MCQ
While doing his experiment, Millikan one day observed the following charges on a single drop
(i) $6.563 \times 10^{-19} \mathrm{C}$
(ii) $8.204 \times 10^{-19} \mathrm{C}$
(iii) $11.50 \times 10^{-19} \mathrm{C}$
(iv) $13.13 \times 10^{-19} \mathrm{C}$
(v) $16.48 \times 10^{-19} \mathrm{C}$
(vi) $18.09 \times 10^{-19} \mathrm{C}$
From this data the value of the elementary charge (e)
was found to be
[MP PMT 1993]
  • (a) $1.641 \times 10^{-19} \mathrm{C}$
  • B
    (b) $1.630 \times 10^{-19} \mathrm{C}$
  • C
    (c) $1.648 \times 10^{-19} \mathrm{C}$
  • D
    (d) $1.602 \times 10^{-19} \mathrm{C}$

Answer

Correct option: A.
(a) $1.641 \times 10^{-19} \mathrm{C}$
(a) Any charge in the universe is given by$q=n e \Rightarrow e=\frac{q}{n}$ (where $n$ is an integer)$\begin{array}{r}q_1: q_2: q_3: q_4: q_5: q_6:: n_1: n_2: n_3: n_4: n_5: n_6 \\6.563: 8.204: 11.5: 13.13: 16.48: 18.09 \\:: n_1: n_2: n_3: n_4: n_5: n_6\end{array}$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$\begin{aligned}& 4: 5: 7: 8: 10: 11::n_1: n_2: n_3: n_4: n_5: n_6 \\& e=\frac{q_1+q_2+q_3+q_4+q_5+q_6}{n_1+n_2+n_3+n_4+n_5+n_6}=\frac{73.967 \times 10^{-19}}{45} \\& =1.641 \times 10^{-19} \mathrm{C}\end{aligned}$(Note : If you take 45.0743 in place of 45 , you will get the exact value)

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