+ 0/1 points ! Pre ous Answers oscoIPhys2016 19.1.WA006. A bare helium nucleus h

Question

+ 0/1 points ! Pre ous Answers oscoIPhys2016 19.1.WA006. A bare helium nucleus has two positive charges and a mass of 6.64 x 10-27 kg. (a) Calculate its kinetic energy in joules at 6.90% of the speed of light. .13e-11X What is the basic definition of kinetic energy? (b) What is this in electron volts? 0.08e8X What is the definition of an electron volt? eV (c) What voltage would be needed to obtain this energy? 406 Did you consider applying the conservation of energy principle? V Additional Materials Reading

Explanation / Answer

Solution :

(a) The speed of the helium nucleus is = v = (0.0690)(3.00*10^8m/s) = 2.07*10^7 m/s

KE = 1/2 mv2 = 1/2 (6.64*10^-27kg)(2.07*10^7)^2 = 1.422*10^-12 J

(b) We use the conversion factor 1 eV = 1.601019J to get

KE= (1.422*10^-12J)*(1eV/1.60*10^-19J) = 8.88*10^6eV

(c) The charge of the helium nucleus is qHe = 2q whereqis the magnitude of the charge on anelectron.
The nucleus has to undergo a change in electric potential energy that is equal in magnitude to the gain in kinetic energy

V = PE/qHe = (1.422*10^-12J)/(2*(1.60*10^-19) = 4.44*10^6V

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