Atoms And Nuclei Question 300
Question: In Rutherford scattering experiment, the number of a-particles scattered at $ 60{}^\circ $ is $ 5\times 10^{6} $ . The number of a-particles scattered at $ 120{}^\circ $ will be
Options:
A) $ 15\times 10^{6} $
B) $ \frac{3}{5}\times 10^{6} $
C) $ \frac{5}{9}\times 10^{6} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- $ N\propto \frac{1}{{{\sin }^{4}}\theta /2};\frac{N_{2}}{N_{1}}=\frac{{{\sin }^{4}}( {\theta_{1}}/2 )}{{{\sin }^{4}}( {\theta_{2}}/2 )} $ $ \text{or }N_{2}=5\times 10^{6}\times {{( \frac{1}{2} )}^{4}}{{( \frac{2}{\sqrt{3}} )}^{4}}=\frac{5}{9}\times 10^{6} $
