Chapter 37: Q 43 (page 1083)

In a head-on collision, the closest approach of a 6.24 MeV alpha particle to the center of a nucleus is 6.00 fm. The nucleus is in an atom of what element? Assume the nucleus remains at rest.

Short Answer

Expert verified

The element is Aluminum

Step by step solution

Step 1. Given information is :Kinetic energy of alpha particle = 6.24 MeVClosest approach = 6.00 fm

We need to find out the element used.

Step 2. Comparing Kinetic and potential energy for the closest approach

Let Z be the atomic number of the element.

$C h a r g e o n e l e m e n t = Z \times 1.6 \times 10^{- 19} C C h a r g e o n a l p h a p a r t i c l e = 2 \times 1.6 \times 10^{- 19} C D i s \tan c e o f c l o s e s t a p p r o a c h = 6.00 f m = 6.00 \times 10^{- 15} m K i n e t i c e n e r g y o f a l p h a p a r t i c l e = 6.24 M e V = 6.24 \times 10^{6} \times 1.6 \times 10^{- 19} C$

At the closest approach, Kinetic energy will be equal to the potential energy. Therefore,

$\frac{1}{4 π \in_{o}} \frac{q_{1} q_{2}}{r} = K 9 \times 10^{9} \times \frac{Z \times 1.6 \times 10^{- 19} \times 2 \times 1.6 \times 10^{- 19}}{6.0 \times 10^{- 15}} = 6.24 \times 10^{6} \times 1.6 \times 10^{- 19} Z = \frac{6.24 \times 10^{6} \times 1.6 \times 10^{- 19} \times 6.0 \times 10^{- 15}}{1.6 \times 10^{- 19} \times 2 \times 1.6 \times 10^{- 19} \times 9 \times 10^{9}} = 13$