Fluoride ions are produced by the addition of a single electron to an atom of fluorine \(^{19}F\) - AQA - A-Level Physics - Question 9 - 2018 - Paper 1

To determine the specific charge of the fluoride ion ( ext{F}^{-}), we first need to understand that fluoride ions are formed by adding an electron to fluorine. The specific charge can be calculated using the formula:

$ext{Specific Charge} = \frac{Q}{m}$

where (Q) is the charge of the ion and (m) is its mass. The charge of the electron is approximately (-1.6 \times 10^{-19}) C. The mass of a fluoride ion can be approximated using the mass of (^{19}F), which is around (3.19 \times 10^{-26}) kg.

Now, substituting these values in:

$ext{Specific Charge} = \frac{1.6 \times 10^{-19}\, C}{3.19 \times 10^{-26}\, kg} \approx 5.0 \times 10^{6}\, C\, kg^{-1}$

Thus, the correct answer for the magnitude of specific charge of the fluoride ion is:

C. 5.0 x 10^6 C kg⁻¹