Answer
$\lambda=9.5199\times10^{-12}m=0.0095199nm$
Work Step by Step
Magnitude of charge on point charge $q=8.3\mu C=8.3\times10^{-6}C$
Charge on proton $q_{p}=1.6\times10^{-19}C$
initial distance $r_{i}=0.420m$
Potential energy of two charge system where two charges $q_{1}$& $q_{2}$ are separated by distance $r$ is given by
$V=-k\frac{q_{1}\times q_{2}}{r} $
so initial energy of system $V_{i}=-k\frac{q\times q_{p}}{r_{i}} $
putting $q=8.3\times10^{-6}C$, $q_{p}=1.6\times10^{-19}C$, $r_{i}=0.420m$ and $k= 9.0\times10^{9}N.m^2/C^2$
$V_{i}=- 9.0\times10^{9}N.m^2/C^2\frac{8.3\times10^{-6}C\times 1.6\times10^{-19}C}{0.420m} $
$V_{i}=-284.5714\times10^{-16}J=-2.845714\times10^{-14}J$
$V_{i}=-2.845714\times10^{-14}J$ .........equation(1)
similarly potential energy at a distance $r_{f}=1.58m$
so final energy of system $V_{f}=-k\frac{q\times q_{p}}{r_{f}} $
putting $q=8.3\times10^{-6}C$, $q_{p}=1.6\times10^{-19}C$, $r_{f}=1.58m$ and $k= 9.0\times10^{9}N.m^2/C^2$
$V_{f}=- 9.0\times10^{9}N.m^2/C^2\frac{8.3\times10^{-6}C\times 1.6\times10^{-19}C}{1.58m} $
$V_{f}=-75.64556\times10^{-16}J=-0.756455\times10^{-14}J$
$V_{f}=-0.756455\times10^{-14}J$ .........equation(2)
change in energy $\Delta E=V_{f}-V_{i}$
$\Delta E=-0.756455\times10^{-14}J-(-2.845714\times10^{-14}J)$
$\Delta E=2.0893\times10^{-14}J $
this energy is carried out by photon
whose wavelength will be
$\lambda=\frac{hc}{E}$
putting $h=6.63\times10^{-34}J.s$, $E=2.0893\times10^{-14}J$, $c=3\times10^{8}m/s$
$\lambda=\frac{6.63\times10^{-34}J.s\times3\times10^{8}m/s}{2.0893\times10^{-14}J} =9.519934\times10^{-12}m$
$\lambda=0.0095199\times10^{-9}m=0.0095199nm$
