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.

• A. $1.641 \times 10^{- 19}C$
• B. $1.630 \times 10^{- 19}C$
• C. $1.648 \times 10^{- 19}C$
• D. $1.602 \times 10^{- 19}C$

Answer 1

Answer: (A)

$1.641 \times 10^{- 19}C$

Answer 2

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)