A) 117 per minute, 25 per minute
B) 50 per minute, 12.5 per minute
C) 100 per minute, 200 per minute
D) 112 per minute, 12.4 per minute
Correct Answer: D
Solution :
[d] No. of particles scattered through an angle \[\theta =N\left( \theta \right)=\frac{k{{Z}^{2}}}{{{\sin }^{4}}\left( \frac{\theta }{2} \right){{\left( K.E. \right)}^{2}}}\] \[\therefore 28=\frac{4kc{{z}^{2}}}{{{\left( K.E. \right)}^{2}}}\text{ for }\theta \text{=90}{}^\circ \] \[\therefore \frac{kc{{z}^{2}}}{{{\left( K.E. \right)}^{2}}}\text{ =}\frac{28}{4}=7\] \[\therefore N\left( 60{}^\circ \right)=\frac{7}{{{\sin }^{4}}\left( \frac{60{}^\circ }{2} \right)}=16\times 7=112/\min .\] \[N\left( 120{}^\circ \right)=\frac{7}{\sin \left( \frac{120{}^\circ }{2} \right)}=12.4/\min \]
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