The smallest unit of charge is −1.6×10−19 c, which is the charge in coulombs of a single electron. robert millikan was able to measure the charge on small droplets of oil by suspending them between a pair of electrically charged plates. which of the values are possible charges of those oil droplets?

The possible charges of those oil droplets from the given choices are:

Explanation:

Since the minimum amount of charge is - 1.6 × 10⁻¹⁹ C, the charge of any matter must be a multiple of that number.

So, find which of the given choices is a multiple of - 1.6 × 10⁻¹⁹ C.

You do that by dividing each choice by - 1.6 × 10⁻¹⁹ C. If the division is a whole number (zero remainder) then the choice is a multiple of - 1.6 × 10⁻¹⁹ C and is a possible charge of the oil droplets:

Since all the powers are 10⁻¹⁹ you must only divive the mantissas:

Option A. - 1.2 × 10⁻¹⁹ C

Option B: - 3.2 × 10⁻¹⁹ C

Option C. - 4.8 × 10⁻¹⁹ C

Option D. - 5.6 × 10⁻¹⁹ C

Option E. - 8 × 10⁻¹⁹ C

Option F. - 9.4 × 10⁻¹⁹ C

h let me think

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years