Kinetic energy of electrons

K = ${{{p^2}} \over {2m}}$ = ${{{{\left( {{h \over \lambda }} \right)}^2}} \over {2m}}$ = ${{{h^2}} \over {2m{\lambda ^2}}}$

For certain frequency, maximum wavelength that can be emitted is $\lambda$₀ which is cut off wavelength obtained at cut off frequency,

E₀ = ${{hc} \over {{\lambda _0}}}$

Since, E = E₀

${{{h^2}} \over {2m{\lambda ^2}}} = {{hc} \over {{\lambda _0}}}$

$\Rightarrow$ $\lambda$₀ = ${{2mc{\lambda ^2}} \over h}$