Question

For the data of Problem 70, assume that the charge qon the drop is given by$\mathit{q}\mathbf{}\mathbf{=}\mathit{n}\mathit{e}$, where nis an integer and eis the elementary charge.

(a) Findfor each given value of q.

(b) Do a linear regression fit of the values ofversus the values of nand then use that fit to find e.

Short answer

1. The values of n for each given value of q are $4,5,7,8,10,11,12,14,\text{and}16.$.
2. Yes, the linear regression fits of the values of q versus the values of n. The value of e is.$1.63x{10}^{-19}C$

Step-by-step solution

The given data

1. The charge on the drop,$q=ne$
2. The given data table

Understanding the concept of the charge

Using the concept of electric charge, we can get the required values of n for each value of charge, q. Again, by creating the approximation value of the charge, we can get the required value of the charge.

Formula:

The value of charge of number of atoms, $q=ne$ (i)

a) Calculation of the value, n

If we subtract each value from the next larger value in the table, we find a set of numbers that are suggestive of a basic unit of charge:, $1.64x{10}^{-19}$,$3.3x{10}^{-19}$,$1.63x{10}^{-19}$ , $3.35x{10}^{-19}$, $1.6x{10}^{-19}$,$1.63x{10}^{-19}$ , $3.18x{10}^{-19}$,$3.24x{10}^{-19}$where the SI unit Coulomb is understood. These values are either close to a common charge value, $e=1.6x{10}^{-19}C$or are double that. Taking this, then, as a crude approximation to our experimental e, we divide it into all the values in the original data set using equation (i) and round to the nearest integer, obtaining $n=4,5,7,8,10,11,12,14,$and.$16$

b) Calculation of the value of the charge

When we perform a least squares fit of the original data set versus these values for n we obtain the linear equation:

$q=\text{(}7.18x{10}^{-21}+1.633x{10}^{-19}\text{n})\text{C}$

If we dismiss the constant term as unphysical (representing, say, systematic errors in our measurements) then we obtain$e=1.63x{10}^{-19}\text{\hspace{0.17em}C}$when we set$n=1$in this equation.

Hence, the linear regression fits the value and the value of the charge is.$1.63x{10}^{-19}\text{\hspace{0.17em}C}$