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12.Atoms
hard
If in Rutherford’s experiment, the number of particles scattered at ${90^o}$ angle are $28$ per min, then number of scattered particles at an angle ${60^o}$ and ${120^o}$ will be
A
$112/min, 12.5/min$
B
$100/min, 200/min$
C
$50/min, 12.5/min$
D
$117/min, 25/min$
Solution
(a) $N \propto \left[ {\frac{1}{{{{\sin }^4}\theta /2}}} \right]$
==> ${N_1} = 7 \times \frac{1}{{{{(\sin {{30}^o})}^4}}} = 112$
and ${N_2} = 7 \times \frac{1}{{{{(\sin {{60}^o})}^4}}} = 12.5$.
Standard 12
Physics
