A water droplet of radius 1 micron in Millikan oil apparatus is first held stationary under the influence of an electric field of intensity 5-1 - 104 NC -1- How many excess electrons does it carry- Take e -1-6 - 10-19 C - g -9-8 ms -2 and density of water -103 kg m -3

neE = mg = ∫ . 4 π/3r^3.g ⇒ n × 1.6 × 10^ 19× 5.1 × 10^4 =10^3 ×4 × 3.14/3× (10^ 6)^3 × 10 Solving n = 5 ...

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Standard XII

Chemistry

Emission and Absorption Spectra

A water dropl...

Question

A water droplet of radius 1 micron in Millikan oil apparatus is first held stationary under the influence of an electric field of intensity $5.1 \times 10^{4} N C^{- 1}$ . How many excess electrons does it carry? Take $e = 1.6 \times 10^{- 19} C$ , $g = 9.8 m s^{- 2}$ and density of water $= 10^{3} k g m^{- 3}$

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Solution

$n e E = m g = \int . \frac{4 π}{3} r^{3} . g \Rightarrow n \times 1.6 \times 10^{- 19} \times 5.1 \times 10^{4} = 10^{3} \times \frac{4 \times 3.14}{3} \times (10^{- 6})^{3} \times 10$
Solving n = 5

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A water droplet of radius 1 micron in Millikan oil apparatus is first held stationary under the influence of an electric field of intensity 5.1×104 NC−1. How many excess electrons does it carry? Take e=1.6×10−19 C, g=9.8 ms−2 and density of water =103 kg m−3

neE=mg=∫.4π3r3.g⇒n×1.6×10−19×5.1×104=103×4×3.143×(10−6)3×10 Solving n = 5

$n e E = m g = \int . \frac{4 π}{3} r^{3} . g \Rightarrow n \times 1.6 \times 10^{- 19} \times 5.1 \times 10^{4} = 10^{3} \times \frac{4 \times 3.14}{3} \times (10^{- 6})^{3} \times 10$
Solving n = 5