In the following reaction the reason why meta-nitro product also formed is:

When light of wavelength $248 \mathrm{~nm}$ falls on a metal of threshold energy $3.0 \mathrm{eV}$, the de-Broglie wavelength of emitted electrons is  _______\AA. (Round off to the Nearest Integer).

$\left[\right.$ Use: $\sqrt{3}=1.73, \mathrm{~h}=6.63 \times 10^{-34} \mathrm{~J}_{\mathrm{S}}$$\mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg} ; \mathrm{c}=3.0 \times 10^{8} \mathrm{~ms}^{-1}$$\left.\mathrm{leV}=1.6 \times 10^{-19} \mathrm{~J}\right]$