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12.Atoms
medium
An $\alpha$- particle of $5\ MeV$ energy strikes with a nucleus of uranium at stationary at an scattering angle of $180^o$. The nearest distance upto which $\alpha$- particle reaches the nucleus will be of the order of
A
$1 \;\mathring A$
B
${10^{ - 10}}\;cm$
C
${10^{ - 12}}\;cm$
D
${10^{ - 15}}\;cm$
(AIEEE-2004) (IIT-1981)
Solution
At closest distance of approach
Kinetic energy = Potential energy
$ \Rightarrow 5 \times {10^6} \times 1.6 \times {10^{ – 19}} = \frac{1}{{4\pi {\varepsilon _0}}} \times \frac{{(ze)(2e)}}{r}$
For uranium $z= 92$, so $r = 5.3 \times {10^{ – 12}}cm$
Standard 12
Physics
