Question

16. alpha-particles are projected towards the following metals, with the same kinetic energy. Towards which metal, the distance of closest approach is minimum? 1) Cu(Z = 29. 2) Ag (Z = 47) 3) Au(Z = 79) 4) Ca(Z = 20)​

Answer 1

Explanation:

Given that,

α particles are projected towards the nuclei of the different metals, with the same kinetic energy.

For the distance of the closest approach, then

KE=PE....(1)

Where, KE is the kinetic energy of the particle and PE is the potential energy of the particle.

By substituting the potential energy formula in the above equation (1), then the equation (1) is written as,

KE=kq1q2r

By rearranging the terms in the above equation, then the above equation is written as,

r=kq1q2KE

By assuming the k, q1 and KE all are constant, then the above equation is written as,

r∝q2

From the above equation, as the r value decreases then the q2 also decreases, so from the given option, the option (A) is the smaller value compared to the others, so the closest distance is Cu(Z=29).

Hence, the option (A) is the correct answer

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Answer 2

Explanation:

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