Answer
$^{42}_{19}K\rightarrow \,\,^{0}_{-1}e+\,\,^{42}_{20}Ca$
Work Step by Step
There are excess neutrons in $\,^{42}_{19}K$, so beta emission is probable.
In the case of $\beta$ emission, mass number does not change, but the atomic number increases by one unit. Therefore, the resulting nuclide is one with the mass number 42 and atomic number=$19+1=20$. $Ca$(calcium) is the element with atomic number 20.
Thus, we can write the nuclear equation for decay as:
$^{42}_{19}K\rightarrow \,\,^{0}_{-1}e+\,\,^{42}_{20}Ca$
