Atoms And Nuclei Question 85
Question: If scattering particles are 56 for $ 90^{o} $ angle then this will be at $ 60^{o} $ angle [RPMT 2000]
Options:
A) 224
B) 256
C) 98
D) 108
Show Answer
Answer:
Correct Answer: A
Solution:
According to scattering formula $ N\propto \frac{1}{{{\sin }^{4}}(\theta /2)}\Rightarrow \frac{N_{2}}{N_{1}}={{[ \frac{\sin ({\theta_{1}}/2)}{\sin ({\theta_{2}}/2)} ]}^{4}} $
$ \Rightarrow \frac{N_{2}}{N_{1}}={{[ \frac{\sin \frac{90^{o}}{2}}{\sin \frac{60^{o}}{2}} ]}^{4}}={{[ \frac{\sin 45^{o}}{\sin 30^{o}} ]}^{4}} $
$ \Rightarrow N_{2}={{(\sqrt{2})}^{4}}\times N_{1}=4\times 56=224 $
BETA
