Answer
$v = 158ms^{-1}$
Work Step by Step
Momentum of photon is given by
$$p_{photon} = {h\over \lambda}$$
For $N$ photons,
$$p_{Nphotons} = {Nh\over \lambda}$$
As spacecraft is ejecting photons and no external force is acting on it.
$\therefore $ Linear momentum is conserved.
Hence, $\frac{Nh}{\lambda} = p_{spacecraft}$
$\frac{Nh}{\lambda} = mv$, $\qquad ......(i)$
where $m$ is mass of spacecraft and $v$ is the velocity attained by spacecraft.
Spacecraft is ejecting photons at a power of 1.5kW
$P = \frac{E}{t} = \frac{Nhc}{\lambda t}$
$\frac{Nh}{\lambda} = {Pt\over c}$
Hence, equation $(i)$ becomes,
$mv = {Pt\over c}$
$v = {Pt\over mc}$
Given $P = 1.5kW = 1.5\times 10^3W$, $t = 10 years = 10\times 365\times 86400 sec$
$m = 10kg,c = 3\times 10^8$
$\therefore v = {\frac{1.5\times 10^{3}\times 10\times 365\times 86400}{10\times 3\times 10^8}}m/s = 158m/s$
