Chemistry, asked by shivamlaul122, 11 months ago

A beam of specific kind of particles of
velocity 2.1 x 107 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.
(A) 4.84 x 107 C/kg
(B) 4.84 x 10-7c/kg
(C) 2.42 x 10°C/kg
(D) 3 x 10-12c/kg​

Answers

Answered by PragyaBhargav
5

QUESTION:-

A beam of specific kind of particles of

velocity 2.1 x 107 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:-

Answer:q2/m

= 4.84×10^7C/kg

Explanation:

At closest approach ,Kinetic Energy of particle

= potential Energy of Particlethat is,1/2 mv^2

= k q1q2/rq2/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/kgRead more on

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