Dual Nature Of Light Question 35
Question: While doing his experiment, Millikan one day observed the following charges on a single drop (i) $ 6.563\times {{10}^{-19}}C $ (ii) $ 8.204\times {{10}^{-19}}C $ (iii) $ 11.50\times {{10}^{-19}}C $ (iv) $ 13.13\times {{10}^{-19}}C $ (v) $ 16.48\times {{10}^{-19}}C $ (vi) $ 18.09\times {{10}^{-19}}C $ From this data the value of the elementary charge (e) was found to be [MP PMT 1993]
Options:
A) $ 1.641\times {{10}^{-19}}C $
B) $ 1.630\times {{10}^{-19}}C $
C) $ 1.648\times {{10}^{-19}}C $
D) $ 1.602\times {{10}^{-19}}C $
Show Answer
Answer:
Correct Answer: A
Solution:
Any charge in the universe is given by $ q=ne\Rightarrow e=\frac{q}{n} $ (where n is an integer) $ q_{1}:q_{2}:q_{3}:q_{4}:q_{5}:q_{6}::n_{1}:n_{2}:n_{3}:n_{4}:n_{5}:n_{6} $ $ 6.563:8.204:11.5:13.13:16.48:18.09 $ $ ::n_{1}:n_{2}:n_{3}:n_{4}:n_{5}:n_{6} $ Divide by 6.563 $ 1:1.25:1.75:2.0:2.5:2.75 $ $ ::n_{1}:n_{2}:n_{3}:n_{4}:n_{5}:n_{6} $ Multiplied by 4 $ 4:5:7:8:10:11 $ $ ::n_{1}:n_{2}:n_{3}:n_{4}:n_{5}:n_{6} $ $ e=\frac{q_{1}+q_{2}+q_{3}+q_{4}+q_{5}+q_{6}}{n_{1}+n_{2}+n_{3}+n_{4}+n_{5}+n_{6}}=\frac{73.967\times {{10}^{-19}}}{45} $ $ =1.641\times {{10}^{-19}}C $ (Note : If you take 45.0743 in place of 45, you will get the exact value)
BETA
