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Chapter 6: Problem 42

so \(\mathrm{Hg}^{-08}\) nucleus is bombarded by \(\alpha\)-particles with velocity \(10^{7} \mathrm{~m} / \mathrm{s} .\) If the \(\alpha\)-particle is approaching the \(\mathrm{Hg}\) nucleus head-on then the distance of closest approach will be (a) \(1.115 \times 10^{-13} \mathrm{~m}\) (b) \(11.15 \times 10^{-13} \mathrm{~m}\) (c) \(111.5 \times 10^{-13} \mathrm{~m}\) (d) Zero

Short Answer

Expert verified
The correct option is (a) \(1.115 \times 10^{-13} \mathrm{~m}\).

Step by step solution

01

Definition of terms

First, understand the terms and symbols used in the exercise. \(\alpha\)-particle refers to a helium nucleus, with a charge of +2e. \(\mathrm{Hg}^{-08}\) refers to a mercury nucleus with -8 electrons. The distance of closest approach is the point at which the \(\alpha\)-particle comes closest to the nucleus before being repelled.
02

Energy Conservation

Using the principle of conservation of energy, the kinetic energy of the alpha particle at infinity is equal to its potential energy at the point of closest approach. Mathematically, this is expressed as \(\frac{1}{2}mv^2 = \frac{KZe^2}{r}\) where m is mass of \(\alpha\)-particle, v is its velocity, K is Coulomb鈥檚 constant, Z is atomic number of mercury, e is charge of an electron and r is the distance of closest approach.
03

Computing distance of closest approach

We substitute given values and the known values (for K, Z, e) into the equation from Step 2. Solving for r (distance of closest approach), which is the unknown, gives the required answer.

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