An electron accelerated by a potential difference 'V' has de-Broglie wavelength ' $\lambda$ '. If the electron is accelerated by a potential difference ' 9 V ', its de-Broglie wavelength will be

The de Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}}$ where $h$ is Planck’s constant, $m$ is the electron mass, and $e$ is the electron charge.

This implies $\lambda$ is inversely proportional to $\sqrt{V}$, so $\lambda \propto \frac{1}{\sqrt{V}}$ and thus $\lambda = \frac{C}{\sqrt{V}}$ where $C = \frac{h}{\sqrt{2me}}$.

For a potential difference of $9V$, the corresponding wavelength $\lambda'$ satisfies $\lambda' = \frac{C}{\sqrt{9V}} = \frac{C}{3\sqrt{V}} = \frac{\lambda}{3}$.

Final Answer: $\boxed{\frac{\lambda}{3}}$