Answer
$\,^{14}_{7}N$.
Work Step by Step
The nuclear equation can be written as
$^{238}_{92}U+\,^{A}_{Z}X\rightarrow ^{247}_{99}Es+ 5\,^{1}_{0}n$
where $X$ is the symbol of the bombarding particle, $A$ its mass number and $Z$ is the charge.
Since sum of the masses and the charges should be equal on both the sides, we have
$238+A=247+5(1)$ and
$92+Z=99+5(0)$
$\implies A=14$ and $Z=7$
The particle with $Z=7$ is $N$.
Therefore, the bombarding particle is $\,^{14}_{7}N$.
