An electron accelerated through a potential difference V _ 1 has a...

Let initial wavelength be $\lambda$, then after $50$ increase wavelength will become $1 . 5\lambda$.

Now, $KE = \dfrac{{P}^{2}}{2m} = eV \text{ and } P = \dfrac{h}{\lambda}$

Therefore, $eV = \dfrac{{\left(\dfrac{h}{\lambda}\right)}^{2}}{2m}$.

So we can write

$e{V}_{1} = \dfrac{{\left(\dfrac{h}{\lambda}\right)}^{2}}{2m} . . . Eq ( 1 )$ and

$e{V}_{2} = \dfrac{{\left(\dfrac{h}{1 . 5\lambda}\right)}^{2}}{2m} . . . Eq ( 2 )$

From both equations, we get

$\dfrac{{V}_{1}}{{V}_{2}} = {\left(1 . 5\right)}^{2} = \dfrac{9}{4}$