Question

The number of electrons in ${\left[{}_{19}{K}^{40}\right]}^{-1}$ is

• A. 28
• B. 19
• C. 40
• D. 20

Answer 1

Answer: (D)

Atomic No. $=19$

$e^{-}$ present originally $=19$

total no. of $e^{-}$ $=19+1$ (from $-ve$

$=20$ charge $)$