear reaction instead of simply scattering. Imagine a helium ion with a kinetic energy of 3.5 MV heading straight toward an atom at rest in the sample. Assume that the atom stays fixed. What minimum charge can the nucleus of the atom have such that the helium ion gets no closer than 10 fm from the center of the atomic nucleus? (1 fm1 x 10-15 m and eis the magnitude of the charge of an electron or a proton.) Submit Provide Feedback Next > Activate Windows Go to Settings to activate Windows Desktop» ^ ?dp :08 AM 18-Jul-18 19 O Type here to search ?t

Explanation / Answer

A)

KE = 3.5 MeV = 3.5 x 106 x 1.6 x 10-19 = 5.6 x 10-13 J

r = 10 fm = 10 x 10-10 m

qn = charge on nucleus

q = charge on helium = 2 e = 2 x 1.6 x 10-19

Using conservation of energy

electric potential energy = kinetic energy

k q qn/r = KE

(9 x 109) (2e) qn/(10 x 10-10) = 3.5 x 106 e

qn=13 e

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