Electrostatics 2 Question 16

17. An alpha particle of energy 5MeV is scattered through 180 by a fixed uranium nucleus. The distance of closest approach is of the order of

(a) 1\AA

(b) 1010cm

(c) 1012cm

(d) 1015cm

(1981,2M)

Assertion and Reason

Mark your answer as

(a) If Statement I is true, Statement II is true; Statement II is the correct explanation for Statement I.

(b) If Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I.

(c) If Statement I is true; Statement II is false.

(d) If Statement I is false; Statement II is true.

Show Answer

Answer:

Correct Answer: 17. (a, b, c, d)

Solution:

  1. From conservation of mechanical energy decrease in kinetic energy = increase in potential energy

 or 14πε0(Ze)(2e)rmin=5MeV=5×1.6×1013Jrmin=14πε02Ze25×1.6×1013=(9×109)(2)(92)(1.6×1019)25×1.6×1013(+Ze)(Z=92)==5.3×1014m=5.3×1012cm

i.e. rmin is of the order of 1012cm.

Correct option is (c).



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक