Using de-Broglie equation \\\\(\\lambda = \\dfrac{h}{p}\\\\) where \\\\( p =\\sqrt{2mE}\\\\)\\\\(\\implies\\\\) \\\\(\\lambda = \\dfrac{h}{\\sqrt{2mE}}\\\\)Energy of the X-ray emitted \\\\(E = \\dfrac{hc}{\\lambda_o}\\\\)\\\\(\ herefore\\\\) \\\\(\\lambda = \\dfrac{h}{\\sqrt{2m \ imes \\dfrac{hc}{\\lambda_o}}}\\\\) \\\\(\\implies \\lambda_o = \\dfrac{2mc\\lambda^2}{h}\\\\)