Answer
wavelength of light is
$\lambda=2.53698\times10^{-7}m=253.69nm$
approximately $254nm$
Work Step by Step
Work function of the metal is $\phi_{0} =2.4eV$
stopping potential is given as $V_{0} =2.5V$
from equation 27.5 maximum kinetic energy $K_{max}$of photoelectron is defined as
$K_{max}=eV_{0}$
putting the value of $V_{0} =2.5V$ we will get
$K_{max}=e2.5V=2.5eV$
From equation 27.7
$E=hf= K_{max} + \phi_{0}$
where $hf$ is energy of incident photon, $K_{max}$ is maximium kinetic energy of photelectron, $\phi_{0}$ is work function
putting the values of $K_{max}=2.5eV$ and $\phi_{0} =2.4eV$
we will get
$E=hf=2.5eV+2.4eV=4.9 eV$
$1eV$ is equal to $1.6\times10^{-19}J$
so $ 4.9 eV$ is equal to $ 4.9\times1.6\times10^{-19}J=7.84\times10^{-19}J$
$E=hf=\frac{hc}{\lambda}$
so $\lambda= \frac{hc}{E}$
$h=6.63\times10^{-34} J.s$, $c=3\times10^{8}m/s$ and $E=7.84\times10^{-19}J$
we will get $\lambda= \frac{6.63\times10^{-34} J.s\times3\times10^{8}m/s}{7.84\times10^{-19}J}$
$\lambda=2.53698\times10^{-7}m=253.69nm$
approximately $254nm$