In Rutherford\'s scattering experiments, alpha particles (charge = +2 e ) were f

Question

In Rutherford's scattering experiments, alpha particles (charge = +2e) were fired at a gold foil. Consider an alpha particle with an initial kinetic energy K heading directly for the nucleus of a gold atom (charge =+79e). The alpha particle will come to rest when all its initial kinetic energy has been converted to electrical potential energy.

Part A

Find the distance of closest approach between the alpha particle and the gold nucleus for the case K = 2.9MeV .

d=____________fm?

I answered 1.515*10^-13 but its wrong what did i do wrong???

Explanation / Answer

potential energy of the system=9*10^9*2*1.6*10^(-19)*79*1.6*10^(-19)/d

=3.6403*10^(-26)/d

=2.2752*10^(-7)/d eV

so we need to find

2.2752*10^(-7)/d=2.9*10^6

d=7.8455*10^(-14) m

now as we know 1 fm=10^(-15) m

so d=78.455 fm

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