Question

A beam of specific kind of particles of velocity
2.1 x 10^7 m/s is scattered by a gold (Z = 79)
nuclei. Find out specific charge (charge/mass) of
this particle if the distance of closest approach is
2.5 x 10^-14 m.​

Answer 1

Answer:

q2/m = 4.84×10^7C/kg

Explanation:

At closest approach ,

Kinetic Energy of particle = potential Energy of Particle

that is,

1/2 mv^2 = k q1q2/r

q2/m= r*v^2/(2*k q1 Z)

q2/m = (2.5 x 10^-14 * (2.1 x 10^7)^2) /  (2* 9x109 * 79 *1.6 x10^-19 )

q2/m = 4.84×10^7C/kg

Answer 2

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