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12.Atoms
medium
$\alpha - $ particles of energy $400\, KeV$ are bombarded on nucleus of $_{82}Pb$. In scattering of $\alpha - $ particles, its minimum distance from nucleus will be
A$0.59 \,nm$
B$0.59\,\mathring A$
C$5.9 \,pm$
D$0.59\, pm$
Solution
(d) Suppose closest distance is $r$, according to conservation of energy.
$400 \times {10^3} \times 1.6 \times {10^{ – 19}} = 9 \times {10^9}\frac{{(ze)\,(2e)}}{r}$
$ \Rightarrow 6.4 \times {10^{ – 14}}$$ = \frac{{9 \times {{10}^9} \times (82 \times 1.6 \times {{10}^{ – 19}}) \times (2 \times 1.6 \times {{10}^{ – 19}})}}{r}$
$ \Rightarrow r = 5.9 \times {10^{ – 13}}m = 0.59\,pm$.
$400 \times {10^3} \times 1.6 \times {10^{ – 19}} = 9 \times {10^9}\frac{{(ze)\,(2e)}}{r}$
$ \Rightarrow 6.4 \times {10^{ – 14}}$$ = \frac{{9 \times {{10}^9} \times (82 \times 1.6 \times {{10}^{ – 19}}) \times (2 \times 1.6 \times {{10}^{ – 19}})}}{r}$
$ \Rightarrow r = 5.9 \times {10^{ – 13}}m = 0.59\,pm$.
Standard 12
Physics
