c
The correct option is \(D \frac{2 m c \lambda^2}{h}\) Let \(K\) be the kinetic energy of the electron.

Electron de- Broglie wavelength

\(\lambda=\frac{h}{m v}=\frac{h}{\sqrt{2 m K}}\)

\(\Rightarrow K=\frac{h^2}{2 m \lambda^2}\)

Now, cut- off wavelenth for continuous \(X\) - rays,

\(\lambda_c=\frac{h c}{K}=\frac{h c}{\left(h^2 / 2 m \lambda^2\right)}\)

\(\therefore \lambda_c=\frac{2 m c \lambda^2}{h}\)

Hence, option \((C)\) is correct.