Given, mass of electron = 9.1 × 10−31kg
K. E. = 3.0 × 10−25J
\(\because\) K. E. = \(\frac{1}{2}\)mv2
v = \(\sqrt{\frac{2K.E}{m}}\) = \(\sqrt{\frac{2\times3.0\times10^{-25}kg\,m^2s^{-1}}{9.1\times10^{-31}\,kg}}\)
= 812 ms−1
Now, apply de-Broglie equation,
Wavelength, \(\lambda\) = \(\frac{h}{mv}\)
= \(\frac{6.626\times10^{-34}Js}{9.1\times10^{-31}kg\times812\,ms^{-1}}\)
= 8967 × 10−10m
= 896.7 nm