Question
$7.9MeV\alpha$-particle scatters from a target material of atomic number 79. From the given data the estimated diameter of nuclei of the target material is (approximately) $\_\_\_\_$ m.
$\left\lbrack \frac{1}{4\pi\epsilon_{o}} = 9 \times 10^{9}{Nm}^{2}/C^{2} \right.\ $ and electron charge $\left. \ = 1.6 \times 10^{- 19}C \right\rbrack$
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Solution
By mechanical energy conservation
$${(Me)_{i} = (Me)_{f} }{{PE}_{i} + {KE}_{i} = {PE}_{f} + {KE}_{f} }{0 + 7.9 \times 10^{6} \times 1.6 \times 10^{- 19} = \frac{k(2e)(Ze)}{r} + 0 }{r = \frac{9 \times 10^{9} \times 2 \times \left( 1.6 \times 10^{- 19} \right)^{2} \times 79}{7.9 \times 10^{6} \times 1.6 \times 10^{- 19}} = 2.88 \times 10^{- 14}\text{ }m }$$For diameter ⇒ D = $2r = 5.76 \times 10^{- 14}\text{ }m$
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